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BMET Trigonometry

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Fortgang Gloag Gloag Hayes Jordan Landers

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Printed: September 1, 2011

Authors

Art Fortgang, Andrew Gloag, Anne Gloag, Andrea Hayes, Lori Jordan, Mara Landers, Brenda Meery,

Larry Ottman, Eve Rawley

i www.ck12.org

Contents

1 Right Triangles and an

Introduction to Trigonometry 1

1.1 The Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Special Right Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3 Basic Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 Solving Right Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.5 Measuring Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.6 Applying Trig Functions to Angles of Rotation . . . . . . . . . . . . . . . . . . . . . . . . . 39

1.7 Trigonometric Functions of Any Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1.8 Relating Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.9 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

2 Graphing Trigonometric

Functions - 2nd edition 73

2.1 Radian Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

2.2 Applications of Radian Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

2.3 Circular Functions of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

2.4 Translating Sine and Cosine Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

2.5 Amplitude, Period and Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

2.6 General Sinusoidal Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

2.7 Graphing Tangent, Cotangent, Secant, and Cosecant . . . . . . . . . . . . . . . . . . . . . . 149

2.8 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

3 Trigonometric Identities and

Equations - 2nd edition 164

3.1 Fundamental Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

3.2 Proving Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

3.3 Solving Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

3.4 Sum and Difference Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

3.5 Double Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

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3.6 Half-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

3.7 Products, Sums, Linear Combinations, and Applications . . . . . . . . . . . . . . . . . . . . 213

3.8 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

4 Inverse Trigonometric

Functions - 2nd edition 227

4.1 Basic Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

4.2 Graphing Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

4.3 Inverse Trigonometric Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

4.4 Applications & Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

4.5 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

5 Triangles and Vectors 265

5.1 The Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

5.2 Area of a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276

5.3 The Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285

5.4 The Ambiguous Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294

5.5 General Solutions of Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304

5.6 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311

5.7 Component Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

5.8 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333

6 The Polar System - 2nd edition 340

6.1 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340

6.2 Graphing Basic Polar Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346

6.3 Converting Between Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362

6.4 More with Polar Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

6.5 The Trigonometric Form of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . 382

6.6 The Product & Quotient Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388

6.7 De Moivre’s and the nth Root Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392

6.8 Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

7 Solving Systems of Equations

and Inequalities 407

7.1 Linear Systems by Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407

7.2 Solving Linear Systems by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417

7.3 Solving Linear Systems by Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424

7.4 Special Types of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435

7.5 Systems of Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443

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www.ck12.org iv

Chapter 1

Right Triangles and an

Introduction to Trigonometry

1.1 The Pythagorean Theorem

Introduction

Right triangles play an integral part in the study of trigonometry. It is from right triangles that the basic

definitions of the trigonometric functions are formed. In this chapter we will explore right triangles and

their properties. Through this, we will introduce the six basic trig functions and the unit circle.

Learning Objectives

• Recognize and use the Pythagorean Theorem.

• Recognize basic Pythagorean Triples.

• Use the Distance Formula.

The Pythagorean Theorem

From Geometry, recall that the Pythagorean Theorem is a2 + b2 = c2 where a and b are the legs of a right

triangle and c is the hypotenuse. Also, the side opposite the angle is lower case and the angle is upper

case. For example, angle A is opposite side a.

The Pythagorean Theorem is used to solve for the sides of a right triangle.

Example 1: Use the Pythagorean Theorem to find the missing side.

1 www.ck12.org

Solution: a = 8, b = 15, we need to find the hypotenuse.

82 + 152 = c2

64 + 225 = c2

289 = c2

17 = c

Notice, we do not include -17 as a solution because a negative number cannot be a side of a triangle.

Example 2: Use the Pythagorean Theorem to find the missing side.

Solution: Use the Pythagorean Theorem to find the missing leg.(

5

√

7

)2

+ x2 =

(

5

√

13

)2

25 · 7 + x2 = 25 · 13

175 + x2 = 325

x2 = 150

x = 5

√

6

Pythagorean Triples

Pythagorean Triples are sets of whole numbers for which the Pythagorean Theorem holds true. The most

well-known triple is 3, 4, 5. This means, that 3 and 4 are the lengths of the legs and 5 is the hypotenuse.

The largest length is always the hypotenuse. If we were to multiply any triple by a constant, this new triple

would still represent sides of a right triangle. Therefore, 6, 8, 10 and 15, 20, 25, among countless others,

would represent sides of a right triangle.

Example 3: Determine if the following lengths are Pythagorean Triples.

a. 7, 24, 25

b. 9, 40, 41

c. 11, 56, 57

Solution: Plug each set of numbers into the Pythagorean Theorem.

a.

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72 + 242 ?= 252

49 + 576 = 625

625 = 625

Yes, 7, 24, 25 is a Pythagorean Triple and sides of a right triangle.

b.

92 + 402 ?= 412

81 + 1600 = 1681

1681 = 1681

Yes, 9, 40, 41 is a Pythagorean Triple and sides of a right triangle.

c.

112 + 562 ?= 572

121 + 3136 = 3249

3257 , 3249

No, 11, 56, 57 do not represent the sides of a right triangle.

Converse of the Pythagorean Theorem

Using the technique from Example 3, we can determine if sets of numbers are acute, right or obtuse

triangles. Examples 3a and 3b were both right triangles because the two sides equaled each other and

made the Pythagorean Theorem true. However in Example 3c, the two sides were not equal. Because

3257 > 3249, we can say that 11, 56, and 57 are the sides of an acute triangle. To help you visualize this,

think of an equilateral triangle with sides of length 5. We know that this is an acute triangle. If you plug in

5 for each number in the Pythagorean Theorem we get 52+52 = 52 and 50 > 25. Therefore, if a2+b2 > c2,

then lengths a, b, and c make up an acute triangle. Conversely, if a2 + b2 < c2, then lengths a, b, and c

make up the sides of an obtuse triangle. It is important to note that the length “c” is always the longest.

Example 4: Determine if the following lengths make an acute, right or obtuse triangle.

a. 5, 6, 7

b. 5, 10, 14

c. 12, 35, 37

Solution: Plug in each set of lengths into the Pythagorean Theorem.

a.

52 + 62 ? 72

25 + 36 ? 49

61 > 49

Because 61 > 49, this is an acute triangle.

b.

52 + 102 ? 142

25 + 100 ? 196

125 < 196

Because 125 < 196, this is an obtuse triangle.

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c.

122 + 352 ? 372

144 + 1225 ? 1369

1369 = 1369

Because the two sides are equal, this is a right triangle.

NOTE: All of the lengths in Example 4 represent the lengths of the sides of a triangle. Recall the Triangle

Inequality Theorem from geometry which states: The length of a side in a triangle is less than the sum of

the other two sides. For example, 4, 7 and 13 cannot be the sides of a triangle because 4+ 7 is not greater

than 13.

The Distance Formula

An application of the Pythagorean Theorem is to find the distance between two points. Consider the points

(-1, 6) and (5, -3). If we plot them on a grid, they make a diagonal line. Draw a vertical line down from

(-1, 6) and a horizontal line to the left of (5, -3) to make a right triangle.

Now we can find the distance between these two points by using the vertical and horizontal distances that

we determined from the graph.

92 + (−6)2 = d2

81 + 36 = d2

117 = d2√

117 = d

3

√

13 = d

Notice, that the x−values were subtracted from each other to find the horizontal distance and the y−values

were subtracted from each other to find the vertical distance. If this process is generalized for two points

(x1, y1) and (x2, y2), the Distance Formula is derived.

(x1 − x2)2 + (y1 − y2)2 = d2

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This is the Pythagorean Theorem with the vertical and horizontal differences between (x1, y1) and (x2, y2).

Taking the square root of both sides will solve the right hand side for d, the distance.

√

(x1 − x2)2 + (y1 − y2)2 = d

This is the Distance Formula. The following example shows how to apply the distance formula.

Example 5: Find the difference between the two points.

a. (4, 2) and (-9, 5)

b. (-10, 3) and (0, -15)

Solution: Plug each pair of points into the distance formula.

a.

d =

√

(4 − (−9))2 + (2 − 5)2

=

√

132 + (−3)2

=

√

169 + 324

=

√

493

b.

d =

√

(−10 − 0)2 + (3 − (−15))2

=

√

(−10)2 + (18)2

=

√

100 + 324

=

√

424 = 2

√

106

Points to Consider

• Does the Pythagorean Theorem apply to all real numbers?

• Can a Pythagorean Triple have irrational numbers in the set?

• What is the difference between the Distance Formula and the Pythagorean Theorem?

5 www.ck12.org

Review Questions

Determine if the lengths below represent the sides of a right triangle. If not, state if the triangle is acute

or obtuse.

1. 6, 9, 13

2. 9, 10, 11

3. 16, 30, 34

4. 20, 23, 40

5. 11, 16, 29

6. 2

√

6, 6

√

3, 2

√

33

Find the missing side of each right triangle below. Leave the answer in simplest radical form.

7.

8.

9.

10. The general formula for a Pythagorean Triple is n2 − m2, 2nm, n2 + m2 where n and m are natural

numbers. Use the Pythagorean Theorem to prove this is true.

11. Find the distance between the pair of points.

(a) (5, -6) and (18, 3)

(b)

(√

3, −√2

)

and

(

−2√3, 5√2

)

Review Answers

1. 62 + 92?132 → 36 + 81?169→ 117 < 169 The triangle is obtuse.

2. 92 + 102?112 → 81 + 100?121→ 181 > 121 The triangle is acute.

3. 162 + 302?342 → 256 + 900?1156→ 1156 = 1156 This is a right triangle.

4. 202 + 232?402 → 400 + 529?1600→ 929 < 1600 The triangle is obtuse.

5. These lengths cannot make up the sides of a triangle. 11 + 16 < 29

6.

(

2

√

6

)2

+

(

6

√

3

)2

?

(

2

√

33

)2 → (4 · 6) + (36 · 3)?(4 · 33) → 24 + 108?132 → 132 = 132 This is a right

triangle.

www.ck12.org 6

7.

72 + x2 = 182

49 + x2 = 324

x2 = 275

x =

√

275 = 5

√

11

8.

52 +

(

5

√

3

)2

= x2

25 + (25 · 3) = x2

25 + 75 = x2

100 = x2

10 = x

9. Both legs are 11.

112 + 112 = x2

121 + 121 = x2

242 = x2√

242 = x

11

√

2 = x

10. Plug n2 − m2, 2nm, n2 + m2 into the Pythagorean Theorem.

(n2 − m2)2 + (2nm)2 = (n2 + m2)2

n4 − 2n2m2 + m4 + 4n2m2 = n4 + 2n2m2 + m4

−2n2m2 + 4n2m2 = 2n2m2

4n2m2 = 4n2m2

11. (a) (5, -6) and (18, 3)

d =

√

(5 − 18)2 + (−6 − 3)2

=

√

(−13)2 + (−9)2

=

√

169 + 81

=

√

250

= 5

√

10

(b)

(√

3, −√2

)

and

(

−2√3, 5√2

)

d =

√(√

3 −

(

−2√3

))2

+

(

−√2 − 5√2

)2

=

√(

3

√

3

)2

+

(

−6√2

)2

=

√

(9 · 3) + (36 · 2)

=

√

27 + 72

=

√

99 = 3

√

11

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1.2 Special Right Triangles

Learning Objectives

• Recognize special right triangles.

• Use the special right triangle ratios to solve special right triangles.

Special Right Triangle #1: Isosceles Right Triangle

An isosceles right triangle is an isosceles triangle and a right triangle. This means that it has two congruent

sides and one right angle. Therefore, the two congruent sides must be the legs.

Because the two legs are congruent, we will call them both a and the hypotenuse c. Plugging both letters

into the Pythagorean Theorem, we get:

a2 + a2 = c2

2a2 = c2√

2a2 =

√

c2

a

√

2 = c

From this we can conclude that the hypotenuse length is the length of a leg multiplied by

√

2. Therefore,

we only need one of the three lengths to determine the other two lengths of the sides of an isosceles right

triangle. The ratio is usually written x : x : x

√

2, where x is the length of the legs and x

√

2 is the length

of the hypotenuse.

Example 1: Find the lengths of the other two sides of the isosceles right triangles below.

a.

b.

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c.

d.

Solution:

a. If a leg has length 8, by the ratio, the other leg is 8 and the hypotenuse is 8

√

2.

b. If the hypotenuse has length 7

√

2, then both legs are 7.

c. Because the leg is 10

√

2, then so is the other leg. The hypotenuse will be 10

√

2 multiplied by an

additional

√

2.

10

√

2 · √2 = 10 · 2 = 20

d. In this problem set x

√

2 = 9

√

6 because x

√

2 is the hypotenuse portion of the ratio.

x

√

2 = 9

√

6

x =

9

√

6√

2

·

√

2√

2

=

9

√

12

2

=

18

√

3

2

= 9

√

3

So, the length of each leg is 9

√

3.

What are the angle measures in an isosceles right triangle? Recall that the sum of the angles in a triangle

is 180◦ and there is one 90◦ angle. Therefore, the other two angles add up to 90◦. Because this is an

isosceles triangle, these two angles are equal and 45◦ each. Sometimes an isosceles right triangle is also

referred to as a 45 − 45 − 90 triangle.

Special Right Triangle #2: 30-60-90 Triangle

30− 60− 90 refers to each of the angles in this special right triangle. To understand the ratios of the sides,

start with an equilateral triangle with an altitude drawn from one vertex.

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Recall from geometry, that an altitude, h, cuts the opposite side directly in half. So, we know that one

side, the hypotenuse, is 2s and the shortest leg is s. Also, recall that the altitude is a perpendicular and

angle bisector, which is why the angle at the top is split in half. To find the length of the longer leg, use

the Pythagorean Theorem:

s2 + h2 = (2s)2

s2 + h2 = 4s2

h2 = 3s2

h = s

√

3

From this we can conclude that the length of the longer leg is the length of the short leg multiplied by

√

3

or s

√

3. Just like the isosceles right triangle, we now only need one side in order to determine the other

two in a 30 − 60 − 90 triangle. The ratio of the three sides is written x : x√3 : 2x, where x is the shortest

leg, x

√

3 is the longer leg and 2x is the hypotenuse.

Notice, that the shortest side is always opposite the smallest angle and the longest side is always opposite

90◦.

If you look back at the Review Questions from the last section we now recognize #8 as a 30 − 60 − 90

triangle.

Example 2: Find the lengths of the two missing sides in the 30 − 60 − 90 triangles.

a.

www.ck12.org 10

b.

c.

d.

Solution: Determine which side in the 30 − 60 − 90 ratio is given and solve for the other two.

a. 4

√

3 is the longer leg because it is opposite the 60◦. So, in the x : x

√

3 : 2x ratio, 4

√

3 = x

√

3, therefore

x = 4 and 2x = 8. The short leg is 4 and the hypotenuse is 8.

b. 17 is the hypotenuse because it is opposite the right angle. In the x : x

√

3 : 2x ratio, 17 = 2x and so the

short leg is 172 and the long leg is 17

√

3

2 .

c. 15 is the long leg because it is opposite the 60◦. Even though 15 does not have a radical after it, we can

still set it equal to x

√

3.

x

√

3 = 15

x =

15√

3

·

√

3√

3

=

15

√

3

3

= 5

√

3 So, the short leg is 5

√

3.

Multiplying 5

√

3 by 2, we get the hypotenuse length, which is 10

√

3.

d. 24

√

21 is the length of the hypotenuse because it is opposite the right angle. Set it equal to 2x and

solve for x to get the length of the short leg.

2x = 24

√

21

x = 12

√

21

To find the length of the longer leg, we need to multiply 12

√

21 by

√

3.

12

√

21 · √3 = 12√3 · 3 · 7 = 36√7

The length of the longer leg is 36

√

7.

11 www.ck12.org

Be careful when doing these problems. You can always check your answers by finding the decimal ap-

proximations of each side. For example, in 2d, short leg = 12

√

21 ≈ 54.99, long leg = 36√7 ≈ 95.25 and

the hypotenuse = 24

√

21 ≈ 109.98. This is an easy way to double-check your work and verify that the

hypotenuse is the longest side.

Using Special Right Triangle Ratios

Special right triangles are the basis of trigonometry. The angles 30◦, 45◦, 60◦ and their multiples have

special properties and significance in the unit circle (sections 1.5 and 1.6). Students are usually required

to memorize these two ratios because of their importance.

First, let’s compare the two ratios, so that we can better distinguish the difference between the two. For

a 45 − 45 − 90 triangle the ratio is x : x : x√2 and for a 30 − 60 − 90 triangle the ratio is x : x√3 : 2x. An

easy way to tell the difference between these two ratios is the isosceles right triangle has two congruent

sides, so its ratio has the

√

2, whereas the 30 − 60 − 90 angles are all divisible by 3, so that ratio includes

the

√

3. Also, if you are ever in doubt or forget the ratios, you can always use the Pythagorean Theorem.

The ratios are considered a short cut.

Example 3: Determine if the sets of lengths below represent special right triangles. If so, which one?

a. 8

√

3 : 24 : 16

√

3

b.

√

5 :

√

5 :

√

10

c. 6

√

7 : 6

√

21 : 12

Solution:

a. Yes, this is a 30− 60− 90 triangle. If the short leg is x = 8√3, then the long leg is 8√3 · √3 = 8 · 3 = 24

and the hypotenuse is 2 · 8√3 = 16√3.

b. Yes, this is a 45 − 45 − 90 triangle. The two legs are equal and √5 · √2 = √10, which would be the

length of the hypotenuse.

c. No, this is not a special right triangle, nor a right triangle. The hypotenuse should be 12

√

7 in order to

be a 30 − 60 − 90 triangle.

Points to Consider

• What is the difference between Pythagorean triples and special right triangle ratios?

• Why are these two ratios considered “special”?

Review Questions

Solve each triangle using the special right triangle ratios.

1.

www.ck12.org 12

2.

3.

4.

5.

6.

7. A square window has a diagonal of 6 ft. To the nearest hundredth, what is the height of the window?

8. Pablo has a rectangular yard with dimensions 10 ft by 20 ft. He is decorating the yard for a party

and wants to hang lights along both diagonals of his yard. How many feet of lights does he need?

Round your answer to the nearest foot.

9. Can 2 : 2 : 2

√

3 be the sides of a right triangle? If so, is it a special right triangle?

10. Can

√

5 :

√

15 : 2

√

5 be the sides of a right triangle? If so, is it a special right triangle?

Review Answers

1. Each leg is 16√

2

= 16√

2

·

√

2√

2

= 16

√

2

2 = 8

√

2.

2. Short leg is

√

6√

3

=

√

6

3 =

√

2 and hypotenuse is 2

√

2.

3. Short leg is 12√

3

= 12√

3

·

√

3√

3

= 12

√

3

3 = 4

√

3 and hypotenuse is 8

√

3.

4. The hypotenuse is 4

√

10 · √2 = 4√20 = 8√5.

5. Each leg is 5

√

2√

2

= 5.

6. The short leg is 152 and the long leg is 15

√

3

2 .

7. If the diagonal of a square is 6 ft, then each side of the square is 6√

2

or 3

√

2 ≈ 4.24 f t.

8. These are not dimensions for a special right triangle, so to find the diagonal (both are the same

13 www.ck12.org

length) do the Pythagorean Theorem:

102 + 202 = d2

100 + 400 = d2√

500 = d

10

√

5 = d

So, if each diagonal is 10

√

5, two diagonals would be 20

√

5 ≈ 45 f t.

Pablo needs 45 ft of lights for his yard.

9. 2 : 2 : 2

√

3 does not fit into either ratio, so it is not a special right triangle. To see if it is a right

triangle, plug these values into the Pythagorean Theorem:

22 + 22 =

(

2

√

3

)2

4 + 4 = 12

8 < 12

this is not a right triangle, it is an obtuse triangle.

10.

√

5 :

√

15 : 2

√

5 is a 30− 60− 90 triangle. The long leg is √5 · √3 = √15 and the hypotenuse is 2√5.

1.3 Basic Trigonometric Functions

Learning Objectives

• Find the values of the six trigonometric functions for angles in right triangles.

Introduction

Consider a situation in which you are building a ramp for wheelchair access to a building. If the ramp

must have a height of 8 feet, and the angle of the ramp must be about 5◦, how long must the ramp be?

Solving this kind of problem requires trigonometry. The word trigonometry comes from two words meaning

triangle and measure. In this lesson we will define six trigonometric functions. For each of these functions,

the elements of the domain are angles. We will define these functions in two ways: first, using right

triangles, and second, using angles of rotation. Once we have defined these functions, we will be able to

solve problems like the one above.

The Sine, Cosine, and Tangent Functions

The first three trigonometric functions we will work with are the sine, cosine, and tangent functions. As

noted above, the elements of the domains of these functions are angles. We can define these functions

in terms of a right triangle: The elements of the range of the functions are particular ratios of sides of

triangles.

www.ck12.org 14

We define the sine function as follows: For an acute angle x in a right triangle, the sinx is equal to the ratio

of the side opposite of the angle over the hypotenuse of the triangle. For example, using this triangle, we

have: sin A = ac and sin B = bc .

Since all right triangles with the same acute angles are similar, this function will produce the same ratio,

no matter which triangle is used. Thus, it is a well-defined function.

Similarly, the cosine of an angle is defined as the ratio of the side adjacent (next to) the angle over the

hypotenuse of the triangle. Using this triangle, we have: cos A = bc and cos B = ac .

Finally, the tangent of an angle is defined as the ratio of the side opposite the angle to the side adjacent

to the angle. In the triangle above, we have: tan A = ab and tan B = ba .

There are a few important things to note about the way we write these functions. First, keep in mind that

the abbreviations sinx, cosx, and tanx are just like f (x). They simply stand for specific kinds of functions.

Second, be careful when using the abbreviations that you still pronounce the full name of each function.

When we write sinx it is still pronounced sine, with a long “i.” When we write cosx, we still say co-sine.

And when we write tanx, we still say tangent.

We can use these definitions to find the sine, cosine, and tangent values for angles in a right triangle.

Example 1: Find the sine, cosine, and tangent of ∠A:

Solution:

15 www.ck12.org

sin A = opposite sidehypotenuse =

4

5

cos A = adjacent sidehypotenuse =

3

5

tan A = opposite sideadjacent side =

4

3

One of the reasons that these functions will help us solve problems is that these ratios will always be the

same, as long as the angles are the same. Consider for example, a triangle similar to triangle ABC.

If CP has length 3, then side AP of triangle NAP is 6. Because NAP is similar to ABC, side NP has length

8. This means the hypotenuse AN has length 10. (This can be shown either by using Pythagorean Triples

or the Pythagorean Theorem.)

If we use triangle NAP to find the sine, cosine, and tangent of angle A, we get:

sin A = opposite sidehypotenuse =

8

10

=

4

5

cos A = adjacent sidehypotenuse =

6

10

=

3

5

tan A = opposite sideadjacent side =

8

6

=

4

3

Also notice that the tangent function is the same as the slope of the hypotenuse. tan A = 43 , which is the

same as riserun or

change in y

change in x . The tan B does not equal the slope because it is the reciprocal of tanA.

Example 2: Find sin B using triangle ABC and triangle NAP.

Solution:

Using triangle ABC : sin B = 35

Using triangle NAP : sin B = 610 = 35

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An easy way to remember the ratios of the sine, cosine, and tangent functions is SOH-CAH-TOA. Sine =

Opposite

Hypotenuse ,Cosine =

Adjacent

Hypotenuse ,Tangent =

Opposite

Adjacent .

Secant, Cosecant, and Cotangent Functions

We can define three more functions also based on a right triangle. They are the reciprocals of sine, cosine

and tangent.

If sin A = ac , then the definition of cosecant, or csc, is csc A = ca .

If cos A = bc , then the definition of secant, or sec, is sec A = cb .

If tan A = ab , then the definition of cotangent, or cot, is cot A = ba .

Example 3: Find the secant, cosecant, and cotangent of angle B.

Solution:

First, we must find the length of the hypotenuse. We can do this using the Pythagorean Theorem:

52 + 122 = H2

25 + 144 = H2

169 = H2

H = 13

Now we can find the secant, cosecant, and cotangent of angle B:

17 www.ck12.org

sec B = hypotenuseadjacent side =

13

12

csc B = hypotenuseopposite side =

13

5

cot B = adjacent sideopposite side =

12

5

Points to Consider

• Do you notice any similarities between the sine of one angle and the cosine of the other, in the same

triangle?

Review Questions

1. Find the values of the six trig functions of angle A.

2. Consider triangle VET below. Find the length of the hypotenuse and values of the six trig functions

of angle T .

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3. Consider the right triangle below.

(a) Find the hypotenuse.

(b) Find the six trigonometric functions of ∠X.

(c) Find the six trigonometric functions of ∠Z.

4. Looking back at #3, are any functions of ∠X equal to any of the functions of ∠Z? If so, which ones?

Do you think this could be generalized for ANY pair of acute angles in the same right triangle (also

called complements)?

5. Consider an isosceles right triangle with legs of length 2. Find the sine, cosine and tangent of both

acute angles.

6. Consider an isosceles right triangle with legs of length x. Find the sine, cosine and tangent of both

acute angles. Write down any similarities or patterns you notice with #5.

7. Consider a 30 − 60 − 90 triangle with hypotenuse of length 10. Find the sine, cosine and tangent of

both acute angles.

8. Consider a 30 − 60 − 90 triangle with short leg of length x. Find the sine, cosine and tangent of both

acute angles. Write down any similarities or patterns you notice with #7.

9. Consider a right triangle, ABC. If sin A = 941 , find the length of the third side.

Review Answers

1. sin A = 915 = 35 , cos A = 1215 = 45 , tan A = 912 = 34 , csc A = 159 = 53 , sec A = 1512 = 54 , cot A = 129 = 43

2. The hypotenuse is 17

(√

152 + 82 =

√

225 + 64 =

√

289 = 17

)

.

sinT = 15

17

, cos T = 8

17

, tanT = 15

8

, cscT = 17

15

, secT = 17

8

, cotT = 8

15

3. (a) The hypotenuse is 13

(√

52 + 122 =

√

25 + 144 =

√

169 = 13

)

.

(b) sin X = 1213 , cos X = 513 , tan X = 125 , csc X = 1312 , sec X = 135 , cot X = 512

(c) sinZ = 513 , cosZ = 1213 , tanZ = 512 , csc Z = 135 , secZ = 1312 , cotZ = 125

4. From #3, we can conclude that sin X = cosZ, cos X = sin Z, tan X = cotZ, cot X = tan Z, csc X = sec Z

and sec X = cscZ. Yes, this can be generalized for all complements.

5. The hypotenuse is 2

√

2. Each angle is 45◦, so the sine, cosine, and tangent are the same for both

angles.

sin 45◦ = 2

2

√

2

=

1√

2

·

√

2√

2

=

√

2

2

, cos 45◦ = 2

2

√

2

=

1√

2

·

√

2√

2

=

√

2

2

, tan 45◦ = 2

2

= 1

6. If the legs are length x, then the hypotenuse is x

√

2. For 45◦, the sine, cosine, and tangent are:

sin 45◦ = x

x

√

2

=

1√

2

·

√

2√

2

=

√

2

2

, cos 45◦ = x

x

√

2

=

1√

2

·

√

2√

2

=

√

2

2

, tan 45◦ = x

x

= 1

This tells us that regardless of the length of the sides of an isosceles right triangle, the sine, cosine

and tangent of 45◦ are always the same.

19 www.ck12.org

7. If the hypotenuse is 10, then the short leg is 5 and the long leg is 5

√

3. Recall, that 30◦ is opposite

the short side, or 5, and 60◦ is opposite the long side, or 5

√

3.

sin 30◦ = 5

10

=

1

2

, cos 30◦ = 5

√

3

10

=

√

3

2

, tan 30◦ = 5

5

√

3

=

1√

3

·

√

3√

3

=

√

3

3

sin 60◦ = 5

√

3

10

=

√

3

2

, cos 60◦ = 5

10

=

1

2

, tan 60◦ = 5

√

3

5

=

√

3

8. If the short leg is x, then the long leg is x

√

3 and the hypotenuse is 2x. 30◦ is opposite the short side,

or x, and 60◦ is opposite the long side, or x

√

3.

sin 30◦ = x

2x

=

1

2

, cos 30◦ = x

√

3

2x

=

√

3

2

, tan 30◦ = x

x

√

3

=

1√

3

·

√

3√

3

=

√

3

3

sin 60◦ = x

√

3

2x

=

√

3

2

, cos 60◦ = x

2x

=

1

2

, tan 60◦ = x

√

3

x

=

√

3

This tells us that regardless of the length of the sides of a 30 − 60 − 90 triangle, the sine, cosine and

tangent of 30◦ and 60◦ are always the same. Also, sin 30◦ = cos 60◦ and cos 30◦ = sin 60◦.

9. If sin A = 941 , then the opposite side is 9x (some multiple of 9) and the hypotenuse is 41x. Therefore,

working with the Pythagorean Theorem would give us the length of the other leg. Also, we could

notice that this is a Pythagorean Triple and the other leg is 40x.

1.4 Solving Right Triangles

Learning Objectives

• Solve right triangles.

• Find the area of any triangle using trigonometry.

• Solve real-world problems that require you to solve a right triangle.

• Find angle measures using inverse trigonometric functions.

Solving Right Triangles

You can use your knowledge of the Pythagorean Theorem and the six trigonometric functions to solve

a right triangle. Because a right triangle is a triangle with a 90 degree angle, solving a right triangle

requires that you find the measures of one or both of the other angles. How you solve will depend on how

much information is given. The following examples show two situations: a triangle missing one side, and

a triangle missing two sides.

Example 1: Solve the triangle shown below.

www.ck12.org 20

Solution:

We need to find the lengths of all sides and the measures of all angles. In this triangle, two of the three

sides are given. We can find the length of the third side using the Pythagorean Theorem:

82 + b2 = 102

64 + b2 = 100

b2 = 36

b = ±6⇒ b = 6

(You may have also recognized the “Pythagorean Triple,” 6, 8, 10, instead of carrying out the Pythagorean

Theorem.)

You can also find the third side using a trigonometric ratio. Notice that the missing side, b, is adjacent to

∠A, and the hypotenuse is given. Therefore we can use the cosine function to find the length of b:

cos 53.13◦ = adjacent sidehypotenuse =

b

10

0.6 =

b

10

b = 0.6(10) = 6

We could also use the tangent function, as the opposite side was given. It may seem confusing that you

can find the missing side in more than one way. The point is, however, not to create confusion, but to

show that you must look at what information is missing, and choose a strategy. Overall, when you need to

identify one side of the triangle, you can either use the Pythagorean Theorem, or you can use a trig ratio.

To solve the above triangle, we also have to identify the measures of all three angles. Two angles are

given: 90 degrees and 53.13 degrees. We can find the third angle using the Triangle Sum Theorem,

180 − 90 − 53.13 = 36.87◦.

Now let’s consider a triangle that has two missing sides.

Example 2: Solve the triangle shown below.

Solution:

In this triangle, we need to find the lengths of two sides. We can find the length of one side using a trig

ratio. Then we can find the length of the third side by using a trig ratio with the same given information,

not the side we solved for. This is because the side we found is an approximation and would not yield

the most accurate answer for the other missing side. Only use the given information when solving right

triangles.

We are given the measure of angle A, and the length of the side adjacent to angle A. If we want to find

the length of the hypotenuse, c, we can use the cosine ratio:

21 www.ck12.org

cos 40◦ = ad jacent

hypotenuse

=

6

c

cos 40◦ = 6

c

c cos 40◦ = 6

c =

6

cos 40◦ ≈ 7.83

If we want to find the length of the other leg of the triangle, we can use the tangent ratio. This will give

us the most accurate answer because we are not using approximations.

tan 40◦ = opposite

ad jacent

=

a

6

a = 6 tan 40◦ ≈ 5.03

Now we know the lengths of all three sides of this triangle. In the review questions, you will verify the

values of c and a using the Pythagorean Theorem. Here, to finish solving the triangle, we only need to

find the measure of ∠B : 180 − 90 − 40 = 50◦

Notice that in both examples, one of the two non-right angles was given. If neither of the two non-right

angles is given, you will need a new strategy to find the angles.

Inverse Trigonometric Functions

Consider the right triangle below.

From this triangle, we know how to determine all six trigonometric functions for both ∠C and ∠T . From

any of these functions we can also find the value of the angle, using our graphing calculators. If you look

back at #7 from 1.3, we saw that sin 30◦ = 12 . If you type 30 into your graphing calculator and then hit

the SIN button, the calculator yields 0.5. (Make sure your calculator’s mode is set to degrees.)

Conversely, with the triangle above, we know the trig ratios, but not the angle. In this case the inverse

of the trigonometric function must be used to determine the measure of the angle. These functions are

located above the SIN, COS, and TAN buttons on the calculator. To access this function, press 2nd and

the appropriate button and the measure of the angle appears on the screen.

cosT = 2425 → cos−1 2425 = T from the calculator we get

Example 3: Find the angle measure for the trig functions below.

a. sin x = 0.687

b. tan x = 43

Solution: Plug into calculator.

a. sin−1 0.687 = 43.4◦

www.ck12.org 22

b. tan−1 43 = 53.13◦

Example 4: You live on a farm and your chore is to move hay from the loft of the barn down to the stalls

for the horses. The hay is very heavy and to move it manually down a ladder would take too much time

and effort. You decide to devise a make shift conveyor belt made of bed sheets that you will attach to the

door of the loft and anchor securely in the ground. If the door of the loft is 25 feet above the ground and

you have 30 feet of sheeting, at what angle do you need to anchor the sheets to the ground?

Solution:

From the picture, we need to use the inverse sine function.

sin θ = 25 f eet

30 f eet

sin θ = 0.8333

sin−1(sin θ) = sin−1 0.8333

θ = 56.4◦

The sheets should be anchored at an angle of 56.4◦.

Finding the Area of a Triangle

In Geometry, you learned that the area for a triangle is A = 12bh, where b is the base and h is the height,

or altitude. Now, that you know the trig ratios, this formula can be changed around, using sine.

Looking at the triangle above, you can use sine to determine h, sinC = ha . So, solving this equation for h,

we have a sinC = h. Substituting this for h, we now have a new formula for area.

A =

1

2

ab sinC

What this means is you do not need the height to find the area anymore. All you now need is two sides

and the angle between the two sides, called the included angle.

Example 5: Find the area of the triangle.

23 www.ck12.org

a.

Solution: Using the formula, A = 12 ab sinC, we have

A =

1

2

· 8 · 13 · sin 82◦

= 4 · 13 · 0.990

= 51.494

Example 6: Find the area of the parallelogram.

Solution: Recall that a parallelogram can be split into two triangles, so the formula for a parallelogram,

using the new formula would be: A = 2 · 12 ab sinC or A = ab sinC.

A = 7 · 15 · sin 65◦

= 95.162

Angles of Elevation and Depression

You can use right triangles to find distances, if you know an angle of elevation or an angle of depression.

The figure below shows each of these kinds of angles.

The angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object.

For example, if you are standing on the ground looking up at the top of a mountain, you could measure

the angle of elevation. The angle of depression is the angle between the horizontal line of sight and the line

of sight down to an object. For example, if you were standing on top of a hill or a building, looking down

at an object, you could measure the angle of depression. You can measure these angles using a clinometer

www.ck12.org 24

or a theodolite. People tend to use clinometers or theodolites to measure the height of trees and other tall

objects. Here we will solve several problems involving these angles and distances.

Example 7: You are standing 20 feet away from a tree, and you measure the angle of elevation to be 38◦.

How tall is the tree?

Solution:

The solution depends on your height, as you measure the angle of elevation from your line of sight. Assume

that you are 5 feet tall.

The figure shows us that once we find the value of T , we have to add 5 feet to this value to find the total

height of the triangle. To find T , we should use the tangent value:

tan 38◦ = opposite

ad jacent

=

T

20

tan 38◦ = T

20

T = 20 tan 38◦ ≈ 15.63

Height of tree ≈ 20.63 f t

The next example shows an angle of depression.

Example 8: You are standing on top of a building, looking at a park in the distance. The angle of

depression is 53◦. If the building you are standing on is 100 feet tall, how far away is the park? Does your

height matter?

Solution:

If we ignore the height of the person, we solve the following triangle:

25 www.ck12.org

Given the angle of depression is 53◦, ∠A in the figure above is 37◦. We can use the tangent function to find

the distance from the building to the park:

tan 37◦ = opposite

ad jacent

=

d

100

tan 37◦ = d

100

d = 100 tan 37◦ ≈ 75.36 f t.

If we take into account the height if the person, this will change the value of the adjacent side. For example,

if the person is 5 feet tall, we have a different triangle:

tan 37◦ = opposite

ad jacent

=

d

105

tan 37◦ = d

105

d = 105 tan 37◦ ≈ 79.12 f t.

If you are only looking to estimate a distance, then you can ignore the height of the person taking the

measurements. However, the height of the person will matter more in situations where the distances or

lengths involved are smaller. For example, the height of the person will influence the result more in the

tree height problem than in the building problem, as the tree is closer in height to the person than the

building is.

www.ck12.org 26

Right Triangles and Bearings

We can also use right triangles to find distances using angles given as bearings. In navigation, a bearing is

the direction from one object to another. In air navigation, bearings are given as angles rotated clockwise

from the north. The graph below shows an angle of 70 degrees:

It is important to keep in mind that angles in navigation problems are measured this way, and not the

same way angles are measured in trigonometry. Further, angles in navigation and surveying may also be

given in terms of north, east, south, and west. For example, N70◦E refers to an angle from the north,

towards the east, while N70◦W refers to an angle from the north, towards the west. N70◦E is the same as

the angle shown in the graph above. N70◦W would result in an angle in the second quadrant.

Example 9: A ship travels on a N50◦E course. The ship travels until it is due north of a port which is 10

nautical miles due east of the port from which the ship originated. How far did the ship travel?

27 www.ck12.org

Solution: The angle between d and 10 nm is the complement of 50◦, which is 40◦. Therefore we can find

d using the cosine function:

cos 40◦ = ad jacent

hypotenuse

=

10

d

cos 40◦ = 10

d

d cos 40◦ = 10

d =

10

cos 40◦ ≈ 13.05 nm

Other Applications of Right Triangles

In general, you can use trigonometry to solve any problem that involves right triangles. The next few

examples show different situations in which a right triangle can be used to find a length or a distance.

Example 10: The wheelchair ramp

In lesson 3 we introduced the following situation: You are building a ramp so that people in wheelchairs

can access a building. If the ramp must have a height of 8 feet, and the angle of the ramp must be about

5◦, how long must the ramp be?

Given that we know the angle of the ramp and the length of the side opposite the angle, we can use the

sine ratio to find the length of the ramp, which is the hypotenuse of the triangle:

sin 5◦ = 8

L

L sin 5◦ = 8

L =

8

sin 5◦ ≈ 91.8 f t

This may seem like a long ramp, but in fact a 5◦ ramp angle is what is required by the Americans with

Disabilities Act (ADA). This explains why many ramps are comprised of several sections, or have turns.

The additional distance is needed to make up for the small slope.

www.ck12.org 28

Right triangle trigonometry is also used for measuring distances that could not actually be measured. The

next example shows a calculation of the distance between the moon and the sun. This calculation requires

that we know the distance from the earth to the moon. In chapter 5 you will learn the Law of Sines, an

equation that is necessary for the calculation of the distance from the earth to the moon. In the following

example, we assume this distance, and use a right triangle to find the distance between the moon and the

sun.

Example 11: The earth, moon, and sun create a right triangle during the first quarter moon. The distance

from the earth to the moon is about 240,002.5 miles. What is the distance between the sun and the moon?

Solution:

Let d = the distance between the sun and the moon. We can use the tangent function to find the value of

d:

tan 89.85◦ = d

240, 002.5

d = 240, 002.5 tan 89.85◦ = 91, 673, 992.71 miles

Therefore the distance between the sun and the moon is much larger than the distance between the earth

and the moon.

(Source: www.scribd.com, Trigonometry from the Earth to the Stars.)

Points to Consider

• In what kinds of situations do right triangles naturally arise?

• Are there right triangles that cannot be solved?

• Trigonometry can solve problems at an astronomical scale as well as problems at a molecular or

atomic scale. Why is this true?

29 www.ck12.org

Review Questions

1. Solve the triangle.

2. Two friends are writing practice problems to study for a trigonometry test. Sam writes the following

problem for his friend Anna to solve:

In right triangle ABC, the measure of angle C is 90 degrees, and the length of side c is 8 inches. Solve

the triangle.

Anna tells Sam that the triangle cannot be solved. Sam says that she is wrong. Who is right?

Explain your thinking.

3. Use the Pythagorean Theorem to verify the sides of the triangle in example 2.

4. Estimate the measure of angle B in the triangle below using the fact that sin B = 35 and sin 30◦ = 12 .

Use a calculator to find sine values. Estimate B to the nearest degree.

5. Find the area of the triangle.

6. Find the area of the parallelogram below.

7. The angle of elevation from the ground to the top of a flagpole is measured to be 53◦. If the

measurement was taken from 15 feet away, how tall is the flagpole?

8. From the top of a hill, the angle of depression to a house is measured to be 14◦. If the hill is 30 feet

www.ck12.org 30

tall, how far away is the house?

9. An airplane departs city A and travels at a bearing of 100◦. City B is directly south of city A. When

the plane is 200 miles east of city B, how far has the plan traveled? How far apart are City A and

City B?

What is the length of the slanted outer wall, w? What is the length of the main floor, f ?

10. A surveyor is measuring the width of a pond. She chooses a landmark on the opposite side of the

pond, and measures the angle to this landmark from a point 50 feet away from the original point.

How wide is the pond?

11. Find the length of side x:

12. A deck measuring 10 feet by 16 feet will require laying boards with one board running along the

diagonal and the remaining boards running parallel to that board. The boards meeting the side of

the house must be cut prior to being nailed down. At what angle should the boards be cut?

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Review Answers1.

∠A = 50◦

b ≈ 5.83

a ≈ 9.33

2. Anna is correct. There is not enough information to solve the triangle. That is, there are infinitely

many right triangles with hypotenuse 8. For example:

3. 62 + 5.032 = 36 + 25.3009 = 61.3009 = 7.832.

4. ∠B ≈37◦

5. A = 12 · 10 · 12 · sin 104◦ = 58.218

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6. A = 4 · 9 · sin 112◦ = 33.379

7. About 19.9 feet tall

8. About 120.3 feet

9. The plane has traveled about 203 miles.

The two cities are 35 miles apart.

10. About 41.95 feet

11. About 7.4412.

tan θ = opposite

ad jacent

tan θ = 0.625

θ = 32◦

1.5 Measuring Rotation

Learning Objectives

• Identify and draw angles of rotation in standard position.

• Identify quadrantal angles.

• Identify co-terminal angles.

Angles of Rotation in Standard Position

Consider, for example, a game that is played with a spinner. When you spin the spinner, how far has

it gone? You can answer this question in several ways. You could say something like “the spinner spun

around 3 times.” This means that the spinner made 3 complete rotations, and then landed back where it

started.

We can also measure the rotation in degrees. In the previous lesson we worked with angles in triangles,

measured in degrees. You may recall from geometry that a full rotation is 360 degrees, usually written

as 360◦. Half a rotation is then 180◦ and a quarter rotation is 90◦. Each of these measurements will be

important in this lesson, as well as in the remainder of the chapter.

We can use our knowledge of graphing to represent any angle. The figure below shows an angle in what is

called standard position.

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The initial side of an angle in standard position is always on the positive x−axis. The terminal side always

meets the initial side at the origin. Notice that the rotation goes in a counterclockwise direction. This

means that if we rotate clockwise, we will generate a negative angle. Below are several examples of angles

in standard position.

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The 90 degree angle is one of four quadrantal angles. A quadrantal angle is one whose terminal side lies

on an axis. Along with 90◦, 0◦, 180◦ and 270◦ are quadrantal angles.

These angles are referred to as quadrantal because each angle defines a quadrant. Notice that without the

arrow indicating the rotation, 270◦ looks as if it is a −90◦, defining the fourth quadrant. Notice also that

360◦ would look just like 0◦. The difference is in the action of rotation. This idea of two angles actually

being the same angle is discussed next.

Coterminal Angles

Consider the angle 30◦, in standard position.

35 www.ck12.org

Now consider the angle 390◦. We can think of this angle as a full rotation (360◦), plus an additional 30

degrees.

Notice that 390◦ looks the same as 30◦. Formally, we say that the angles share the same terminal side.

Therefore we call the angles co-terminal. Not only are these two angles co-terminal, but there are

infinitely many angles that are co-terminal with these two angles. For example, if we rotate another 360◦,

we get the angle 750◦. Or, if we create the angle in the negative direction (clockwise), we get the angle

−330◦. Because we can rotate in either direction, and we can rotate as many times as we want, we can

continuously generate angles that are co-terminal with 30◦.

Example 1: Which angles are co-terminal with 45◦?

a. −45◦

b. 405◦

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c. −315◦

d. 135◦

Solution: b. 405◦ and c. −315◦ are co-terminal with 45◦.

Notice that terminal side of the first angle, −45◦, is in the 4th quadrant. The last angle, 135◦ is in the 2nd

quadrant. Therefore neither angle is co-terminal with 45◦.

Now consider 405◦. This is a full rotation, plus an additional 45 degrees. So this angle is co-terminal with

45◦. The angle −315◦ can be generated by rotating clockwise. To determine where the terminal side is, it

can be helpful to use quadrantal angles as markers. For example, if you rotate clockwise 90 degrees 3 times

(for a total of 270 degrees), the terminal side of the angle is on the positive y−axis. For a total clockwise

rotation of 315 degrees, we have 315 − 270 = 45 degrees more to rotate. This puts the terminal side of the

angle at the same position as 45◦.

Points to Consider

• How can one angle look exactly the same as another angle?

• Where might you see angles of rotation in real life?

Review Questions

1. Plot the following angles in standard position.

(a) 60◦

(b) −170◦

(c) 365◦

(d) 325◦

(e) 240◦

2. State the measure of an angle that is co-terminal with 90◦.

3. Name a positive and negative angle that are co-terminal with:

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(a) 120◦

(b) 315◦

(c) −150◦

4. A drag racer goes around a 180 degree circular curve in a racetrack in a path of radius 120 m. Its

front and back wheels have different diameters. The front wheels are 0.6 m in diameter. The rear

wheels are much larger; they have a diameter of 1.8 m. The axles of both wheels are 2 m long.

Which wheel has more rotations going around the curve? How many more degrees does the front

wheel rotate compared to the back wheel?

Review Answers

1. (a)

(b)

(c)

(d)

(e)

2. Answers will vary. Examples: 450◦, −270◦

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3. (a) Answers will vary. Examples: −240◦, 480◦

(b) Answers will vary. Examples: −45◦, 675◦

(c) Answers will vary. Examples: 210◦, −510◦, 570◦

4. The front wheel rotates more because it has a smaller diameter. It rotates 200 revolutions versus

66.67 revolutions for the back wheel, which is a 48, 000◦ difference ((200 − 66.6̄) · 360◦).

1.6 Applying Trig Functions to Angles of Rota-

tion

Learning Objectives

• Find the values of the six trigonometric functions for angles of rotation.

• Recognize angles of the unit circle.

Trigonometric Functions of Angles in Standard Position

In section 1.3, we defined the six trigonometric functions for angles in right triangles. We can also define

the same functions in terms of angles of rotation. Consider an angle in standard position, whose terminal

side intersects a circle of radius r. We can think of the radius as the hypotenuse of a right triangle:

The point (x, y) where the terminal side of the angle intersects the circle tells us the lengths of the two legs

of the triangle. Now, we can define the trigonometric functions in terms of x, y, and r:

39 www.ck12.org

cos θ = x

r

sec θ = r

x

sin θ = y

r

csc θ = r

y

tan θ = y

x

cot θ = x

y

And, we can extend these functions to include non-acute angles.

Example 1: The point (-3, 4) is a point on the terminal side of an angle in standard position. Determine

the values of the six trigonometric functions of the angle.

Solution:

Notice that the angle is more than 90 degrees, and that the terminal side of the angle lies in the second

quadrant. This will influence the signs of the trigonometric functions.

cos θ = −3

5

sec θ = 5−3

sin θ = 4

5

csc θ = 5

4

tan θ = 4−3 cot θ =

−3

4

Notice that the value of r depends on the coordinates of the given point. You can always find the value of

r using the Pythagorean Theorem. However, often we look at angles in a circle with radius 1. As you will

see next, doing this allows us to simplify the definitions of the trig functions.

The Unit Circle

Consider an angle in standard position, such that the point (x, y) on the terminal side of the angle is a

point on a circle with radius 1.

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This circle is called the unit circle. With r = 1, we can define the trigonometric functions in the unit

circle:

cos θ = x

r

=

x

1

= x sec θ = r

x

=

1

x

sin θ = y

r

=

y

1

= y csc θ = r

y

=

1

y

tan θ = y

x

cot θ = x

y

Notice that in the unit circle, the sine and cosine of an angle are the x and y coordinates of the point on

the terminal side of the angle. Now we can find the values of the trigonometric functions of any angle of

rotation, even the quadrantal angles, which are not angles in triangles.

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We can use the figure above to determine values of the trig functions for the quadrantal angles. For

example, sin 90◦ = y = 1.

Example 2: Use the unit circle above to find each value:

a. cos 90◦

b. cot 180◦

c. sec 0◦

Solution:

a. cos 90◦ = 0

The ordered pair for this angle is (0, 1). The cosine value is the x coordinate, 0.

b. cot 180◦ is undefined

The ordered pair for this angle is (-1, 0). The ratio xy is −10 , which is undefined.

c. sec 0◦ = 1

The ordered pair for this angle is (1, 0). The ratio 1x is 11 = 1.

There are several important angles in the unit circle that you will work with extensively in your study

of trigonometry, primarily 30◦, 45◦, and 60◦. Recall section 1.2 to find the values of the trigonometric

functions of these angles. First, we need to know the ordered pairs. Let’s begin with 30◦.

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This triangle is identical to #8 from 1.3. If you look back at this problem, you will recall that you found

the sine, cosine and tangent of 30◦ and 60◦. It is no coincidence that the endpoint on the unit circle is the

same as your answer from #8.

The terminal side of the angle intersects the unit circle at the point

( √

3

2 ,

1

2

)

. Therefore we can find the values

of any of the trig functions of 30◦. For example, the cosine value is the x−coordinate, so cos(30◦) =

√

3

2 .

Because the coordinates are fractions, we have to do a bit more work in order to find the tangent value:

tan 30◦ = y

x

=

1

2√

3

2

=

1

2

× 2√

3

=

1√

3

.

In the review exercises you will find the values of the remaining four trig functions of this angle. The table

below summarizes the ordered pairs for 30◦, 45◦, and 60◦ on the unit circle.

Table 1.1:

Angle x−coordinate y−coordinate

30◦

√

3

2

1

2

45◦

√

2

2

√

2

2

60◦ 12

√

3

2

We can use these values to find the values of any of the six trig functions of these angles.

Example 3: Find the value of each function.

a. cos 45◦

b. sin 60◦

c. tan 45◦

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Solution:

a. cos 45◦ =

√

2

2 The cosine value is the x− coordinate of the point.

b. sin 60◦ =

√

3

2 The sine value is the y− coordinate of the point.

c. tan 45◦ = 1 The tangent value is the ratio of the y− coordinate to the x− coordinate. Because the x−

and y− coordinates are the same for this angle, the tangent ratio is 1.

Points to Consider

• How can some values of the trig functions be negative? How is it that some are undefined?

• Why is the unit circle and the trig functions defined on it useful, even when the hypotenuses of

triangles in the problem are not 1?

Review Questions

1. The point (3, -4) is a point on the terminal side of an angle θ in standard position.

(a) Determine the radius of the circle.

(b) Determine the values of the six trigonometric functions of the angle.

2. The point (-5, -12) is a point on the terminal side of an angle θ in standard position.

(a) Determine the radius of the circle.

(b) Determine the values of the six trigonometric functions of the angle.

3. tan θ = −23 and cos θ > 0. Find sin θ.

4. csc θ = −4 and tan θ > 0. Find the exact values of the remaining trigonometric functions.

5. (2, 6) is a point on the terminal side of θ. Find the exact values of the six trigonometric functions.

6. The terminal side of the angle 270◦ intersects the unit circle at (0, -1). Use this ordered pair to find

the six trig functions of 270◦.

7. In the lesson you learned that the terminal side of the angle 30◦ intersects the unit circle at the point( √

3

2 ,

1

2

)

. Here you will prove that this is true.

(a) Explain why Triangle ABD is an equiangular triangle. What is the measure of angle DAB?

(b) What is the length of BD? How do you know?

(c) What is the length of BC and CD? How do you know?

(d) Now explain why the ordered pair is

( √

3

2 ,

1

2

)

.

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(e) Why does this tell you that the ordered pair for 60◦ is

(

1

2 ,

√

3

2

)

?

8. In the lesson you learned that the terminal side of the angle 45◦ is

( √

2

2 ,

√

2

2

)

. Use the figure below

and the Pythagorean Theorem to show that this is true.

9. In what quadrants will an angle in standard position have a positive tangent value? Explain your

thinking.

10. Sketch the angle 150◦ on the unit circle is. How is this angle related to 30◦? What do you think the

ordered pair is?

11. We now know that sin θ = y, cos θ = x, and tan θ = yx . First, explain how it looks as though sine,

cosine, and tangent are related. Second, can you rewrite tangent in terms of sine and cosine?

Review Answers

1. The radius of the circle is 5.cos θ = 3

5

sec θ = 5

3

sin θ = −4

5

csc θ = 5−4

tan θ = −4

3

cot θ = 3−4

2. The radius of the circle is 13.cos θ = −5

13

sec θ = 13−5

sin θ = −12

13

csc θ = 13−12

tan θ = −12−5 =

12

5

cot θ = −5−12 =

5

12

3. If tan θ = −23 , it must be in either Quadrant II or IV. Because cos θ > 0, we can eliminate Quadrant

IV. So, this means that the 3 is negative. (All Students Take Calculus) From the Pythagorean

Theorem, we find the hypotenuse: 22 + (−32) = c2

4 + 9 = c2

13 = c2√

13 = c

So, sin θ = 2√

13

or 2

√

13

13 (Rationalize the denominator)

4. If csc θ = −4, then sin θ = −14 , sine is negative, so θ is in either Quadrant III or IV. Because tan θ > 0,

we can eliminate Quadrant IV, therefore θ is in Quadrant III. From the Pythagorean Theorem, we

can find the other leg:

45 www.ck12.org

a2 + (−1)2 = 42

a2 + 1 = 16

a2 = 15

a =

√

15

So, cos θ = −

√

15

4

, sec θ = − 4√

15

or − 4

√

15

15

tan θ = − 1√

15

or

√

15

15

, cot θ =

√

15

5. If the terminal side of θ is on (2, 6) it means θ is in Quadrant I, so sine, cosine and tangent are all

positive. From the Pythagorean Theorem, the hypotenuse is:

22 + 62 = c2

4 + 36 = c2

40 = c2√

40 = 2

√

10 = c

Therefore, sin θ = 6

2

√

10

= 3√

10

= 3

√

10

10 , cos θ = 22√10 =

1√

10

=

√

10

10 and tan θ = 62 = 36.

cos 270 = 0 sec 270 = unde f ined

sin 270 = −1 csc 270 = 1−1 = −1

tan 270 = unde f ined cot 270 = 0

7. (a) The triangle is equiangular because all three angles measure 60 degrees. Angle DAB measures

60 degrees because it is the sum of two 30 degree angles.

(b) BD has length 1 because it is one side of an equiangular, and hence equilateral, triangle.

(c) BC and CD each have length 12 , as they are each half of BD. This is the case because Triangle

ABC and ADC are congruent.

(d) We can use the Pythagorean theorem to show that the length of AC is

√

3

2 . If we place angle

BAC as an angle in standard position, then AC and BC correspond to the x and y coordinates

where the terminal side of the angle intersects the unit circle. Therefore the ordered pair is( √

3

2 ,

1

2

)

.

(e) If we draw the angle 60◦ in standard position, we will also obtain a 30 − 60 − 90 triangle, but

the side lengths will be interchanged. So the ordered pair for 60◦ is

(

1

2 ,

√

3

2

)

.

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8.

n2 + n2 = 12

2n2 = 1

n2 =

1

2

n = ±

√

1

2

n = ± 1√

2

n = ± 1√

2

×

√

2√

2

= ±

√

2

2

Because the angle is in the first quadrant, the x and y coordinates are positive.

9. An angle in the first quadrant, as the tangent is the ratio of two positive numbers. And, angle in the

third quadrant, as the tangent in the ratio of two negative numbers, which will be positive.

10. The terminal side of the angle is a reflection of the terminal side of 30◦. From this, students should

see that the ordered pair is

(

−

√

3

2 ,

1

2

)

.

11. Students should notice that tangent is the ratio of sincos , which is

y

x , which is also slope.

1.7 Trigonometric Functions of Any Angle

Learning objectives

• Identify the reference angles for angles in the unit circle.

• Identify the ordered pair on the unit circle for angles whose reference angle is 30◦, 45◦, and 60◦, or a

quadrantal angle, including negative angles, and angles whose measure is greater than 360◦.

• Use these ordered pairs to determine values of trig functions of these angles.

• Use calculators to find values of trig functions of any angle.

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Reference Angles and Angles in the Unit Circle

In the previous lesson, one of the review questions asked you to consider the angle 150◦. If we graph this

angle in standard position, we see that the terminal side of this angle is a reflection of the terminal side of

30◦, across the y−axis.

Notice that 150◦ makes a 30◦ angle with the negative x−axis. Therefore we say that 30◦ is the reference

angle for 150◦. Formally, the reference angle of an angle in standard position is the angle formed with

the closest portion of the x−axis. Notice that 30◦ is the reference angle for many angles. For example, it

is the reference angle for 210◦ and for −30◦.

In general, identifying the reference angle for an angle will help you determine the values of the trig

functions of the angle.

Example 1: Graph each angle and identify its reference angle.

a. 140◦

b. 240◦

c. 380◦

Solution:

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a. 140◦ makes a 40◦ angle with the x−axis. Therefore the reference angle is 40◦.

b. 240◦ makes a 60◦ with the x−axis. Therefore the reference angle is 60◦.

c. 380◦ is a full rotation of 360◦, plus an additional 20◦. So this angle is co-terminal with 20◦, and 20◦ is

its reference angle.

If an angle has a reference angle of 30◦, 45◦, or 60◦, we can identify its ordered pair on the unit circle, and

so we can find the values of the six trig functions of that angle. For example, above we stated that 150◦

has a reference angle of 30◦. Because of its relationship to 30◦, the ordered pair for is 150◦ is

(

−

√

3

2 ,

1

2

)

.

Now we can find the values of the six trig functions of 150◦:

cos 150 = x = −

√

3

2

sec 150 = 1

x

=

1

−√3

2

=

−2√

3

sin 150 = y = 1

2

csc 150 = 1

y

=

1

1

2

= 2

tan 150 = y

x

=

1

2

−√3

2

=

1

−√3 cot 150 =

x

y

=

−√3

2

1

2

= −√3

Example 2: Find the ordered pair for 240◦ and use it to find the value of sin 240◦.

Solution: sin 240◦ = −

√

3

2

As we found in example 1, the reference angle for 240◦ is 60◦. The figure below shows 60◦ and the three

other angles in the unit circle that have 60◦ as a reference angle.

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The terminal side of the angle 240◦ represents a reflection of the terminal side of 60◦ over both axes. So

the coordinates of the point are

(

−12 ,−

√

3

2

)

. The y−coordinate is the sine value, so sin 240◦ = −

√

3

2 .

Just as the figure above shows 60◦ and three related angles, we can make similar graphs for 30◦ and 45◦.

Knowing these ordered pairs will help you find the value of any of the trig functions for these angles.

Example 3: Find the value of cot 300

Solution: cot 300 − 1√

3

Using the graph above, you will find that the ordered pair is

(

1

2 ,−

√

3

2

)

. Therefore the cotangent value is

cot 300 = xy =

1

2

−

√

3

2

= 12 × − 2√3 = −

1√

3

We can also use the concept of a reference angle and the ordered pairs we have identified to determine the

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values of the trig functions for other angles.

Trigonometric Functions of Negative Angles

Recall that graphing a negative angle means rotating clockwise. The graph below shows −30◦.

Notice that this angle is coterminal with 330◦. So the ordered pair is

( √

3

2 ,−12

)

. We can use this ordered

pair to find the values of any of the trig functions of −30◦. For example, cos(−30◦) = x =

√

3

2 .

In general, if a negative angle has a reference angle of 30◦, 45◦, or 60◦, or if it is a quadrantal angle, we

can find its ordered pair, and so we can determine the values of any of the trig functions of the angle.

Example 4: Find the value of each expression.

a. sin(−45◦)

b. sec(−300◦)

c. cos(−90◦)

Solution:

a. sin(−45◦) = −

√

2

2

−45◦ is in the 4th quadrant, and has a reference angle of 45◦. That is, this angle is coterminal with 315◦.

Therefore the ordered pair is

( √

2

2 ,−

√

2

2

)

and the sine value is −

√

2

2 .

b. sec(−300◦) = 2

The angle −300◦ is in the 1st quadrant and has a reference angle of 60◦. That is, this angle is coterminal

with 60◦. Therefore the ordered pair is

(

1

2 ,

√

3

2

)

and the secant value is 1x = 11

2

= 2.

c. cos(−90◦) = 0

The angle −90◦ is coterminal with 270◦. Therefore the ordered pair is (0, -1) and the cosine value is 0.

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We can also use our knowledge of reference angles and ordered pairs to find the values of trig functions of

angles with measure greater than 360 degrees.

Trigonometric Functions of Angles Greater than 360 Degrees

Consider the angle 390◦. As you learned previously, you can think of this angle as a full 360 degree rotation,

plus an additional 30 degrees. Therefore 390◦ is coterminal with 30◦. As you saw above with negative

angles, this means that 390◦ has the same ordered pair as 30◦, and so it has the same trig values. For

example,

cos 390◦ = cos 30◦ =

√

3

2

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In general, if an angle whose measure is greater than 360 has a reference angle of 30◦, 45◦, or 60◦, or if it is

a quadrantal angle, we can find its ordered pair, and so we can find the values of any of the trig functions

of the angle. Again, determine the reference angle first.

Example 5: Find the value of each expression.

a. sin 420◦

b. tan 840◦

c. cos 540◦

Solution:

a. sin 420◦ =

√

3

2

420◦ is a full rotation of 360 degrees, plus an additional 60 degrees. Therefore the angle is coterminal with

60◦, and so it shares the same ordered pair,

(

1

2 ,

√

3

2

)

. The sine value is the y−coordinate.

b. tan 840◦ = −√3

840◦ is two full rotations, or 720 degrees, plus an additional 120 degrees:

840 = 360 + 360 + 120

Therefore 840◦ is coterminal with 120◦, so the ordered pair is

(

−12 ,

√

3

2

)

. The tangent value can be found

by the following:

tan 840◦ = tan 120◦ = y

x

=

√

3

2

−12

=

√

3

2

× −2

1

= −√3

c. cos 540◦ = −1

540◦ is a full rotation of 360 degrees, plus an additional 180 degrees. Therefore the angle is coterminal

with 180◦, and the ordered pair is (-1, 0). So the cosine value is -1.

So far all of the angles we have worked with are multiples of 30, 45, 60, and 90. Next we will find

approximate values of the trig functions of other angles.

Using a Calculator to Find Values

If you have a scientific calculator, you can determine the value of any trig function for any angle. Here we

will focus on using a TI graphing calculator to find values.

First, your calculator needs to be in the correct “mode.” In chapter 2 you will learn about a different system

for measuring angles, known as radian measure. In this chapter, we are measuring angles in degrees. We

need to make sure that the calculator is in degrees. To do this, press MODE . In the third row, make

sure that Degree is highlighted. If Radian is highlighted, scroll down to this row, scroll over to Degree,

and press ENTER . This will highlight Degree. Then press 2nd MODE to return to the main screen.

Now you can calculate any value. For example, we can verify the values from the table above. To find

sin 130◦, press Sin 130 ENTER . The calculator should return the value .7660444431.

Example 6: Find the approximate value of each expression. Round your answer to 4 decimal places.

a. sin 130◦

b. cos 15◦

53 www.ck12.org

c. tan 50◦

Solution:

a. sin 130◦ ≈ 0.7660

b. cos 15◦ ≈ 0.9659

c. tan 50◦ ≈ 1.1918

You may have noticed that the calculator provides a “(“ after the SIN. In the previous calculations, you

can actually leave off the “)”. However, in more complicated calculations, leaving off the closing “)” can

create problems. It is a good idea to get in the habit of closing parentheses.

You can also use a calculator to find values of more complicated expressions.

Example 7: Use a calculator to find an approximate value of sin 25◦ + cos 25◦. Round your answer to 4

decimal places.

Solution: sin 25◦ + cos 25◦ ≈ 1.3289

∗ This is an example where you need to close the parentheses.

Points to Consider

• What is the difference between the measure of an angle, and its reference angle? In what cases are

these measures the same value?

• Which angles have the same cosine value, or the same sine value? Which angles have opposite cosine

and sine values?

Review Questions

1. State the reference angle for each angle.

(a) 190◦

(b) −60◦

(c) 1470◦

(d) −135◦

2. State the ordered pair for each angle.

(a) 300◦

(b) −150◦

(c) 405◦

3. Find the value of each expression.

(a) sin 210◦

(b) tan 270◦

(c) csc 120◦

4. Find the value of each expression.

(a) sin 510◦

(b) cos 930◦

(c) csc 405◦

5. Find the value of each expression.

(a) cos−150◦

(b) tan−45◦

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(c) sin−240◦

6. Use a calculator to find each value. Round to 4 decimal places.

(a) sin 118◦

(b) tan 55◦

(c) cos 100◦

7. Recall, in lesson 1.4, we introduced inverse trig functions. Use your calculator to approximate the

measure of an angle whose sine value is 0.2.

8. In example 6c, we found that tan 50◦ ≈ 1.1918. Use your knowledge of a special angle to explain why

this value is reasonable. HINT: You will need to change the tangent of this angle into a decimal.

9. Use the table below or a calculator to explore sum and product relationships among trig functions.

Consider the following functions:

f (x) = sin(x+ x) and g(x) = sin(x) + sin(x)

h(x) = sin(x) ∗ sin(x) and j(x) = sin(x2)

Do you observe any patterns in these functions? Are there any equalities among the functions? Can

you make a general conjecture about sin(a) + sin(b) and sin(a+ b) for all values of a, b?

What about sin(a) sin(a) and sin(a2)?

Table 1.2:

a◦ b◦ sin a+ sin b sin(a+ b)

10 30 .6736 .6428

20 60 1.2080 .9848

55 78 1.7973 .7314

122 25 1.2707 .5446

200 75 .6239 -.9962

10. Use a calculator or your knowledge of special angles to fill in the values in the table, then use

the values to make a conjecture about the relationship between (sin a)2 and (cos a)2. If you use a

calculator, round all values to 4 decimal places.

Table 1.3:

a (sin a)2 (cos a)2

0

25

45

80

90

120

250

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Review Answers

1. (a) 10◦

(b) 60◦

(c) 30◦

(d) 45◦

2. (a)

(

1

2 ,−

√

3

2

)

(b)

(

−

√

3

2 ,−12

)

(c)

( √

2

2 ,

√

2

2

)

3. (a) −12

(b) 0

(c) 2√

3

4. (a) 12

(b) −

√

3

2

(c)

√

2

5. (a) −

√

3

2

(b) -1

(c)

√

3

2

6. (a) 0.8828

(b) 1.4281

(c) -0.1736

7. Between 165 and 160 degrees.

8. This is reasonable because tan 45◦ = 1 and the tan 60◦ =

√

3 ≈ 1.732, and the tan 50◦ should fall

between these two values.

9. Conjecture: sin a+ sin b , sin(a+ b)

10.

Table 1.4:

a (sin a)2 (cos a)2

0 0 1

25 .1786 .8216

45 12 12

80 .9698 .0302

90 1 0

120 .75 .25

235 .6710 .3290

310 .5898 .4132

Conjecture: (sin a)2 + (cos a)2 = 1.

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1.8 Relating Trigonometric Functions

Learning objectives

• State the reciprocal relationships between trig functions, and use these identities to find values of

trig functions.

• State quotient relationships between trig functions, and use quotient identities to find values of trig

functions.

• State the domain and range of each trig function.

• State the sign of a trig function, given the quadrant in which an angle lies.

• State the Pythagorean identities and use these identities to find values of trig functions.

Reciprocal identities

The first set of identities we will establish are the reciprocal identities. A reciprocal of a fraction ab

is the fraction ba . That is, we find the reciprocal of a fraction by interchanging the numerator and the

denominator, or flipping the fraction. The six trig functions can be grouped in pairs as reciprocals.

First, consider the definition of the sine function for angles of rotation: sin θ = yr . Now consider the

cosecant function: csc θ = ry . In the unit circle, these values are sin θ =

y

1 = y and csc θ = 1y . These two

functions, by definition, are reciprocals. Therefore the sine value of an angle is always the reciprocal of the

cosecant value, and vice versa. For example, if sin θ = 12 , then csc θ = 21 = 2.

Analogously, the cosine function and the secant function are reciprocals, and the tangent and cotangent

function are reciprocals:

sec θ = 1cos θ or cos θ =

1

sec θ

cot θ = 1tan θ or tan θ =

1

cot θ

Example 1: Find the value of each expression using a reciprocal identity.

a. cos θ = .3, sec θ =?

b. cot θ = 43 , tan θ =?

Solution:

a. sec θ = 103

These functions are reciprocals, so if cos θ = .3, then sec θ = 1.3 . It is easier to find the reciprocal if we

express the values as fractions: cos θ = .3 = 310 ⇒ sec θ = 103 .

b. tan θ = 34

These functions are reciprocals, and the reciprocal of 43 is 34 .

We can also use the reciprocal relationships to determine the domain and range of functions.

Domain, Range, and Signs of Trig Functions

While the trigonometric functions may seem quite different from other functions you have worked with,

they are in fact just like any other function. We can think of a trig function in terms of “input” and

“output.” The input is always an angle. The output is a ratio of sides of a triangle. If you think about the

trig functions in this way, you can define the domain and range of each function.

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Let’s first consider the sine and cosine functions. The input of each of these functions is always an angle,

and as you learned in the previous sections, these angles can take on any real number value. Therefore the

sine and cosine function have the same domain, the set of all real numbers, R. We can determine the range

of the functions if we think about the fact that the sine of an angle is the y−coordinate of the point where

the terminal side of the angle intersects the unit circle. The cosine is the x−coordinate of that point. Now

recall that in the unit circle, we defined the trig functions in terms of a triangle with hypotenuse 1.

In this right triangle, x and y are the lengths of the legs of the triangle, which must have lengths less than

1, the length of the hypotenuse. Therefore the ranges of the sine and cosine function do not include values

greater than one. The ranges do, however, contain negative values. Any angle whose terminal side is in

the third or fourth quadrant will have a negative y−coordinate, and any angle whose terminal side is in

the second or third quadrant will have a negative x−coordinate.

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In either case, the minimum value is -1. For example, cos 180◦ = −1 and sin 270◦ = −1. Therefore the sine

and cosine function both have range from -1 to 1.

The table below summarizes the domains and ranges of these functions:

Table 1.5:

Domain Range

Sine θ = R −1 ≤ y ≤ 1

Cosine θ = R −1 ≤ y ≤ 1

Knowing the domain and range of the cosine and sine function can help us determine the domain and range

of the secant and cosecant function. First consider the sine and cosecant functions, which as we showed

above, are reciprocals. The cosecant function will be defined as long as the sine value is not 0. Therefore

the domain of the cosecant function excludes all angles with sine value 0, which are 0◦, 180◦, 360◦, etc.

In Chapter 2 you will analyze the graphs of these functions, which will help you see why the reciprocal

relationship results in a particular range for the cosecant function. Here we will state this range, and in

the review questions you will explore values of the sine and cosecant function in order to begin to verify

this range, as well as the domain and range of the secant function.

Table 1.6:

Domain Range

Cosecant θ◦, θ , 0.180, 360 . . . csc θ ≤ −1 or csc θ ≥ 1

Secant θ◦, θ , 90, 270, 450 . . . sec θ ≤ −1 or sec θ ≥ 1

Now let’s consider the tangent and cotangent functions. The tangent function is defined as tan θ = yx .

Therefore the domain of this function excludes angles for which the ordered pair has an x−coordinate of

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0 : 90◦, 270◦, etc. The cotangent function is defined as cot θ = xy , so this function’s domain will exclude

angles for which the ordered pair has a y−coordinate of 0 : 0◦, 180◦, 360◦, etc.

Table 1.7:

Function Domain Range

Tangent θ◦, θ , 90, 270, 450 . . . All reals

Cotangent θ◦, θ , 0, 180, 360 . . . All reals

Knowing the ranges of these functions tells you the values you should expect when you determine the value

of a trig function of an angle. However, for many problems you will need to identify the sign of the function

of an angle: Is it positive or negative?

In determining the ranges of the sine and cosine functions above, we began to categorize the signs of these

functions in terms of the quadrants in which angles lie. The figure below summarizes the signs for angles

in all 4 quadrants.

An easy way to remember this is “All Students Take Calculus.” Quadrant I: All values are positive,

Quadrant II: Sine is positive, Quadrant III: Tangent is positive, and Quadrant IV: Cosine is positive. This

simple analogy will help you remember which trig functions are positive and where.

Example 2: State the sign of each expression.

a. cos 100◦

b. csc 220◦

c. tan 370◦

Solution:

a. The angle 100◦ is in the second quadrant. Therefore the x−coordinate is negative and so cos 100◦ is

negative.

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b. The angle 220◦ is in the third quadrant. Therefore the y−coordinate is negative. So the sine, and the

cosecant are negative.

c. The angle 370◦ is in the first quadrant. Therefore the tangent value is positive.

So far we have considered relationships between pairs of functions: the six trig functions can be grouped

in pairs as reciprocals. Now we will consider relationships among three trig functions.

Quotient Identities

The definitions of the trig functions led us to the reciprocal identities above. They also lead us to another

set of identities, the quotient identities.

Consider first the sine, cosine, and tangent functions. For angles of rotation (not necessarily in the unit

circle) these functions are defined as follows:

sin θ = y

r

cos θ = x

r

tan θ = y

x

Given these definitions, we can show that tan θ = sin θcos θ , as long as cos θ , 0:

sin θ

cos θ =

y

r

x

r

=

y

r

× r

x

=

y

x

= tan θ.

The equation tan θ = sin θcos θ is therefore an identity that we can use to find the value of the tangent function,

given the value of the sine and cosine.

Example 3: If cos θ = 513 and sin θ = 1213 , what is the value of tan θ?

Solution: tan θ = 125

tan θ = sin θcos θ =

12

13

5

13

=

12

13

× 13

5

=

12

5

Example 4: Show that cot θ = cos θsin θ

Solution:

cos θ

sin θ =

x

r

y

r

=

x

r

× r

y

=

x

y

= cot θ

This is also an identity that you can use to find the value of the cotangent function, given values of sine

and cosine. Both of the quotient identities will also be useful in chapter 3, in which you will prove other

identities.

Cofunction Identities and Reflection

These identities relate to the problems you did in section 1.3. Recall, #3 and #4 from the review questions,

where sin X = cosZ and cos X = sinZ, where X and Z were complementary angles. These are called

cofunction identities because the functions have common values. These identities are summarized below.

61 www.ck12.org

sin θ = cos(90◦ − θ) cos θ = sin(90◦ − θ)

tan θ = cot(90◦ − θ) cot θ = tan(90◦ − θ)

Example 5: Find the value of each trig function.

a. cos 120◦

b. cos(−120◦)

c. sin 135◦

d. sin(−135◦)

Solution: Because these angles have reference angles of 60◦ and 45◦, the values are:

a. cos 120◦ = −12

b. cos(−120◦) = cos 240◦ = −12

c. sin 135◦ =

√

2

2

d. sin(−135◦) = sin 225◦ = −

√

2

2

These values show us that sine and cosine also reflect over the x axis. This allows us to generate three

more identities.

sin(−θ) = − sin θ cos(−θ) = cos θ tan(−θ) = − tan θ

Pythagorean Identities

The final set of identities are called the Pythagorean Identities because they rely on the Pythagorean

Theorem. In previous lessons we used the Pythagorean Theorem to find the sides of right triangles.

Consider once again the way that we defined the trig functions in 1.3. Let’s look at the unit circle:

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The legs of the right triangle are x, and y. The hypotenuse is 1. Therefore the following equation is true

for all x and y on the unit circle:

x2 + y2 = 1

Now remember that on the unit circle, cos θ = x and sin θ = y. Therefore the following equation is an

identity:

cos2 θ + sin2 θ = 1

Note: Writing the exponent 2 after the cos and sin is the standard way of writing exponents. Just keeping

mind that cos2 θ means (cos θ)2 and sin2 θ means (sin θ)2.

We can use this identity to find the value of the sine function, given the value of the cosine, and vice versa.

We can also use it to find other identities.

Example 6: If cos θ = 14 what is the value of sin θ? Assume that θ is an angle in the first quadrant.

Solution: sin θ =

√

15

4

cos2 θ + sin2 θ = 1(1

4

)2

+ sin2 θ = 1

1

16

+ sin2 θ = 1

sin2 θ = 1 − 1

16

sin2 θ = 15

16

sin θ = ±

√

15

16

sin θ = ±

√

15

4

Remember that it was given that θ is an angle in the first quadrant. Therefore the sine value is positive,

so sin θ =

√

15

4 .

Example 7: Use the identity cos2 θ + sin2 θ = 1 to show that cot2 θ + 1 = csc2 θ

Solution:

63 www.ck12.org

cos2 θ + sin2 θ = 1 Divide both sides by sin2 θ.

cos2 θ + sin2 θ

sin2 θ =

1

sin2 θ

cos2 θ

sin2 θ +

sin2 θ

sin2 θ =

1

sin2 θ

sin2 θ

sin2 θ = 1

cos2 θ

sin2 θ + 1 =

1

sin2 θ

cos θ

sin θ ×

cos θ

sin θ + 1 =

1

sin θ ×

1

sin θ Write the squared functions in terms

of their factors.

cot θ × cot θ + 1 = csc θ × csc θ Use the quotient and reciprocal

identities.

cot2 θ + 1 = csc2 θ Write the functions as squared

functions.

Points to Consider

1. How do you know if an equation is an identity? HINT: you could consider using a the calculator and

graphing a related function, or you could try to prove it mathematically.

2. How can you verify the domain or range of a function?

Review Questions

1. Use reciprocal identities to give the value of each expression.

(a) sec θ = 4, cos θ =?

(b) sin θ = 13 , csc θ =?

2. In the lesson, the range of the cosecant function was given as: csc θ ≤ −1 or csc θ ≥ 1.

(a) Use a calculator to fill in the table below. Round values to 4 decimal places.

(b) Use the values in the table to explain in your own words what happens to the values of the

cosecant function as the measure of the angle approaches 0 degrees.

(c) Explain what this tells you about the range of the cosecant function.

(d) Discuss how you might further explore values of the sine and cosecant to better understand the

range of the cosecant function.

Table 1.8:

Angle Sin Csc

10

5

1

0.5

0.1

0

-.1

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Table 1.8: (continued)

Angle Sin Csc

-.5

-1

-5

-10

3. In the lesson the domain of the secant function were given:

Domain: θ◦, θ , 90, 270, 450 . . .

Explain why certain values are excluded from the domain.

4. State the quadrant in which each angle lies, and state the sign of each expression

(a) sin 80◦

(b) cos 200◦

(c) cot 325◦

(d) tan 110◦

5. If cos θ = 610 and sin θ = 810 , what is the value of tan θ?

6. Use quotient identities to explain why the tangent and cotangent function have positive values for

angles in the third quadrant.

7. If sin θ = 0.4, what is the value of cos θ? Assume that θ is an angle in the first quadrant.

8. If cot θ = 2, what is the value of csc θ? Assume that θ is an angle in the first quadrant.

9. Show that 1 + tan2 θ = sec2 θ.

10. Explain why it is necessary to state the quadrant in which the angle lies for problems such as #7.

Review Answers

1. (a) 14

(b) 31 = 3

2. (a)

Table 1.9:

Angle Sin Csc

10 .1737 5.759

5 .0872 11.4737

1 .0175 57.2987

0.5 .0087 114.5930

0.1 .0018 572.9581

0 0 undefined

-.1 -.0018 -572.9581

-.5 -.0087 -114.5930

-1 -.0175 -57.2987

-5 -.0872 -11.4737

-10 -.1737 -5.759

(b) As the angle gets smaller and smaller, the cosecant values get larger and larger.

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(c) The range of the cosecant function does not have a maximum, like the sine function. The values get

larger and larger.

(d) Answers will vary. For example, if we looked at values near 90 degrees, we would see the cosecant

values get smaller and smaller, approaching 1.

3. The values 90, 270, 450, etc, are excluded because they make the function undefined.

4. (a) Quadrant 1; positive

(b) Quadrant 3; negative

(c) Quadrant 4; negative

(d) Quadrant 2; negative

5. 86 = 43

6. The ratio of sine and cosine will be positive in the third quadrant because sine and cosine are both

negative in the third quadrant.

7. cos θ ≈ .92

8. csc θ =

√

59.

cos2 θ + sin2 θ = 1

cos2 θ + sin2 θ

cos2 θ =

1

cos2 θ

1 +

sin2 θ

cos2 θ =

1

cos2 θ

1 + tan2 θ = sec2 θ

10. Using the Pythagorean identities results in a quadratic equation and will have two solutions. Stat-

ing that the angle lies in a particular quadrant tells you which solution is the actual value of the

expression. In #7, the angle is in the first quadrant, so both sine and cosine must be positive.

1.9 Chapter Review

Chapter Summary

In this chapter students learned about right triangles and special right triangles. Through the special right

triangles and the Pythagorean Theorem, the study of trigonometry was discovered. Sine, cosine, tangent,

secant, cosecant, and cotangent are all functions of angles and the result is the ratio of the sides of a right

triangle. We learned that only our special right triangles generate sine, cosine, tangent values that can be

found without the use of a scientific calculator. When incorporating the trig ratios and the Pythagorean

Theorem, we discovered the first of many trig identities. This concept will be explored further in Chapter

3.

Vocabulary

Adjacent A side adjacent to an angle is the side next to the angle. In a right triangle, it is the leg that

is next to the angle.

Angle of depression The angle between the horizontal line of sight, and the line of sight down to a

given point.

Angle of elevation The angle between the horizontal line of sight, and the line of sight up to a given

point.

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Bearings The direction from one object to another, usually measured as an angle.

Clinometer A device used to measure angles of elevation or depression.

Coterminal angles Two angles in standard position are coterminal if they share the same terminal side.

Distance Formula d =

√

(x1 − x2)2 + (y1 − y2)2

Hypotenuse The hypotenuse is the longest side in a right triangle, opposite the right angle.

Identity An identity is an equation that is always true, as long as the variables and expressions involved

are defined.

Included Angle The angle inbetween two sides of a polygon.

Leg The legs of a right triangle are the two shorter sides.

Nautical Mile A nautical mile is a unit of length that corresponds approximately to one minute of

latitude along any meridian. A nautical mile is equal to 1.852 meters.

Pythagorean Theorem a2 + b2 = c2

Pythagorean Triple A set whole numbers for which the Pythagorean Theorem holds true.

Quadrantal angle A quadrantal angle is an angle in standard position whose terminal side lies on an

axis.

Radius The radius of a circle is the distance from the center of the circle to the edge. The radius defines

the circle.

Reference angle The reference angle of an angle in standard position is the measure of the angle between

the terminal side and the closest portion of the x−axis.

Standard position An angle in standard position has its initial side on the positive x−axis, its vertex

at the origin, and its terminal side anywhere in the plane. A positive angle means a counterclockwise

rotation. A negative angle means a clockwise rotation.

Theodolite A device used to measure angles of elevation or depression.

Unit Circle The unit circle is the circle with radius 1 and center (0, 0). The equation of the unit circle

is x2 + y2 = 1

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Review Questions

1. One way to prove the Pythagorean Theorem is through the picture below. Determine the area of the

square two different ways and set each equal to each other.

2. A flute is resting diagonally, d, in the rectangular box (prism) below. Find the length of the flute.

3. Solve the right triangle.

4. Solve the right triangle.

5. Find the exact value of the area of the parallelogram below.

6. The modern building shown below is built with an outer wall (shown on the left) that is not at a

90-degree angle with the floor. The wall on the right is perpendicular to both the floor and ceiling.

Find the length of w.

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7. Given that cos(90◦ − x) = 27 , find the sin x.

8. If cos(−x) = 34 and tan x =

√

7

3 , find sin(−x).

9. If sin y = 13 , what is cos y?

10. sin θ = 13 find the value(s) of cos θ.

11. cos θ = −25 , and θ is a second quadrant angle. Find the exact values of remaining trigonometric

functions.

12. (3, -4) is a point on the terminal side of θ. Find the exact values of the six trigonometric functions.

13. Determine reference angle and two coterminal angles for 165◦. Plot the angle in standard position.

14. It is very helpful to have the unit circle with all the special values on one circle. Fill out the unit

circle below with all of the endpoints for each special value and quadrantal value.

Review Answers

1. Area 1:

(a+ b)2

(a+ b)(a+ b)

a2 + 2ab+ b2

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Area 2: Add up 4 triangles and inner square.

4 · 1

2

ab+ c2

2ab+ c2

Set the two equal to each other:

a2 + 2ab+ b2 = 2ab+ c2

a2 + b2 = c2

2. First, find the diagonal of the base. This is a Pythagorean Triple, so the base diagonal is 25 (you

could have also done the Pythagorean Theorem if you didn’t see this). Now, do the Pythagorean

Theorem with the height and the diagonal to get the three-dimensional diagonal.

72 + 252 = d2

49 + 225 = d2

274 = d2√

274 = d ≈ 16.553.

∠C = 90◦ − 23.6◦ = 66.4◦

sin 23.6 = CA

25

cos 23.6 = AT

25

25 · sin 23.6 = CA 25 · cos 23.6 = AT

10.01 ≈ CA 22.9 ≈ AT

4. First do the Pythagorean Theorem to get the third side.

72 + x2 = 182

49 + x2 = 324

x2 = 275

x =

√

275 = 5

√

11

Second, use one of the inverse functions to find the two missing angles.

sinG = 7

18

sin−1

( 7

18

)

= G We can subtract ∠G from 90 to get 67.11◦.

G ≈ 22.89◦5.

A = ab sinC

= 16 · 22 · sin 60◦

= 352 ·

√

3

2

= 176

√

3

6. Make a right triangle with 165 as the opposite leg and w is the hypotenuse.

sin 85◦ = 165

w

w sin 85◦ = 165

w =

165

sin 85◦

w ≈ 165.63

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7.

cos(90◦ − x) = sin x

sin x = 2

7

8. If cos(−x) = 34 , then cos x = 34 . With tan x =

√

7

3 , we can conclude that sin x =

√

7

4 and sin(−x) = −

√

7

4 .

9. If sin y = 13 , then we know the opposite side and the hypotenuse. Using the Pythagorean Theorem, we

get that the adjacent side is 2

√

2

(

12 + b2 = 32 → b = √9 − 1→ b = √8 = 2√2

)

. Thus, cos y = ±2

√

2

3

because we don’t know if the angle is in the second or third quadrant.

10. sin θ = 13 , sine is positive in Quadrants I and II. So, there can be two possible answers for the cos θ.

Find the third side, using the Pythagorean Theorem:

12 + b2 = 32

1 + b2 = 9

b2 = 8

b =

√

8 = 2

√

2

In Quadrant I, cos θ = 2

√

2

3

In Quadrant II, cos θ = −2

√

2

3

11. cos θ = −25 and is in Quadrant II, so from the Pythagorean Theorem :

a2 + (−2)2 = 52

a2 + 4 = 25

a2 = 21

a =

√

21

So, sin θ =

√

21

5 and tan θ = −

√

21

2

12. If the terminal side of θ is on (3, -4) means θ is in Quadrant IV, so cosine is the only positive function.

Because the two legs are lengths 3 and 4, we know that the hypotenuse is 5. 3, 4, 5 is a Pythagorean

Triple (you can do the Pythagorean Theorem to verify). Therefore, sin θ = 35 , cos θ = −45 , tan θ = −43

13. Reference angle = 15◦. Possible coterminal angles = −195◦, 525◦

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14.

Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator

activities designed to supplement the objectives for some of the lessons in this chapter. See

http://www.ck12.org/flexr/chapter/9699.

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Chapter 2

Graphing Trigonometric

Functions - 2nd edition

2.1 Radian Measure

Introduction

Now that we know how to find the sine, cosine and tangent of any angle, we can extend this concept to the

x − y plane. First, we need to derive a different way to measure angles, called radians. Radians are much

like arc length. This way, we can take the “length” of a degree measurement and plot it like x. Then, the

value of the function is the y value on the graph. In this manner we will be able to see the six trigonometric

functions in a whole new way.

Learning Objectives

• Define radian measure.

• Convert angle measure from degrees to radians and radians to dgrees.

• Calculate the values of the 6 trigonometric functions for special angles in terms of radians or degrees.

Understanding Radian Measure

Until now, we have used degrees to measure angles. But, what exactly is a degree? A degree is 1

360th

of

a complete rotation around a circle. Radians are alternate units used to measure angles in trigonometry.

Just as it sounds, a radian is based on the radius of a circle. One radian (abbreviated rad) is the angle

created by bending the radius length around the arc of a circle. Because a radian is based on an actual

part of the circle rather than an arbitrary division, it is a much more natural unit of angle measure for

upper level mathematics.

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What if we were to rotate all the way around the circle? Continuing to add radius lengths, we find that it

takes a little more than 6 of them to complete the rotation.

Recall from geometry that the arc length of a complete rotation is the circumference, where the formula is

equal to 2pi times the length of the radius. 2pi is approximately 6.28, so the circumference is a little more

than 6 radius lengths. Or, in terms of radian measure, a complete rotation (360 degrees) is 2pi radians.

360 degrees = 2pi radians

With this as our starting point, we can find the radian measure of other angles. Half of a rotation, or 180

degrees, must therefore be pi radians, and 90 degrees must be 12pi, written pi2 .

Example 1: Find the radian measure of these angles.

Table 2.1:

Angle in Degrees Angle in Radians

90 pi2

45

30

60

75

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Solution: Because 45 is half of 90, half of 12pi is 14pi. 30 is one-third of a right angle, so multiplying gives:

pi

2

× 1

3

=

pi

6

and because 60 is twice as large as 30:

2 × pi

6

=

2pi

6

=

pi

3

Here is the completed table:

Table 2.2:

Angle in Degrees Angle in Radians

90 pi2

45 pi4

30 pi6

60 pi3

There is a formula to convert between radians and degrees that you may already have discovered while

doing this example. However, many angles that are commonly used can be found easily from the values

in this table. For example, most students find it easy to remember 30 and 60. 30 is pi over 6 and 60 is pi

over 3. Knowing these angles, you can find any of the special angles that have reference angles of 30 and

60 because they will all have the same denominators. The same is true of multiples of pi4 (45 degrees) and

pi

2 (90 degrees).

Critial Angles in Radians

Extending the radian measure past the first quadrant, the quadrantal angles have been determined, except

270◦. Because 270◦ is halfway between 180◦, pi, and 360◦, 2pi, it must be 1.5pi, usually written 3pi2 .

For the 45◦ angles, the radians are all multiples of pi4 . For example, 135◦ is 3 · 45◦. Therefore, the radian

measure should be 3 · pi4 , or 3pi4 . Here are the rest of the multiples of 45◦, in radians:

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Notice that the additional angles in the drawing all have reference angles of 45 degrees and their radian

measures are all multiples of pi4 . All of the even multiples are the quadrantal angles and are reduced, just

like any other fraction.

Example 2: Complete the following radian measures by counting in multiples of pi3 and pi6 :

Solution:

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Notice that all of the angles with 60-degree reference angles are multiples of pi3 , and all of those with

30-degree reference angles are multiples of pi6 . Counting in these terms based on this pattern, rather than

converting back to degrees, will you better understand radians.

Converting Any Degree to Radians

For all examples there is a conversion formula. This formula works for all degrees and radians. Remember

that: pi radians = 180◦. If you divide both sides of this equation by pi, you will have the conversion formula:

radians × 180

pi

= degrees

If we have a degree measure and wish to convert it to radians, then manipulating the equation above gives:

degrees × pi

180

= radians

Example 3: Convert 11pi3 to degree measure

From the last section, you should recognize that this angle is a multiple of pi3 (or 60 degrees), so there are

11, pi3 ’s in this angle, pi3 × 11 = 60 × 11 = 660◦.

Here is what it would look like using the formula:

radians × 180pi = degrees

Example 4: Convert −120◦ to radian measure. Leave the answer in terms of pi.

degrees × pi

180

= radians

−120 × pi

180

=

−120pi

180

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and reducing to lowest terms gives us −2pi3

You could also have noticed that 120 is 2 × 60. Since 60◦ is pi3 radians, then 120 is 2, pi3 ’s, or 2pi3 . Make it

negative and you have the answer, −2pi3 .

Example 5: Express 11pi12 radians terms of degrees.

radians × 180pi = degrees

Note: Sometimes students have trouble remembering if it is 180pi or pi180 . It might be helpful to remember

that radian measure is almost always expressed in terms of pi. If you want to convert from radians to

degrees, you want the pi to cancel out when you multiply, so it must be in the denominator.

The Six Trig Functions and Radians

Eventhough you are used to performing the trig functions on degrees, they still will work on radians. The

only difference is the way the problem looks. If you see sin pi6 , that is still sin 30◦ and the answer is still 12 .

Example 6: Find tan 3pi4 .

Solution: If needed, convert 3pi4 to degrees. Doing this, we find that it is 135◦. So, this is tan 135◦, which

is -1.

Example 7: Find the value of cos 11pi6 .

Solution: If needed, convert 11pi6 to degrees. Doing this, we find that it is 330◦. So, this is cos 330◦, which

is

√

3

2 .

Example 8: Convert 1 radian to degree measure.

Solution: Many students get so used to using pi in radian measure that they incorrectly think that 1

radian means 1pi radians. While it is more convenient and common to express radian measure in terms

of pi, don’t lose sight of the fact that pi radians is a number. It specifies an angle created by a rotation of

approximately 3.14 radius lengths. So 1 radian is a rotation created by an arc that is only a single radius

in length.

radians × 180

pi

= degrees

So 1 radian would be 180pi degrees. Using any scientific or graphing calculator will give a reasonable

approximation for this degree measure, approximately 57.3◦.

Example 9: Find the radian measure of an acute angle, θ, with sin θ =

√

2

2 .

Solution: Here, we are working backwards. From last chapter, you may recognize that

√

2

2 goes with 45◦.

Because the example is asking for an acute angle, we just need to convert 45◦ to radians. 45◦ in radians is

pi

4 .

Check the Mode

Most scientific and graphing calculators have aMODE setting that will allow you to either convert between

the two, or to find approximations for trig functions using either measure. It is important that if you are

using your calculator to estimate a trig function that you know which mode you are using. Look at the

following screen:

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If you entered this expecting to find the sine of 30 degrees you would realize based on the last chapter

that something is wrong because it should be 12 . In fact, as you may have suspected, the calculator is

interpreting this as 30 radians. In this case, changing the mode to degrees and recalculating will give the

expected result.

Scientific calculators will usually have a 3-letter display that shows either DEG or RAD to tell you which

mode the calculator is in.

Points to Consider

• In certain cases, why are radians more useful than degrees?

• Think about the steps you would take to solve sin 11pi6 . Are these step similar for finding any trig

function for any angle in radians?

Review Questions

1. The following picture is a sign for a store that sells cheese.

(a) Estimate the degree measure of the angle of the circle that is missing.

(b) Convert that measure to radians.

(c) What is the radian measure of the part of the cheese that remains?

2. Convert the following degree measures to radians. All answers should be in terms of pi.

(a) 240◦

(b) 270◦

(c) 315◦

(d) −210◦

(e) 120◦

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Figure 2.1

(f) 15◦

(g) −450◦

(h) 72◦

(i) 720◦

(j) 330◦

3. Convert the following radian measures to degrees:

(a) pi2

(b) 11pi5

(c) 2pi3

(d) 5pi

(e) 7pi2

(f) 3pi10

(g) 5pi12

(h) −13pi6

(i) 8pi

(j) 4pi15

4. The drawing shows all the quadrant angles as well as those with reference angles of 30◦, 45◦, and 60◦.

On the inner circle, label all angles with their radian measure in terms of pi and on the outer circle,

label all the angles with their degree measure.

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5. Using a calculator, find the approximate degree measure (to the nearest tenth) of each angle expressed

in radians.

(a) 6pi7

(b) 1 radian

(c) 3 radian

(d) 20pi11

6. Gina wanted to calculate the cos 210◦ and got the following answer on her calculator:

Fortunately, Kylie saw her answer and told her that it was obviously incorrect.

(a) Write the correct answer, in simplest radical form.

(b) Explain what she did wrong.

7. Complete the following chart. Write your answers in simplest radical form.

Table 2.3:

x Sin(x) Cos(x) Tan(x)

5pi

4

11pi

6

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Table 2.3: (continued)

x Sin(x) Cos(x) Tan(x)

2pi

3

pi

2

7pi

2

Review Answers

1. (a) Answer may vary, but 120◦ seems reasonable.

(b) Based on the answer in part a., the ration masure would be 2pi3

(c) Again, based on part a., 4pi3

2. (a) 4pi3

(b) 3pi2

(c) 7pi4

(d) −7pi6

(e) 2pi3

(f) pi12

(g) −5pi2

(h) pi5

(i) 4pi

(j) 11pi6

3. (a) 90◦

(b) 396◦

(c) 120◦

(d) 540◦

(e) 630◦

(f) 54◦

(g) 75◦

(h) −210◦

(i) 1440◦

(j) 48◦

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4.

5. (a) 154.3◦

(b) 57.3◦

(c) 171.9◦

(d) 327.3◦

6. (a) The correct answer is −12

(b) Her calculator was is the wrong mode and she calculated the sine of 210 radians.

7.

Table 2.4:

x Sin(x) Cos(x) Tan(x)

5pi

4 −

√

2

2 −

√

2

2 1

11pi

6 −12

√

3

2 −

√

3

3

2pi

3

√

3

2 −12 −

√

3

pi

2 1 0 undefined

7pi

2 −1 0 undefined

2.2 Applications of Radian Measure

Learning Objectives

• Solve problems involving angles of rotation using radian measure.

• Calculate the length of an arc and the area of a sector.

• Approximate the length of a chord given the central angle and radius.

• Solve problems about angular speed.

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Rotations

Example 1: The hands of a clock show 11:20. Express the obtuse angle formed by the hour and minute

hands in radian measure.

Solution: The following diagram shows the location of the hands at the specified time.

Because there are 12 increments on a clock, the angle between each hour marking on the clock is 2pi12 = pi6 (or

30◦). So, the angle between the 12 and the 4 is 4 × pi6 = 2pi3 (or 120◦). Because the rotation from 12 to 4 is

one-third of a complete rotation, it seems reasonable to assume that the hour hand is moving continuously

and has therefore moved one-third of the distance between the 11 and the 12. So, 13 × pi6 = pi18 , and the

total measure of the angle is therefore pi18 + 2pi3 = pi18 + 12pi18 = 13pi18 .

Length of Arc

The length of an arc on a circle depends on both the angle of rotation and the radius length of the circle.

If you recall from the last lesson, the measure of an angle in radians is defined as the length of the arc cut

off by one radius length. What if the radius is 4 cm? Then, the length of the half-circle arc would be pi

multiplied by the radius length, or 4pi cm in length.

This results in a formula that can be used to calculate the length of any arc.

s = rθ,

where s is the length of the arc, r is the radius, and θ is the measure of the angle in radians.

Solving this equation for θ will give us a formula for finding the radian measure given the arc length and

the radius length:

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θ =

s

r

Example 2: The free-throw line on an NCAA basketball court is 12 ft wide. In international competition,

it is only about 11.81 ft. How much longer is the half circle above the free-throw line on the NCAA court?

Solution: Find both arc lengths.

NCAA INTERNATIONAL

s1 = rθ s2 = rθ

s1 = 12(pi) s2 ≈ 11.81(pi)

s1 = 12pi s2 ≈ 11.81pi

So the answer is approximately 12pi − 11.81pi ≈ 0.19pi

This is approximately 0.6 ft, or about 7.2 inches longer.

Example 3: Two connected gears are rotating. The smaller gear has a radius of 4 inches and the larger

gear’s radius is 7 inches. What is the angle through which the larger gear has rotated when the smaller

gear has made one complete rotation?

Solution: Because the blue gear performs one complete rotation, the length of the arc traveled is:

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s = rθ

s = 4 × 2pi

So, an 8pi arc length on the larger circle would form an angle as follows:

θ =

s

r

θ =

8pi

7

θ ≈ 3.6

So the angle is approximately 3.6 radians.

3.6 × 180

pi

≈ 206◦

Area of a Sector

One of the most common geometric formulas is the area of a circle:

A = pir2

In terms of angle rotation, this is the area created by 2pi radians.

2pi rad = pir2 area

A half-circle, or pi radian rotation would create a section, or sector of the circle equal to half the area or:

1

2

pir2

So an angle of 1 radian would define an area of a sector equal to:

1 = 12r

2

From this we can determine the area of the sector created by any angle, θ radians, to be:

A =

1

2

r2θ

Example 4: Crops are often grown using a technique called center pivot irrigation that results in circular

shaped fields.

Here is a satellite image taken over fields in Kansas that use this type of irrigation system.

If the irrigation pipe is 450 m in length, what is the area that can be irrigated after a rotation of 2pi3 radians?

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Figure 2.2

Figure 2.3

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Solution: Using the formula:

A =

1

2

r2θ

A =

1

2

(450)2

(2pi

3

)

The area is approximately 212,058 square meters.

Length of a Chord

You may recall from your Geometry studies that a chord is a segment that begins and ends on a circle.

AB is a chord in the circle.

We can calculate the length of any chord if we know the angle measure and the length of the radius.

Because each endpoint of the chord is on the circle, the distance from the center to A and B is the same as

the radius length.

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Next, if we bisect angle, the angle bisector must be perpendicular to the chord and bisect it (we will leave

the proof of this to your Geometry class). This forms a right triangle.

We can now use a simple sine ratio to find half the chord, called c here, and double the result to find the

length of the chord.

sin θ

2

=

c

r

c = r × sin θ

2

So the length of the chord is:

2c = 2r sin θ

2

Example 5: Find the length of the chord of a circle with radius 8 cm and a central angle of 110◦.

Approximate your answer to the nearest mm.

Solution: We must first convert the angle measure to radians:

110 × pi

180

=

11pi

18

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Using the formula, half of the chord length should be the radius of the circle times the sine of half the

angle.

11pi

18

× 1

2

=

11pi

36

8 × sin 11pi

36

Multiply this result by 2.

So, the length of the arc is approximately 13.1 cm.

Angular Velocity

What about objects that are traveling on a circular path? Do you remember playing on a merry-go-round

when you were younger?

Figure 2.4

If two people are riding on the outer edge, their velocities should be the same. But, what if one person is

close to the center and the other person is on the edge? They are on the same object, but their speed is

actually not the same.

Look at the following drawing.

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Imagine the point on the larger circle is the person on the edge of the merry-go-round and the point on

the smaller circle is the person towards the middle. If the merry-go-round spins exactly once, then both

individuals will also make one complete revolution in the same amount of time.

However, it is obvious that the person in the center did not travel nearly as far. The circumference (and of

course the radius) of that circle is much smaller and therefore the person who traveled a greater distance in

the same amount of time is actually traveling faster, even though they are on the same object. So the person

on the edge has a greater linear velocity (recall that linear velocity is found using distance = rate · time).

If you have ever actually ridden on a merry-go-round, you know this already because it is much more fun

to be on the edge than in the center! But, there is something about the two individuals traveling around

that is the same. They will both cover the same rotation in the same period of time. This type of speed,

measuring the angle of rotation over a given amount of time is called the angular velocity.

The formula for angular velocity is:

ω =

θ

t

ω is the last letter in the Greek alphabet, omega, and is commonly used as the symbol for angular velocity.

θ is the angle of rotation expressed in radian measure, and t is the time to complete the rotation.

In this drawing, θ is exactly one radian, or the length of the radius bent around the circle. If it took point

A exactly 2 seconds to rotate through the angle, the angular velocity of A would be:

ω =

θ

t

ω =

1

2

radians per second

In order to know the linear speed of the particle, we would have to know the actual distance, that is, the

length of the radius. Let’s say that the radius is 5 cm.

If linear velocity is v = dt then, v = 52 or 2.5 cm per second.

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If the angle were not exactly 1 radian, then the distance traveled by the point on the circle is the length

of the arc, s = rθ, or, the radius length times the measure of the angle in radians.

Substituting into the formula for linear velocity gives: v = rθt or v = r · θt .

Look back at the formula for angular velocity. Substituting ω gives the following relationship between

linear and angular velocity, v = rω. So, the linear velocity is equal to the radius times the angular velocity.

Remember in a unit circle, the radius is 1 unit, so in this case the linear velocity is the same as the angular

velocity.

v = rω

v = 1 × ω

v = ω

Here, the distance traveled around the circle is the same for a given unit of time as the angle of rotation,

measured in radians.

Example 6: Lindsay and Megan are riding on a Merry-go-round. Megan is standing 2.5 feet from the

center and Lindsay is riding on the outside edge 7 feet from the center. It takes them 6 seconds to complete

a rotation. Calculate the linear and angular velocity of each girl.

Solution: We are told that it takes 6 seconds to complete a rotation. A complete rotation is the same as

2pi radians. So the angular velocity is:

ω = θt =

2pi

6 =

pi

3 radians per second, which is slightly more than 1 (about 1.05), radian per second. Because

both girls cover the same angle of rotation in the same amount of time, their angular speed is the same.

In this case they rotate through approximately 60 degrees of the circle every second.

As we discussed previously, their linear velocities are different. Using the formula, Megan’s linear velocity

is:

v = rω = (2.5)

(

pi

3

)

≈ 2.6 ft per sec

Lindsay’s linear velocity is:

v = rω = (7)

(

pi

3

)

≈ 7.3 ft per sec

Points to Consider

• What is the difference between finding arc length and the area of a sector?

• What is the difference between linear velocity and angular velocity?

• How are linear and angular velocity related?

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Review Questions

1. The following image shows a 24-hour clock in Curitiba, Paraná, Brasil.

Figure 2.5

(a) What is the angle between each number of the clock expressed in:

i. exact radian measure in terms of pi ?

ii. to the nearest tenth of a radian?

iii. in degree measure?

(b) Estimate the measure of the angle between the hands at the time shown in:

i. to the nearest whole degree

ii. in radian measure in terms of pi

2. The following picture is a window of a building on the campus of Princeton University in Princeton,

New Jersey.

(a) What is the exact radian measure in terms of pi between two consecutive circular dots on the

small circle in the center of the window?

(b) If the radius of this circle is about 0.5 m, what is the length of the arc between the centers of

each consecutive dot? Round your answer to the nearest cm.

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Figure 2.6

3. Now look at the next larger circle in the window.

(a) Find the exact radian measure in terms of pi between two consecutive dots in this window.

(b) The radius of the glass portion of this window is approximately 1.20 m. Calculate an estimate

of the length of the highlighted chord to the nearest cm. Explain the reasoning behind your

solution.

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4. The state championship game is to be held at Ray Diaz Memorial Arena. The seating forms a perfect

circle around the court. The principal of Archimedes High School is sent the following diagram

showing the seating allotted to the students at her school.

It is 55 ft from the center of the court to the beginning of the stands and 110 ft from the center

to the end. Calculate the approximate number of square feet each of the following groups has been

granted:

(a) the students from Archimedes.

(b) general admission.

(c) the press and officials.

5. Doris and Lois go for a ride on a carousel. Doris rides on one of the outside horses and Lois rides

on one of the smaller horses near the center. Lois’ horse is 3 m from the center of the carousel, and

Doris’ horse is 7 m farther away from the center than Lois’. When the carousel starts, it takes them

12 seconds to complete a rotation.

(a) Calculate the linear velocity of each girl.

(b) Calculate the angular velocity of the horses on the carousel.

6. The Large Hadron Collider near Geneva, Switerland began operation in 2008 and is designed to

perform experiments that physicists hope will provide important information about the underlying

structure of the universe. The LHC is circular with a circumference of approximately 27,000 m.

Protons will be accelerated to a speed that is very close to the speed of light (≈ 3 × 108 meters per

second).

(a) How long does it take a proton to make a complete rotation around the collider?

(b) What is the approximate (to the nearest meter per second) angular speed of a proton traveling

around the collider?

(c) Approximately how many times would a proton travel around the collider in one full second?

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Review Answers

1. (a) i. pi12

ii. ≈ 0.3 radians

iii. 15◦

(b) i. 20◦. Answers may vary, anything above 15◦ and less than 25◦ is reasonable.

ii. pi9 Again, answers may vary

2. (a) pi6

(b) ≈ 26 cm

3. (a) pi6

(b) Let’s assume, to simplify, that the chord stretches to the center of each of the dots. We need to

find the measure of the central angle of the circle that connects those two dots.

Since there are 13 dots, this angle is 13pi16 . The length of the chord then is:

= 2r sin θ

2

= 2 × 1.2 × sin 1

2

× 13pi

16

The chord is approximately 2.30 cm.

4. Each section is pi6 radians. The area of one section of the stands is therefore the area of the outer

sector minus the area of the inner sector:

A = Aouter − Ainner

A =

1

2

(router)2 × pi6 −

1

2

(rinner)2 × pi6

A =

1

2

(110)2 × pi

6

− 1

2

(55)2 × pi

6

The area of each section is approximately 2376 f t2.

(a) The students have 4 sections or ≈ 9503 f t2

(b) There are 3 general admission sections or ≈ 7127 f t2

(c) There is only one press and officials section or ≈ 2376 f t2

5. It is actually easier to calculate the angular velocity first. ω = 2pi12 = pi6 , so the angular velocity is

pi

6 rad, or 0.524. Because the linear velocity depends on the radius,each girl has her own.

Lois: v = rω = 3 · pi6 = pi2 or 1.57 m/sec

Doris: v = rω = 10 · pi6 = 5pi3 or 5.24 m/sec

6. (a) v = dt → 3 × 108 = 27,000t → t = 2.7×10

4

3×108 = 0.9 × 10−4 = 9 × 10−5 or 0.00009 seconds.

(b) ω = θt = 2pi0.00009 = 22, 222.22 rad/sec

(c) The proton rotates around once in 0.00009 seconds. So, in one second it will rotate around the

LHC 1 ÷ 0.00009 = 11, 111.11 times, or just over 11,111 rotations.

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2.3 Circular Functions of Real Numbers

Learning Objectives

• Graph the six trigonometric ratios as functions on the Cartesian plane.

• Identify the domain and range of these six trigonometric functions.

• Identify the radian and degree measure, as well as the coordinates of points on the unit circle and

graph for the critical angles.

The Sine Graph

By now, you have become very familiar with the specific values of sine, cosine, and tangents for certain

angles of rotation around the coordinate grid. In mathematics, we can often learn a lot by looking at how

one quantity changes as we consistently vary another. We will be looking at the sine value as a function of

the angle of rotation around the coordinate plane. We refer to any such function as a circular function,

because they can be defined using the unit circle. Recall from earlier sections that the sine of an angle

in standard position is the ratio of yr , where y is the y−coordinate of any point on the circle and r is the

distance from the origin to that point.

Because the ratios are the same for a given angle, regardless of the length of the radius r, we can use the

unit circle as a basis for all calculations.

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The denominator is now 1, so we have the simpler expression, sin x = y. The advantage to this is that

we can use the y−coordinate of the point on the unit circle to trace the value of sin θ through a complete

rotation. Imagine if we start at 0 and then rotate counter-clockwise through gradually increasing angles.

Since the y−coordinate is the sine value, watch the height of the point as you rotate.

Through Quadrant I that height gets larger, starting at 0, increasing quickly at first, then slower until the

angle reaches 90◦, at which point, the height is at its maximum value, 1.

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As you rotate into the second quadrant, the height starts to decrease towards zero.

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When you start to rotate into the third and fourth quadrants, the length of the segment increases, but this

time in a negative direction, growing to -1 at 270◦ and heading back toward 0 at 360◦.

After one complete rotation, even though the angle continues to increase, the sine values will repeat

themselves. The same would have been true if we chose to rotate clockwise to investigate negative angles,

and this is why the sine function is a periodic function. The period is 2pi because that is the angle measure

before the sine of the angle will repeat its values.

Let’s translate this circular motion into a graph of the sine value vs. the angle of rotation. The following

sequence of pictures demonstrates the connection. These pictures plot (θ, sin θ) on the coordinate plane as

(x, y).

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After we rotate around the circle once, the values start repeating. Therefore, the sine curve, or “wave,”

also continues to repeat. The easiest way to sketch a sine curve is to plot the points for the quadrant

angles. The value of sin θ goes from 0 to 1 to 0 to -1 and back to 0. Graphed along a horizontal axis, it

would look like this:

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Filling in the gaps in between and allowing for multiple rotations as well as negative angles results in the

graph of y = sin x where x is any angle of rotation, in radians.

As we have already mentioned, sin x has a period of 2pi. You should also note that the y−values never go

above 1 or below -1, so the range of a sine curve is {−1 ≤ y ≤ 1}. Because angles can be any value and will

continue to rotate around the circle infinitely, there is no restriction on the angle x, so the domain of sin x

is all reals.

The Cosine Graph

In chapter 1, you learned that sine and cosine are very closely related. The cosine of an angle is the same

as the sine of its complementary angle. So, it should not be a surprise that sine and cosine waves are very

similar in that they are both periodic with a period of 2pi, a range from -1 to 1, and a domain of all real

angles.

The cosine of an angle is the ratio of xr , so in the unit circle, the cosine is the x−coordinate of the point of

rotation. If we trace the x−coordinate through a rotation, notice the change in the distance is similar to

sin x, but cos x starts at one instead of zero. The x−coordinate at 0◦ is 1 and the x−coordinate for 90◦ is

0, so the cosine value is decreasing from 1 to 0 through the 1st quadrant.

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Here is a similar sequence of rotations to the one used for sine. This time compare the x− coordinate of

the point of rotation with the height of the point as it traces along the horizontal. These pictures plot

(θ, cos θ) on the coordinate plane as (x, y).

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Plotting the quadrant angles and filling in the in-between values shows the graph of y = cos x

The graph of y = cos x has a period of 2pi. Just like sin x, the range of a cosine curve is {−1 ≤ y ≤ 1} and

the domain of cos x is all reals. Notice that the shape of the curve is exactly the same, but shifted by pi2 .

The Tangent Graph

The name of the tangent function comes from the tangent line of a circle. This is a line that is perpendicular

to the radius at a point on the circle so that the line touches the circle at exactly one point.

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If we extend angle θ through the unit circle so that it intersects with the tangent line, the tangent function

is defined as the length of the red segment.

The dashed segment is 1 because it is the radius of the unit circle. Recall that the tan θ = yx , and it can

be verified that this segment is the tangent by using similar triangles.

tan θ = y

x

=

t

1

= t

tan θ = t

So, as we increase the angle of rotation, think about how this segment changes. When the angle is zero,

the segment has no length. As we rotate through the first quadrant, it will increase very slowly at first

and then quickly get very close to one, but never actually touch it.

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As we get very close to the y−axis the segment gets infinitely large, until when the angle really hits 90◦,

at which point the extension of the angle and the tangent line will actually be parallel and therefore never

intersect.

This means there is no finite length of the tangent segment, or the tangent segment is infinitely large.

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Let’s translate this portion of the graph onto the coordinate plane. Plot (θ, tan θ) as (x, y).

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In fact as we get infinitely close to 90◦, the tangent value increases without bound, until when we actually

reach 90◦, at which point the tangent is undefined. Recall there are some angles (90◦ and 270◦, for example)

for which the tangent is not defined. Therefore, at these points, there are going to be vertical asymptotes.

Rotating past 90◦, the intersection of the extension of the angle and the tangent line is actually below the

x−axis. This fits nicely with what we know about the tangent for a 2nd quadrant angle being negative. At

first, it will have very large negative values, but as the angle rotates, the segment gets shorter, reaches 0,

then crosses back into the positive numbers as the angle enters the 3rd quadrant. The segment will again

get infinitely large as it approaches 270◦. After being undefined at 270◦, the angle crosses into the 4th

quadrant and once again changes from being infinitely negative, to approaching zero as we complete a full

rotation.

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The graph y = tan x over several rotations would look like this:

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Notice the x−axis is measured in radians. Our asymptotes occur every pi radians, starting at pi2 . The period

of the graph is therefore pi radians. The domain is all reals except for the asymptotes at pi2 , 3pi2 ,−pi2 , etc. and

the range is all real numbers.

The Three Reciprocal Functions

For the three reciprocal functions, it gets increasingly difficult to show the segment representation on the

unit circle. Instead of going through all of this, we will show the cot x, csc x, and sec x through the graphs

of their reciprocal functions, tan x, sin x, and cos x.

Cotangent

Cotangent is the reciprocal of tangent, xy , so it would make sense that where ever the tangent had an

asymptote, now the cotangent will be zero. The opposite of this is also true. When the tangent is zero,

now the cotangent will have an asymptote. The shape of the curve is generally the same, so the graph

looks like this:

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When you overlap the two functions, notice that the graphs consistently intersect at 1 and -1. These are

the angles that have 45◦ as reference angles, which always have tangents and cotangents equal to 1 or -1.

It makes sense that 1 and -1 are the only values for which a function and it’s reciprocal are the same. Keep

this in mind as we look at cosecant and secant compared to their reciprocals of sine and cosine.

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The cotangent function has a domain of all real angles except multiples of pi {. . . − 2pi,−pi, 0, pi, 2pi . . .} The

range is all real numbers.

Cosecant

Cosecant is the reciprocal of sine, or 1y . Therefore, whenever the sine is zero, the cosecant is going to have

a vertical asymptote because it will be undefined. It also has the same sign as the sine function in the

same quadrants. Here is the graph.

The period of the function is 2pi, just like sine. The domain of the function is all real numbers, except

multiples of pi {. . . − 2pi,−pi, 0, pi, 2pi . . .}. The range is all real numbers greater than or equal to 1, as well as

all real numbers less than or equal to -1. Notice that the range is everything except where sine is defined

(other than the points at the top and bottom of the sine curve).

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Notice again the reciprocal relationships at 0 and the asymptotes. Also look at the intersection points of

the graphs at 1 and -1. Many students are reminded of parabolas when they look at the half-period of

the cosecant graph. While they are similar in that they each have a local minimum or maximum and they

have the same beginning and ending behavior, the comparisons end there. Parabolas are not restricted by

asymptotes, whereas the cosecant curve is.

Secant

Secant is the reciprocal of cosine, or 1x . Therefore, whenever the cosine is zero, the secant is going to have

a vertical asymptote because it will be undefined. It also has the same sign as the cosine function in the

same quadrants. Here is the graph.

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The period of the function is 2pi, just like cosine. The domain of the function is all real numbers, except

multiples of pi starting at pi2 .

{

. . . −pi2 ,

pi

2 , 0,

3pi

2 ,

5pi

2 . . .

}

. The range is all real numbers greater than or equal to 1

as well as all real numbers less than or equal to -1. Notice that the range is everything except where cosine

is defined (other than the tops and bottoms of the cosine curve).

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Notice again the reciprocal relationships at 0 and the asymptotes. Also look at the intersection points of

the graphs at 1 and -1. Again, this graph looks parabolic, but it is not.

The table below summarizes the functions with their domains and ranges:

Table 2.5:

Function Domain Range

sin x all reals {y : −1 ≤ y ≤ 1}

cos x all reals {y : −1 ≤ y ≤ 1}

tan x

{

x : x , n × pi2 , where n is any odd integer

}

all reals

csc x {x : x , npi, where n is any integer} {y : y ≥ 1 or y ≤ −1}

sec x

{

x : x , n × pi2 ,where n is any odd interger

}

{y : y ≥ 1 or y ≤ −1}

cot x {x : x , npi,where n is any integer} all reals

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Points to Consider

• How are all the reciprocal functions’ graphs related to sine, cosine and tangent?

• What would the inverse function of y = sin x look like?

Review Questions

1. Show that side A (in orange) in this drawing is equal to sec θ. Use similar triangles in your proof.

2. In Chapter 1, you learned that tan2 θ + 1 = sec2 θ. Use the drawing and results from question 1 to

demonstrate this identity.

3. This diagram shows a unit circle with all the angles that have reference angles of 30◦, 45◦, and 60◦,

as well as the quadrant angles. Label the coordinates of all points on the unit circle. On the smallest

circle, label the angles in degrees, and on the middle circle, label the angles in radians.

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4. Which of the following shows functions that are both increasing as x increases from 0 to pi2?

(a) sin x and cos x

(b) tan x and csc x

(c) sec x and cot x

(d) csc x and sec x

5. Which of the following statements are true as x increases from 3pi2 to 2pi?

(a) cos x approaches 0

(b) tan x gets infinitely large

(c) cos x < sin x

(d) cot x gets infinitely small

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Review Answers

1. Use similar triangles:

So:

x

1

=

1

A

→ Ax = 1→ A = 1

x

cos θ = x→ 1cos θ =

1

x

→ 1cos θ = sec θ =

1

x

∴ sec θ = A

2. Using the Pythagorean theorem, tan2 θ + 1 = sec2 θ.

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3.

4. b

5. d

2.4 Translating Sine and Cosine Functions

Learning Objectives

• Translate sine and cosine functions vertically and horizontally.

• Identify the vertical and horizontal translations of sine and cosine from a graph and an equation.

Vertical Translations

When you first learned about vertical translations in a coordinate grid, you started with simple shapes.

Here is a rectangle:

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To translate this rectangle vertically, move all points and lines up by a specified number of units. We do

this by adjusting the y−coordinate of the points. So to translate this rectangle 5 units up, add 5 to every

y−coordinate.

This process worked the same way for functions. Since the value of a function corresponds to the y−value

on its graph, to move a function up 5 units, we would increase the value of the function by 5. Therefore,

to translate y = x2 up five units, you would increase the y−value by 5. Because y is equal to x2, then the

equation y = x2 + 5, will show this translation.

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Hence, for any graph, adding a constant to the equation will move it up, and subtracting a constant will

move it down. From this, we can conclude that the graphs of y = sin x and y = cos x will follow the same

rules. That is, the graph of y = sin(x) + 2 will be the same as y = sin x, only it will be translated, or

shifted, 2 units up.

To avoid confusion, this translation is usually written in front of the function: y = 2 + sin x.

Various texts use different notation, but we will use D as the constant for vertical translations. This would

lead to the following equations: y = D ± sin x and y = D ± cos x where D is the vertical translation. D can

be positive or negative.

Another way to think of this is to view sine or cosine curves “wrapped” around a horizontal line. For

y = sin x and y = cos x, the graphs are wrapped around the x−axis, or the horizontal line, y = 0.

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For y = 3 + sin x, we know the curve is translated up 3 units. In this context, think of the sine curve as

being “wrapped” around the line, y = 3.

Either method works for the translation of a sine or cosine curve. Pick the thought process that works

best for you.

Example 1: Find the minimum and maximum of y = −6 + cos x

Solution: This is a cosine wave that has been shifted down 6 units, or is now wrapped around the line

y = −6. Because the graph still rises and falls one unit in either direction, the cosine curve will extend one

unit above the “wrapping line” and one unit below it. The minimum is -7 and the maximum is -5.

Example 2: Graph y = 4 + cos x.

Solution: This will be the basic cosine curve, shifted up 4 units.

Horizontal Translations or Phase Shifts

Horizontal translations are a little more complicated. If we return to the example of the parabola, y = x2,

what change would you make to the equation to have it move to the right or left? Many students guess

that if you move the graph vertically by adding to the y−value, then we should add to the x−value in order

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to translate horizontally. This is correct, but behaves in the opposite way than what you may think.

Here is the graph of y = (x+ 2)2.

Notice that adding 2 to the x−value shifted the graph 2 units to the left, or in the negative direction.

To compare, the graph y = (x − 2)2 moves the graph 2 units to the right or in the positive direction.

We will use the letter C to represent the horizontal shift value. Therefore, subtracting C from the x−value

will shift the graph to the right and adding C will shift the graph C units to the left.

Adding to our previous equations, we now have y = D ± sin(x ± C) and y = D ± cos(x ± C) where D is the

vertical translation and C is the opposite sign of the horizontal shift.

Example 3: Sketch y = sin

(

x − pi2

)

Solution: This is a sine wave that has been translated pi2 units to the right.

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Horizontal translations are also referred to as phase shifts. Two waves that are identical, but have been

moved horizontally are said to be “out of phase” with each other. Remember that cosine and sine are

really the same waves with this phase variation.

y = sin x can be thought of as a cosine wave shifted horizontally to the right by pi2 radians.

Alternatively, we could also think of cosine as a sine wave that has been shifted pi2 radians to the left.

Example 4: Draw a sketch of y = 1 + cos(x − pi)

Solution: This is a cosine curve that has been translated up 1 unit and pi units to the right. It may help

you to use the quadrant angles to draw these sketches. Plot the points of y = cos x at 0, pi2 , pi, 3pi2 , 2pi (as well

as the negatives), and then translate those points before drawing the translated curve. The blue curve

below is the final answer.

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Example 5: Graph y = −2 + sin

(

x+ 3pi2

)

Solution: This is a sine curve that has been translated 2 units down and moved 3pi2 radians to the left.

Again, start with the quadrant angles on y = sin x and translate them down 2 units.

Then, take that result and shift it 3pi2 to the left. The blue graph is the final answer.

Example 6: Write the equation of the following sinusoid

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Solution: Notice that you have been given some points to help identify the curve properly. Remember

that sine and cosine are essentially the same wave so you can choose to model the sinusoid with either one.

Think of the function as a cosine curve because the maximum value of a cosine function is on the y−axis,

which makes cosine easier to visualize. From the points on the curve, the first maximum point to the right

of the y−axis occurs at halfway between pi and 2pi, or 3pi2 . Because the next maximum occurs 2pi units to

the right of that, or at 7pi2 , there is no change in the period of this function. This means that the cosine

curve has been translated 3pi2 units to the right, or y = cos

(

x − 3pi2

)

. The vertical translation value can be

found by locating the center of the wave. If it is not obvious from the graph, you can find the center by

averaging the minimum and maximum values.

This center is the wrapping line of the translated function and is therefore the same as D. In this example,

the maximum value is 1.5 and the minimum is -0.5. So,

1.5 + (−0.5)

2

=

1

2

Placing these two values into our equation, y = D ± cos(x ±C), gives:

y =

1

2

+ cos

(

x − 3pi

2

)

Because the cosine graph is periodic, there are an infinite number of possible answers for the horizontal

translation. If we keep going in either direction to the next maximum and translate the wave back that

far, we will obtain the same graph. Some other possible answers are:

y =

1

2

+ cos

(

x+

pi

2

)

, y =

1

2

+ cos

(

x − 5pi

2

)

, and y = 1

2

+ cos

(

x − 7pi

2

)

.

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Because sine and cosine are essentially the same function, we could also have modeled the curve with a

sine function. Instead of looking for a maximum peak though, for sine we need to find the middle of an

increasing part of the wave to consider as a starting point because sine starts at zero.

The coordinates of this point may not always be obvious from the graph. It this case, the drawing shows

that the point just to the right of the y−axis is

(

pi, 12

)

. So the horizontal, or C value would be pi. The

vertical shift, amplitude, and frequency are all the same as the were for the cosine wave because it is the

same graph. So the equation would become y = 12 + sin(x − pi).

Once again, there are an infinite number of other possible answers if you extend away from the C value

multiples of 2pi in either direction, such as y = 12 + sin(x − 3pi) or y = 12 + sin(x − pi).

Points to Consider

• Amplitude is the “stretching” of a sine or cosine curve. Where do you think that would go in the

equation?

• Do you think the other four trig functions are translated, vertically and horizontally, in the same

way as sine and cosine?

• Why is there an infinitely many number of equations that can represent a sine or cosine curve?

Review Questions

For problems 1-5, find the equation that matches each description.

1. ____the minimum value is 0 – A. y = sin

(

x − pi2

)

2. ____the maximum value is 3 – B. y = 1 + sin x

3. ____the y−intercept is -2 – C. y = cos(x − pi)

4. ____the y−intercept is -1 – D. y = −1 + sin

(

x − 3pi2

)

5. ____the same graph as y = cos(x) – E. y = 2 + cos x

6. Express the equation of the following graph as both a sine and a cosine function. Several points have

been plotted at the quadrant angles to help.

For problems 7-10, match the graph with the correct equation.

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7. ____ y = 1 + sin

(

x − pi2

)

8. ____ y = −1 + cos

(

x+ 3pi2

)

9. ____ y = 1 + cos

(

x − pi2

)

10. ____ y = −1 + sin(x − pi)

A.

B.

C.

D.

11. Sketch the graph of y = 1 + sin

(

x − pi4

)

on the axes below.

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Review Answers

1. B

2. E

3. D

4. C

5. A6.

y = −2 + sin(x − pi) or y = −2 + sin(x+ pi)

y = −2 + cos

(

x+

pi

2

)

or y = −2 + cos

(

x − 3pi

2

)

Note: this list is not exhaustive, there are other possible answers.

7. C

8. D

9. A

10. B

11.

2.5 Amplitude, Period and Frequency

Learning Objectives

• Calculate the amplitude and period of a sine or cosine curve.

• Calculate the frequency of a sine or cosine wave.

• Graph transformations of sine and cosine waves involving changes in amplitude and period (fre-

quency).

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Amplitude

The amplitude of a wave is basically a measure of its height. Because that height is constantly changing,

amplitude is defined as the farthest distance the wave gets from its center. In a graph of f (x) = sin x, the

wave is centered on the x−axis and the farthest away it gets (in either direction) from the axis is 1 unit.

So the amplitude of f (x) = sin x (and f (x) = cos x) is 1.

Recall how to transform a linear function, like y = x. By placing a constant in front of the x value, you

may remember that the slope of the graph affects the steepness of the line.

The same is true of a parabolic function, such as y = x2. By placing a constant in front of the x2, the

graph would be either wider or narrower. So, a function such as y = 18 x2, has the same parabolic shape

but it has been “smooshed,” or looks wider, so that it increases or decreases at a lower rate than the graph

of y = x2.

No matter the basic function; linear, parabolic, or trigonometric, the same principle holds. To dilate

(flatten or steepen, wide or narrow) the function, multiply the function by a constant. Constants greater

than 1 will stretch the graph vertically and those less than 1 will shrink it vertically.

Look at the graphs of y = sin x and y = 2 sin x.

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Notice that the amplitude of y = 2 sin x is now 2. An investigation of some of the points will show that each

y−value is twice as large as those for y = sin x. Multiplying values less than 1 will decrease the amplitude

of the wave as in this case of the graph of y = 12 cos x:

Example 1: Determine the amplitude of f (x) = 10 sin x.

Solution: The 10 indicates that the amplitude, or height, is 10. Therefore, the function rises and falls

between 10 and -10.

Example 2: Graph g(x) = −5 cos x

Solution: Even though the 5 is negative, the amplitude is still positive 5. The amplitude is always the

absolute value of the constant A. However, the negative changes the appearance of the graph. Just like a

parabola, the sine (or cosine) is flipped upside-down. Compare the blue graph, g(x) = −5 cos x, to the red

parent graph, f (x) = cos x.

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So, in general, the constant that creates this stretching or shrinking is the amplitude of the sinusoid.

Continuing with our equations from the previous section, we now have y = D ± A sin(x ± C) or y =

D ± A cos(x ± C). Remember, if 0 < |A| < 1, then the graph is shrunk and if |A| > 1, then the graph is

stretched. And, if A is negative, then the graph is flipped.

Period and Frequency

The period of a trigonometric function is the horizontal distance traversed before the y−values begin to

repeat. For both graphs, y = sin x and y = cos x, the period is 2pi. As we learned earlier in the chapter,

after completing one rotation of the unit circle, these values are the same.

Frequency is a measurement that is closely related to period. In science, the frequency of a sound or light

wave is the number of complete waves for a given time period (like seconds). In trigonometry, because all

of these periodic functions are based on the unit circle, we usually measure frequency as the number of

complete waves every 2pi units. Because y = sin x and y = cos x cover exactly one complete wave over this

interval, their frequency is 1.

Period and frequency are inversely related. That is, the higher the frequency (more waves over 2pi units),

the lower the period (shorter distance on the x−axis for each complete cycle).

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After observing the transformations that result from multiplying a number in front of the sinusoid, it

seems natural to look at what happens if we multiply a constant inside the argument of the function, or

in other words, by the x value. In general, the equation would be y = sin Bx or y = cos Bx. For example,

look at the graphs of y = cos 2x and y = cos x.

Notice that the number of waves for y = cos 2x has increased, in the same interval as y = cos x. There are

now 2 waves over the interval from 0 to 2pi. Consider that you are doubling each of the x values because

the function is 2x. When pi is plugged in, for example, the function becomes 2pi. So the portion of the

graph that normally corresponds to 2pi units on the x−axis, now corresponds to half that distance—so the

graph has been “scrunched” horizontally. The frequency of this graph is therefore 2, or the same as the

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constant we multiplied by in the argument. The period (the length for each complete wave) is pi.

Example 3: What is the frequency and period of y = sin 3x?

Solution: If we follow the pattern from the previous example, multiplying the angle by 3 should result in

the sine wave completing a cycle three times as often as y = sin x. So, there will be three complete waves

if we graph it from 0 to 2pi. The frequency is therefore 3. Similarly, if there are 3 complete waves in 2pi

units, one wave will be a third of that distance, or 2pi3 radians. Here is the graph:

This number that is multiplied by x, called B, will create a horizontal dilation. The larger the value of B,

the more compressed the waves will be horizontally. To stretch out the graph horizontally, we would need

to decrease the frequency, or multiply by a number that is less than 1. Remember that this dilation factor

is inversely related to the period of the graph.

Adding, one last time to our equations from before, we now have: y = D ± A sin(B(x ± C)) or y = D ±

A cos(B(x±C)), where B is the frequency, the period is equal to 2piB , and everything else is as defined before.

Example 4: What is the frequency and period of y = cos 14 x?

Solution: Using the generalization above, the frequency must be 14 and therefore the period is

2pi

1

4 , which

simplifies to: 2pi1

4

=

2pi

1

1

4

· 414

1

= 8pi1 = 8pi

Thinking of it as a transformation, the graph is stretched horizontally. We would only see 14 of the curve if

we graphed the function from 0 to 2pi. To see a complete wave, therefore, we would have to go four times

as far, or all the way from 0 to 8pi.

Combining Amplitude and Period

Here are a few examples with both amplitude and period.

Example 5: Find the period, amplitude and frequency of y = 2 cos 12 x and sketch a graph from 0 to 2pi.

Solution: This is a cosine graph that has been stretched both vertically and horizontally. It will now

reach up to 2 and down to -2. The frequency is 12 and to see a complete period we would need to graph

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the interval [0, 4pi]. Since we are only going out to 2pi, we will only see half of a wave. A complete cosine

wave looks like this:

So, half of it is this:

This means that this half needs to be stretched out so it finishes at 2pi, which means that at pi the graph

should cross the x−axis:

The final sketch would look like this:

amplitude = 2, frequency = 12 , period = 2pi1

2

= 4pi

Example 6: Identify the period, amplitude, frequency, and equation of the following sinusoid:

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Solution: The amplitude is 1.5. Notice that the units on the x−axis are not labeled in terms of pi. This

appears to be a sine wave because the y−intercept is 0.

One wave appears to complete in 1 unit (not 1pi units!), so the period is 1. If one wave is completed in 1

unit, how many waves will be in 2pi units? In previous examples, you were given the frequency and asked

to find the period using the following relationship:

p =

2pi

B

Where B is the frequency and p is the period. With just a little bit of algebra, we can transform this

formula and solve it for B:

p =

2pi

B

→ Bp = 2pi→ B = 2pi

p

Therefore, the frequency is:

B =

2pi

1

= 2pi

If we were to graph this out to 2pi we would see 2pi (or a little more than 6) complete waves.

Replacing these values in the equation gives: f (x) = 1.5 sin 2pix.

Points to Consider

• Are the graphs of the other four trigonometric functions affected in the same way as sine and cosine

by amplitude and period?

• We saw what happens to a graph when A is negative. What happens when B is negative?

Review Questions

1. Using the graphs from section 3, identify the period and frequency of y = sec x, y = cot x and y = csc x.

2. Identify the minimum and maximum values of these functions.

(a) y = cos x

(b) y = 2 sin x

(c) y = − sin x

(d) y = tan x

(e) y = 12 cos 2x

(f) y = −3 sin 4x

3. How many real solutions are there for the equation 4 sin x = sin x over the interval 0 ≤ x ≤ 2pi?

(a) 0

(b) 1

(c) 2

(d) 3

4. For each equation, identify the period, amplitude, and frequency.

(a) y = cos 2x

(b) y = 3 sin x

(c) y = 2 sin pix

www.ck12.org 138

(d) y = 2 cos 3x

(e) y = 12 cos 12 x

(f) y = 3 sin 12 x

5. For each of the following graphs; 1) identify the period, amplitude, and frequency and 2) write the

equation.

(a)

(b)

(c)

(d)

6. For each equation, draw a sketch from 0 to 2pi.

(a) y = 3 sin 2x

(b) y = 2.5 cos pix

(c) y = 4 sin 12 x

139 www.ck12.org

Review Answers

1. y = sec x: period = 2pi, frequency = 1

y = cot x: period = pi, frequency = 2

y = csc x: period = 2pi, frequency = 1

Because these are reciprocal functions, the periods are the same as cosine, tangent, and sine, repec-

tively.

2. (a) min: -1, max: 1

(b) min: -2, max: 2

(c) min: -1, max: 1

(d) there is no minimum or maximum, tangent has a range of all real numbers

(e) min: −12 , max: 12

(f) min: -3, max: 3

3. d.

4. (a) period: pi, amplitude: 1, frequency: 2

(b) period: 2pi, amplitude: 3, frequency: 1

(c) period: 2, amplitude: 2, frequency: pi

(d) period: 2pi3 , amplitude: 2, frequency: 3

(e) period: 4pi, amplitude: 12 , frequency: 12

(f) period: 4pi, amplitude: 3, frequency: 12

5. (a) period: pi, amplitude: 1, frequency: 2, y = 3 cos 2x

(b) period: 4pi, amplitude: 2, frequency: 12 , y = 2 sin 12 x

(c) period: 3, amplitude: 2, frequency: 2pi3 , y = 2 cos 2pi3 x

(d) period: pi3 , amplitude: 12 , frequency: 6, y = 12 sin 6x

6. (a)

(b)

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(c)

2.6 General Sinusoidal Graphs

Learning Objectives

• Given any sinusoid in the form: y = D ± A cos(B(x ± C)) or y = D ± A sin(B(x ± C)) identify the

transformations performed by A, B, C, and D.

• Graph any sinusoid given an equation in the form y = D± A cos(B(x±C)) or y = D± A sin(B(x±C)).

• Identify the equation of any sinusoid given a graph and critical values.

The Generalized Equations

In the previous two sections, you learned how to translate and dilate sine and cosine waves both horizontally

and vertically. Combining all the information learned, the general equations are: y = D ± A cos(B(x ± C))

or y = D ± A sin(B(x ± C)), where A is the amplitude, B is the frequency, C is the horizontal translation,

and D is the vertical translation.

Recall the relationship between period, p, and frequency, B.

p =

2pi

B

and B = 2pi

p

With this knowledge, we should be able to sketch any sine or cosine function as well as write an equation

given its graph.

Drawing Sketches/Identifying Transformations from the Equa-

tion

Example 1: Given the function: f (x) = 1 + 2 sin(2(x+ pi))

a. Identify the period, amplitude, and frequency.

b. Explain any vertical or horizontal translations present in the equation.

c. Sketch the graph from −2pi to 2pi.

Solution: a. From the equation, the amplitude is 2 and the frequency is also 2. To find the period we

use:

p =

2pi

B

→ p = 2pi

2

= pi

141 www.ck12.org

So, there are two complete waves from [0, 2pi] and each individual wave requires pi radians to complete.

b. D = 1 and C = −pi, so this graph has been translated 1 unit up, and pi units to the left.

c. To sketch the graph, start with the graph of y = sin(x)

Translate the graph pi units to the left (the C value).

Next, move the graph 1 unit up (D value)

Now we can add the dilations. Remember that the “starting point” of the wave is −pi because of the

horizontal translation. A normal sine wave takes 2pi units to complete a cycle, but this wave completes one

cycle in pi units. The first wave will complete at 0, then we will see a second wave from 0 to pi and a third

from pi to 2pi. Start by placing points at these values:

Using symmetry, each interval needs to cross the line y = 1 through the center of the wave.

www.ck12.org 142

One sine wave contains a “mountain” and a “valley”. The mountain “peak” and the valley low point must

occur halfway between the points above.

Extend the curve through the domain.

Finally, extend the minimum and maximum points to match the amplitude of 2.

Example 2: Given the function: f (x) = 3 + 3 cos

(

1

2(x − pi2

)

)

a. Identify the period, amplitude, and frequency.

b. Explain any vertical or horizontal translations present in the equation.

c. Sketch the graph from −2pi to 2pi.

Solution: a. From the equation, the amplitude is 3 and the frequency is 12 . To find the period we use:

period =

2pi

1

2

= 4pi

So, there is only one half of a cosine curve from 0 to 2pi and each individual wave requires 4pi radians to

complete.

b. D = 3 and C = pi2 , so this graph has been translated 3 units up, and pi2 units to the right.

c. To sketch the graph, start with the graph of y = cos(x)

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Adjust the amplitude so the cosine wave reaches up to 3 and down to negative three. This affects the

maximum points, but the points on the x−axis remain the same. These points are sometimes called

nodes.

According to the period, we should see one of these shapes every 4pi units. Because the interval specified

is [−2pi, 2pi] and the cosine curve “starts” at the y−axis, at (0, 3) and at 2pi the value is -3. Conversely, at

−2pi, the function is also -3.

Now, shift the graph pi2 units to the right.

Finally, we need to adjust for the vertical shift by moving it up 3 units.

www.ck12.org 144

Writing the Equation from a Sketch

In order to write the equation from a graph, you need to be provided with enough information to find the

four constants.

Example 3: Find the equation of the sinusoid graphed here.

Solution: First of all, remember that either sine or cosine could be used to model these graphs. However,

it is usually easier to use cosine because the horizontal shift is easier to locate in most cases. Therefore,

the model that we will be using is y = D ± A cos(B(x ±C)) .

First, if we think of the graph as a cosine function, it has a horizontal translation of zero. The maximum

point is also the y−intercept of the graph, so there is no need to shift the graph horizontally and therefore,

C = 0. The amplitude is the height from the center of the wave. If you can’t find the center of the wave

by sight, you can calculate it. The center should be halfway between the highest and the lowest points,

which is really the average of the maximum and minimum. This value will actually be the vertical shift,

or D value.

145 www.ck12.org

D = center =

60 + −20

2

=

40

2

= 20

The amplitude is the height from the center line, or vertical shift, to either the minimum or the maximum.

So, A = 60 − 20 = 40.

The last value to find is the frequency. In order to do so, we must first find the period. The period is the

distance required for one complete wave. To find this value, look at the horizontal distance between two

consecutive maximum points.

On our graph, from maximum to maximum is 3.

Therefore, the period is 3, so the frequency is B = 2pi3 .

We have now calculated each of the four parameters necessary to write the equation. Replacing them in

the equation gives:

y = 20 + 40 cos 2pi

3

x

If we had chosen to model this curve with a sine function instead, the amplitude, period and frequency,

as well as the vertical shift would all be the same. The only difference would be the horizontal shift. The

sine wave starts in the middle of an upward sloped section of the curve as shown by the red circle.

www.ck12.org 146

This point intersects with the vertical translation line and is a third of the distance back to -3. So, in this

case, the sine wave has been translated 1 unit to the left. The equation using a sine function instead would

have been: y = 20 + 40 sin

(

2pi

3 (x+ 1)

)

Points to Consider

• When using either sine or cosine to model a graph, why is only the phase shift different?

• How would you write y = sin x in the form y = D ± A sin(B(x ±C))? What are A, B, C, and D?

• Is it possible to solve y = D ± A sin(B(x ±C)) for x?

Review Questions

For problems 1-5, identify the amplitude, period, frequency, maximum and minimum points, vertical shift,

and horizontal shift.

1. y = 2 + 3 sin(2(x − 1))

2. y = −1 + sin

(

pi(x+ pi3

)

)

3. y = cos(40(x − 120)) + 5

4. y = − cos

(

1

2(x+

5pi

4

)

)

5. y = 2 cos(−x) + 3

For problems 6-10, write the equation of each graph. Recall that cosine might be an easier model, but you

may write your answer in terms of cosine or sine.

6.

7.

147 www.ck12.org

8.

9.

10.

Review Answers

1. This is a sine wave that has been translated 1 unit to the right and 2 units up. The amplitude is 3

and the frequency is 2. The period of the graph is pi. The function reaches a maximum point of 5

and a minimum of -1.

2. This is a sine wave that has been translated 1 unit down and pi3 radians to the left. The amplitude

www.ck12.org 148

is 1 and the period is 2. The frequency of the graph is pi. The function reaches a maximum point of

0 and a minimum of -2.

3. This is a cosine wave that has been translated 5 units up and 30 radians to the right. The amplitude

is 1 and the frequency is 40. The period of the graph is pi20 . The function reaches a maximum point

of 6 and a minimum of 4.

4. This is a cosine wave that has not been translated vertically. It has been translated 5pi4 radians to the

left. The amplitude is 1 and the frequency is 12 . The period of the graph is 4pi. The function reaches

a maximum point of 1 and a minimum of -1. The negative in front of the cosine function does not

change the amplitude, it simply reflects the graph across the x−axis.

5. This is a cosine wave that has been translate up 3 units and has an amplitude of 2. The frequency is

1 and the period is 2pi. There is no horizontal translation. Putting a negative in front of the x−value

reflects the function across the y−axis. A cosine wave that has not been translated horizontally is

symmetric to the y−axis so this reflection will have no visible effect on the graph. The function

reaches a maximum of 5 and a minimum of 1.

∗∗∗other answers are possible given different horizontal translations of sine/cosine

6. y = 3 + 2 cos

(

4(x − pi6

)

)

7. y = 2 + sin x or y = 2 + cos

(

x − pi2

)

8. y = 10 + 20 cos(6(x − 30))

9. y = 3 + 34 cos

(

1

2(x+ pi

)

)

10. y = 3 + 7 cos

(

1

3(x − pi4

)

)

2.7 Graphing Tangent, Cotangent, Secant, and

Cosecant

Learning Objectives

• Apply transformations to the remaining four trigonometric functions.

• Identify the equation, given a basic graph.

Tangent and Cotangent

From Section 2.3, the graph of tangent looks like the picture below, where the period is pi and vertical

asymptotes are at 2pin ± pi2 and 2pin ± 3pi2 , where n is any integer. Notice that the period is only pi and the

function repeats after every asymptote. The x−intercepts are . . . ,−pi, 0, pi, 2pi, . . . The general equation of a

tangent function is just like sine and cosine, f (x) = D ± A tan(B(x ± C)), where A, B, C, and D represent

the same transformations as they did before.

149 www.ck12.org

Cotangent also has a period of pi, but the asymptotes and x−intercepts are reversed. What this means is that

the vertical asymptotes are now at 0 and ±npi, and the x−intercepts are at 2pin± pi2 and 2pin± 3pi2 , where n is an

integer. The general equation of a cotangent function is just like sine and cosine, f (x) = D±A cot(B(x±C)),

where A, B, C, and D represent the same transformations as they did before.

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One important difference: the period of sine and cosine is defined as 2piB . The period of tangent and

cotangent is only pi, so the period would be piB .

Example 1: Sketch the graph of g(x) = −2 + cot 13 x over the interval [0, 6pi].

Solution: Starting with y = cot x, g(x) would be shifted down two and frequency is 13 , which means the

period would be 3pi, instead of pi. So, in our interval of [0, 6pi] there would be two complete repetitions.

The red graph is y = cot x .

Example 2: Sketch the graph of y = −3 tan

(

x − pi4

)

over the interval [−pi, 2pi].

Solution: If you compare this graph to y = tan x, it will be stretched and flipped. It will also have a phase

shift of pi4 to the right. The red graph is y = tan x.

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Secant and Cosecant

Because secant is the reciprocal of cosine, it will have the same period, 2pi. Notice that an entire period

encompasses an upward ⋃ and downward ⋂ and the asymptote between them. There are no x−intercepts

and only one y−intercept at (0, 1). The vertical asymptotes are everywhere cosine is zero, so pin ± pi2

and pin ± 3pi2 , where n is any integer. The general equation of a secant function is just like the others,

f (x) = D ± A sec(B(x ±C)), where A, B, C, and D represent the same transformations as they did before.

The cosecant is the reciprocal of sine and it has the same period, 2pi. Notice that an entire period

encompasses an upward ⋃ and downward ⋂ and the asymptote between them, just like secant. There are

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no x−intercepts and no y−intercepts. The vertical asymptotes are everywhere sine is zero, so ±npi, where n

is any integer. The general equation of a cosecant function is just like the others, f (x) = D±A csc(B(x±C)),

where A, B, C, and D represent the same transformations as they did before.

Recall that the period of sine and cosine is defined as 2piB . The period of secant and cosecant will also be

defined this way.

Example 3: Sketch a graph of h(x) = 5 − 12 sec 4x over the interval [0, 2pi].

Solution: If you compare this example to f (x) = sec x, it will be translated 5 units up, flipped upside

down, with amplitude of 12 and frequency of 4. This means, in our interval of 2pi, there will be 4 secant

curves.

153 www.ck12.org

Graphing Calculator Note

For the two examples above, it might seem difficult to graph these on a graphing calculator. Most graphing

calculators do not have sec, csc, or cot buttons. However, we do know that these three functions are

reciprocals of cosine, sine, and tangent, respectively. So, you must enter them into the calculator in this

way. For example, the equation f (x) = 2 + 3 csc

(

3

4(x − 2)

)

would be entered like 2+3(sin(( 34)(x−2))) in the y =menu.

Find the Equation from a Graph

For tangent, cotangent, secant, and cosecant it can be difficult to determine the equation from a graph, so

to simplify this section amplitude changes will not be included.

Example 3: Find the equation of the graph below.

Solution: From the graph, we can see this is tangent. Usually tangent intercepts the origin, but here

it intercepts at (0, 2). Therefore, we know that there is no horizontal shift and the vertical shift is up 2.

Because we have eliminated amplitude from this section, the only thing left to find is the period. Normally,

the period of tangent is pi, but as you can see from the graph, there are three curves from [0, pi]. So, the

frequency is 3. The equation is y = 2 + tan 3x.

Example 4: Find the equation for the graph below.

Solution: First of all, this could be either a secant or cosecant function. Let’s say this is a secant function.

Secant usually intersects the y−axis at (0, 1) at a minimum. Now, that corresponding minimum is

(

pi

2 ,−2

)

.

Because there is no amplitude change, we can say that the vertical shift is the difference between the two

y−values, -3. It looks like there is a phase shift and a period change. From minimum to minimum is one

period, which is 9pi2 − pi2 = 8pi2 = 4pi and B = 2pi4pi = 12 . Lastly, we need to find the horizontal shift. The easiest

way to find this, is to work backwards. We know that the period is 4pi and normally it is 2pi, which is half of

4pi. So, if we multiply pi2 · 12 , we will get the horizontal shift, or pi4 . Our equation is f (x) = −3+ sec

(

1

2(x − pi4

)

)

NOTE: The technique used above can be used to find most phase shifts. First, determine the period

www.ck12.org 154

(and B) of the graph. Second, determine the y−intercept of the parent graph and then the point that

corresponds with it in the transformed graph (minimums or maximum are usually good points to pick).

Find the difference between the two x−values and then multiply by B. This would be the phase shift.

Points to Consider

• How can you shift or change tangent to make it look like cotangent?

• Are secant and cosecant “out of phase” like sine and cosine?

• Why do tangent and cotangent have a different period than the other four trig functions?

Review Questions

For questions 1-6, graph the following functions. Determine the amplitude, period, frequency, phase shift

and vertical translation.

1. y = −1 + 13 cot 2x

2. g(x) = 5 csc

(

1

4(x+ pi

)

)

3. f (x) = 4 + tan(0.5(x − pi))

4. y = −2 + 12 sec(4(x − 1))

5. y = −2 tan 2x

6. h(x) = − cot 13 x+ 1

7. We know that sine and cosine (and secant and cosecant) would be the same graph with a shift of pi2 .

How can we manipulate the graph of y = cot x to match up with y = tan x?

For problems 8 and 9, determine the equation of the trig functions below. All amplitudes are 1.

8.

9.

155 www.ck12.org

Review Answers

1. y = −1 + 13 cot 2x

2. g(x) = 5 csc

(

1

4(x+ pi)

)

3. f (x) = 4 + tan(0.5(x − pi))

4. y = −2 + 12 sec(4(x − 1))

www.ck12.org 156

5. y = −2 tan 2x

6. h(x) = − cot

(

1

3 x

)

+ 1

7. To make cotangent match up with tangent, it is helpful to graph the two on the same set of axis.

First, cotangent needs to be flipped, which would make the amplitude of -1. Once cotangent is

flipped, it also needs a phase shift of pi2 . So, tan x = − cot

(

x − pi2

)

.

8. This is a tangent graph. From the two points we are given, we can determine the phase shift, vertical

shift and frequency. There is no phase shift, the vertical shift is 3 and the frequency is 6. y = 3+tan 6x

9. This could be either a secant or cosecant function. We will use a cosecant model. First, the vertical

shift is -1. The period is the difference between the two given x−values, 7pi4 − 3pi4 = pi, so the frequency

is 2pipi = 2. The horizontal shift incorporates the frequency, so in y = csc x the corresponding x−value

to

(

3pi

4 , 0

)

is

(

pi

2 , 1

)

. The difference between the x−values is 3pi4 − pi2 = 3pi4 − 2pi4 = pi4 and then multiply it

by the frequency, 2 · pi4 = pi2 . The equation is y = −1 + csc

(

2(x − pi2

)

).

2.8 Chapter Review

Chapter Summary

In this chapter we learned about another way to measure degrees, called radians. Radians are based off

of the unit circle, which is a circle with a radius of one. Because all circles are similar, it doesn’t matter

which one we use to measure radians, the ratios will always reduce to be the same. Therefore, we can

now graph the six trigonometric functions on the x − y plane. All the trigonometric functions are periodic

and, wherever the trig value is undefined the graph has a vertical asymptote. Finally, we learned that trig

functions can be transformed, much like a quadratic or cubic function. Every trig function has amplitude,

phase shift, vertical shift and a period, or frequency.

157 www.ck12.org

Vocabulary

Amplitude A dilation on the y−value of a trigonometric function. A is multiplied by the function, to

make it stretch or shorten.

Angular Velocity The velocity of rotation, measured in radians.

Arc Length The length of a portion of the circumference of a circle. The formula is θ · 2pir, where θ is

the corresponding central angle, in radians.

Circular Function The collective term for a function that can be defined by the unit circle.

Critical Angle(s) Any angle that is a multiple of 30◦ or 45◦.

Dilation A transformation that changes the size of an object or function.

Frequency The number of times the graph repeats in 2pi or pi for tangent and cotangent.

Period The distance it takes a graph to complete one phase.

Phase Shift The shift, or translation, in the x−direction of a trigonometric function. Also called a

horizontal translation.

Radian An alternate way to measure degrees, defined by the arc length on a circle that is equal to the

radius. 360◦ = 2pi radians.

Sector The area of a portion of a circle. The formula is θ · pir2, where θ is the central angle, measured in

radians.

Transformation Any change made to an object or graph. Transformations can either be dilations or

translations.

Translation Either a vertical or horizontal movement of an object or function.

Vertical Shift The vertical translation of a function.

Review Questions

1. Convert 160◦ to radians.

2. Convert 11pi12 to degrees.

3. Find the exact value of cos 3pi4 .

4. Find all possible answers in radians, between 0 < θ < 2pi : tan θ =

√

3

5. This is an image of the state flag of Colorado

It turns out that the diameter of the gold circle is 13 the total height of the flag (the same width as

the white stripe) and the outer diameter of the red circle is 23 of the total height of the flag. The

angle formed by the missing portion of the red band is pi4 radians. In a flag that is 33 inches tall,

what is the area of the red portion of the flag to the nearest square inch?

www.ck12.org 158

Figure 2.7

Figure 2.8

6. Suppose the radius of the dial of an electric meter on a house is 7 cm.

(a) How fast is a point on the outside edge of the dial moving if it completes a revolution in 9

seconds?

(b) Find the angular velocity of a point on the dial.

7. The cotangent (orange) can be represented by the picture below. How can you show this segment is

equal to xy? Use similar triangles. You may assume that the radii shown (pink) are one unit.

8. Graph y = sin x and y = cos x on the same set of axes over the interval [0, 2pi]. Where do they

intersect?

For questions 9-12, determine the amplitude, period, frequency, vertical shift, and phase shift. Then, graph

each function over the interval [0, 2pi].

159 www.ck12.org

9. y = −2 + 4 sin 5x

10. f (x) = 14 cos

(

1

2(x − pi3

)

)

11. g(x) = 4 + tan

(

2(x+ pi2

)

)

12. h(x) = 3 − 6 cos(pix)

For questions 13 and 14, find the equation of the graph below. Only sine and cosine functions will have an

amplitude other than 1.

13.

14.

Review Answers

1. 160◦ · pi180◦ = 16pi18 = 8pi9

2. 11pi12 · 180

◦

pi = 11 · 15◦ = 165◦

3. cos 3pi4 = cos 135◦ = −

√

2

2

4. For tan θ =

√

3, θ must equal 60◦ or 240◦. In radians, pi3 or 4pi3 .

5. There are many difference approaches to the problem. Here is one possibility:

First, calculate the area of the red ring as if it went completely around the circle:

A = Atotal − Agold

A = pi

(2

3

× 33

)2

− pi

(1

3

× 33

)2

A = pi × 222 − pi × 112

A = 484pi − 121pi = 363pi

A ≈ 1140.4 in2

www.ck12.org 160

Next, calculate the area of the total sector that would form the opening of the “c”

A =

1

2

r2θ

A =

1

2

(22)2

(

pi

4

)

A ≈ 190.1 in2

Then, calculate the area of the yellow sector and subtract it from the previous answer.

A =

1

2

r2θ → A = 1

2

(11)2

(

pi

4

)

→ A ≈ 47.5 in2

190.1 − 47.5 = 142.6 in2

Finally, subtract this answer from the first area calculated. The area is approximately 998 in2

6. (a) First find the circumfrence: 2pi · 7 = 14pi. This will be the distance for the linear velocity.

v = dt = 14pi · 9 = 126pi ≈ 395.84 cm/sec

(b) ω = θt = 2pi9 ≈ 0.698 rad/sec

7. Using similar triangles

1

t =

cot

1 , which means that the cotangent is equal to 1tangent . t =

y

x , so 1t = 1y

x

= xy .

Therefore, cotangent = xy .

161 www.ck12.org

8.

The intersections are

(

pi

4 ,

√

2

2

)

and

(

5pi

4 ,−

√

2

2

)

.

9. y = −2 + 4 sin 5x, A = 4, B = 5, p = 2pi5 ,C = 0,D = −2

10. f (x) = 14 cos

(

1

2(x − pi3

)

), A = 14 , B =

1

2 , p = 4pi,C =

pi

3 ,D = 0

11. g(x) = 4 + tan

(

2(x+ pi2

)

), A = 1, B = 2, p = pi2 ,C =

−pi

2 ,D = 4

12. h(x) = 3 − 6 cos(pix), A = −6, B = pi,C = 0,D = 3

13. y = −1 + 12 cos 3x

14. y = tan 6x

www.ck12.org 162

Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator

activities designed to supplement the objectives for some of the lessons in this chapter. See

http://www.ck12.org/flexr/chapter/9700.

Image Sources

(1) http://commons.wikimedia.org/wiki/Image:Flag of Colorado.svg. Public Domain.

(2) http://upload.wikimedia.org/wikipedia/commons/e/e4/Crops_Kansas_AST_20010624.jpg.

PD-USGov-NOAA.

(3) [Derived from http://commons.wikimedia.org/wiki/File:Nealsyarddairy1.jpg ]. GNU Free

Documentation License.

(4) http://en.wikipedia.org/wiki/Image:PivotIrrigationOnCotton.jpg. PD-USGov-NOAA.

(5) http://en.wikipedia.org/wiki/Image:

W.F._Mangels_Kiddie_Galloping_Horse_Carrousel.jpg. Public Domain.

(6) http://commons.wikimedia.org/wiki/File:

24_hour_analog_clock_rua_24_horas_curitiba_brasil.jpg,. GNU-Free Documentation

License.

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163 www.ck12.org

Chapter 3

Trigonometric Identities and

Equations - 2nd edition

3.1 Fundamental Identities

Introduction

We now enter into the proof portion of trigonometry. Starting with the basic definitions of sine, cosine,

and tangent, identities (or fundamental trigonometric equations) emerge. Students will learn how to prove

certain identities, using other identities and definitions. Finally, students will be able solve trigonometric

equations for theta, also using identities and definitions.

Learning Objectives

• use the fundamental identities to prove other identities.

• apply the fundamental identities to values of θ and show that they are true.

Quotient Identity

In Chapter 1, the three fundamental trigonometric functions sine, cosine and tangent were introduced. All

three functions can be defined in terms of a right triangle or the unit circle.

www.ck12.org 164

sin θ = opposite

hypotenuse

=

y

r

=

y

1

= y

cos θ = ad jacent

hypotenuse

=

x

r

=

x

1

= x

tan θ = opposite

ad jacent

=

y

x

=

sin θ

cos θ

The Quotient Identity is tan θ = sin θcos θ . We see that this is true because tangent is equal to

y

x , in the unit

circle. We know that y is equal to the sine values of θ and x is equal to the cosine values of θ. Substituting

sin θ for y and cos θ for x and we have a new identity.

Example 1: Prove tan θ = sin θcos θ by using θ = 45◦.

Solution: Plugging in 45◦, we have: tan 45◦ = sin 45◦cos 45◦ . Then, substitute each function with its actual value

and simplify both sides.

sin 45◦

cos 45◦ =

√

2

2√

2

2

=

√

2

2 ÷

√

2

2 =

√

2

2 · 2√2 = 1 and we know that tan 45◦ = 1, so this is true.

Example 2: Show that tan 90◦ is undefined using the Quotient Identity.

Solution: tan 90◦ = sin 90◦cos 90◦ = 10 , because you cannot divide by zero, the tangent at 90◦ is undefined.

Reciprocal Identities

Chapter 1 also introduced us to the idea that the three fundamental reciprocal trigonometric functions are

cosecant (csc), secant (sec) and cotangent (cot) and are defined as:

csc θ = 1sin θ sec θ =

1

cos θ cot θ =

1

tan θ

If we apply the Quotient Identity to the reciprocal of tangent, an additional quotient is created:

cot θ = 1tan θ =

1

sin θ

cos θ

=

cos θ

sin θ

Example 3: Prove tan θ = sin θ sec θ

Solution: First, you should change everything into sine and cosine. Feel free to work from either side, as

long as the end result from both sides ends up being the same.

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tan θ = sin θ sec θ

= sin θ · 1cos θ

=

sin θ

cos θ

Here, we end up with the Quotient Identity, which we know is true. Therefore, this identity is also true

and we are finished.

Example 4: Given sin θ = −

√

6

5 , find sec θ.

Solution: Secant is the reciprocal of cosine, so we need to find the adjacent side. We are given the opposite

side,

√

6 and the hypotenuse, 5. Because θ is in the fourth quadrant, cosine will be positive. From the

Pythagorean Theorem, the third side is:

(√

6

)2

+ b2 = 52

6 + b2 = 25

b2 = 19

b =

√

19

from this we can now find cos θ =

√

19

5 . Since secant is the reciprocal of cosine, sec θ = 5√19 .

Pythagorean Identity

Using the fundamental trig functions, sine and cosine and some basic algebra can reveal some interesting

trigonometric relationships. Note when a trig function such as sin θ is multiplied by itself, the mathematical

convention is to write it as sin2 θ. (sin θ2 can be interpreted as the sine of the square of the angle, and is

therefore avoided.)

sin2 θ = y2r2 and cos2 θ = x

2

r2 or sin2 θ + cos2 θ =

y2

r2 +

x2

r2 =

x2+y2

rr

Using the Pythagorean Theorem for the triangle above: x2 + y2 = r2

Then, divide both sides by r2, x2+y2r2 = r

2

r2 = 1. So, because

x2+y2

r2 = 1, sin2 θ + cos2 θ also equals 1.

This is known as the Trigonometric Pythagorean Theorem or the Pythagorean Identity and is written

sin2 θ + cos2 θ = 1. Alternative forms of the Theorem are: 1 + cot2 θ = csc2 θ and tan2 θ + 1 = sec2 θ. The

second form is found by taking the original form and dividing each of the terms by sin2 θ, while the third

form is found by dividing all the terms of the first by cos2 θ.

Example 5: Use 30◦ to show that sin2 θ + cos2 θ = 1 holds true.

Solution: Plug in 30◦ and find the values of sin 30◦ and cos 30◦.

sin2 30◦ + cos2 30◦(1

2

)2

+

√32

2

1

4

+

3

4

= 1

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Even and Odd Identities

Functions are even or odd depending on how the end behavior of the graphical representation looks. For

example, y = x2 is considered an even function because the ends of the parabola both point in the same

direction and the parabola is symmetric about the y−axis. y = x3 is considered an odd function for the

opposite reason. The ends of a cubic function point in opposite directions and therefore the parabola is

not symmetric about the y−axis. What about the trig functions? They do not have exponents to give us

the even or odd clue (when the degree is even, a function is even, when the degree is odd, a function is

odd).

Even Function Odd Function

y = (−x)2 = x2 y = (−x)3 = −x3

Let’s consider sine. Start with sin(−x). Will it equal sin x or − sin x? Plug in a couple of values to see.

sin(−30◦) = sin 330◦ = −1

2

= − sin 30◦

sin(−135◦) = sin 225◦ = −

√

2

2

= − sin 135◦

From this we see that sine is odd. Therefore, sin(−x) = − sin x, for any value of x. For cosine, we will plug

in a couple of values to determine if it’s even or odd.

cos(−30◦) = cos 330◦ =

√

3

2

= cos 30◦

cos(−135◦) = cos 225◦ = −

√

2

2

= cos 135◦

This tells us that the cosine is even. Therefore, cos(−x) = cos x, for any value of x. The other four

trigonometric functions are as follows:

tan(−x) = − tan x

csc(−x) = − csc x

sec(−x) = sec x

cot(−x) = − cot x

Notice that cosecant is odd like sine and secant is even like cosine.

Example 6: If cos(−x) = 34 and tan(−x) = −

√

7

3 , find sin x.

Solution: We know that sine is odd. Cosine is even, so cos x = 34 . Tangent is odd, so tan x =

√

7

3 .

Therefore, sine is positive and sin x =

√

7

4 .

Cofunction Identities

Recall that two angles are complementary if their sum is 90◦. In every triangle, the sum of the interior

angles is 180◦ and the right angle has a measure of 90◦. Therefore, the two remaining acute angles of

the triangle have a sum equal to 90◦, and are complementary. Let’s explore this concept to identify the

relationship between a function of one angle and the function of its complement in any right triangle, or

the cofunction identities. A cofunction is a pair of trigonometric functions that are equal when the variable

in one function is the complement in the other.

In 4ABC, ∠C is a right angle, ∠A and ∠B are complementary.

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Chapter 1 introduced the cofunction identities (section 1.8) and because ∠A and ∠B are complementary, it

was found that sin A = cos B, cos A = sin B, tan A = cot B, cot A = tan B, csc A = sec B and sec A = csc B. For

each of the above ∠A = pi2 − ∠B. To generalize, sin

(

pi

2 − θ

)

= cos θ and cos

(

pi

2 − θ

)

= sin θ, tan

(

pi

2 − θ

)

= cot θ

and cot

(

pi

2 − θ

)

= tan θ, csc

(

pi

2 − θ

)

= sec θ and sec

(

pi

2 − θ

)

= csc θ.

The following graph represents two complete cycles of y = sin x and y = cos θ.

Notice that a phase shift of pi2 on y = cos x, would make these graphs exactly the same. These cofunction

identities hold true for all real numbers for which both sides of the equation are defined.

Example 7: Use the cofunction identities to evaluate each of the following expressions:

a. If tan

(

pi

2 − θ

)

= −4.26 determine cot θ

b. If sin θ = 0.91 determine cos

(

pi

2 − θ

)

.

Solution:

a. tan

(

pi

2 − θ

)

= cot θ therefore cot θ = −4.26

b. cos

(

pi

2 − θ

)

= sin θ therefore cos

(

pi

2 − θ

)

= 0.91

Example 8: Show sin

(

pi

2 − x

)

= cos(−x) is true.

Solution: Using the identities we have derived in this section, sin

(

pi

2 − x

)

= cos x, and we know that cosine

is an even function so cos(−x) = cos x. Therefore, each side is equal to cos x and thus equal to each other.

Points to Consider

• Why do you think secant is even like cosine?

• How could you show that tangent is odd?

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Review Questions

1. Use the Quotient Identity to show that the tan 270◦ is undefined.

2. If cos

(

pi

2 − x

)

= 45 , find sin(−x).

3. If tan(−x) = − 512 and sin x = − 513 , find cos x.

4. Simplify sec x cos

(

pi

2 − x

)

.

5. Verify sin2 θ + cos2 θ = 1 using:

(a) the sides 5, 12, and 13 of a right triangle, in the first quadrant

(b) the ratios from a 30 − 60 − 90 triangle

6. Prove 1 + tan2 θ = sec2 θ using the Pythagorean Identity

7. If csc z = 178 and cos z = −1517 , find cot z.

8. Factor:

(a) sin2 θ − cos2 θ

(b) sin2 θ + 6 sin θ + 8

9. Simplify sin4 θ−cos4 θsin2 θ−cos2 θ using the trig identities10. Rewrite cos xsec x−1 so that it is only in terms of cosine. Simplify completely.

11. Prove that tangent is an odd function.

Review Answers

1. tan 270◦ = sin 270◦cos 270◦ = −10 , you cannot divide by zero, therefore tan 270◦ is undefined.

2. If cos

(

pi

2 − x

)

= 45 , then, by the cofunction identities, sin x = 45 . Because sine is odd, sin(−x) = −45 .

3. If tan(−x) = − 512 , then tan x = 512 . Because sin x = − 513 , cosine is also negative, so cos x = −1213 .

4. Use the reciprocal and cofunction identities to simplifysec x cos

(

pi

2

− x

)

1

cos x · sin x

sin x

cos x

tan x

5. (a) Using the sides 5, 12, and 13 and in the first quadrant, it doesn’t really matter which is cosine

or sine. So, sin2 θ + cos2 θ = 1 becomes

(

5

13

)2

+

(

12

13

)2

= 1. Simplifying, we get: 25169 + 144169 = 1, and

finally 169169 = 1.

(b) sin2 θ + cos2 θ = 1 becomes

(

1

2

)2

+

( √

3

2

)2

= 1. Simplifying we get: 14 + 34 = 1 and 44 = 1.

6. To prove tan2 θ + 1 = sec2 θ, first use sin θcos θ = tan θ and change sec2 θ = 1cos2 θ .tan2 θ + 1 = sec2 θ

sin2 θ

cos2 θ + 1 =

1

cos2 θ

sin2 θ

cos2 θ +

cos2 θ

cos2 θ =

1

cos2 θ

sin2 θ + cos2 θ = 1

7. If csc z = 178 and cos z = −1517 , then sin z = 817 and tan z = − 815 . Therefore cot z = −158 .

8. (a) Factor sin2 θ − cos2 θ using the difference of squares.sin2 θ − cos2 θ = (sin θ + cos θ)(sin θ − cos θ)

(b) sin2 θ + 6 sin θ + 8 = (sin θ + 4)(sin θ + 2)

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9. You will need to factor and use the sin2 θ + cos2 θ = 1 identity.sin4 θ − cos4 θ

sin2 θ − cos2 θ

=

(sin2 θ − cos2 θ)(sin2 θ + cos2 θ)

sin2 θ − cos2 θ

= sin2 θ + cos2 θ

= 1

10. To rewrite cos xsec x−1 so it is only in terms of cosine, start with changing secant to cosine.cos x

sec x − 1 =

cos x

1

cos x − 1

Now, simplify the denominator.

cos x

1

cos x − 1

=

cos x

1−cos x

cos x

Multiply by the reciprocal cos x1−cos x

cos x

= cos x ÷ 1−cos xcos x = cos x · cos x1−cos x = cos

2 x

1−cos x

11. The easiest way to prove that tangent is odd to break it down, using the Quotient Identity.

tan(−x) = sin(−x)cos(−x) from this statement, we need to show that tan(−x) = − tan x

=

− sin x

cos x because sin(−x) = − sin x and cos(−x) = cos x

= − tan x

3.2 Proving Identities

Learning Objectives

• Prove identities using several techniques.

Working with Trigonometric Identities

During the course, you will see complex trigonometric expressions. Often, complex trigonometric expres-

sions can be equivalent to less complex expressions. The process for showing two trigonometric expressions

to be equivalent (regardless of the value of the angle) is known as validating or proving trigonometric

identities.

There are several options a student can use when proving a trigonometric identity.

Option One: Often one of the steps for proving identities is to change each term into their sine and cosine

equivalents:

Example 1: Prove the identity: csc θ × tan θ = sec θ

Solution: Reducing each side separately. It might be helpful to put a line down, through the equals sign.

Because we are proving this identity, we don’t know if the two sides are equal, so wait until the end to

include the equality.

csc x × tan x sec x

1

sin x × sin xcos x 1cos x

1

sin x ×

sin x

cos x

1

cos x

1

cos x

1

cos x

At the end we ended up with the same thing, so we know that this is a valid identity.

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Notice when working with identities, unlike equations, conversions and mathematical operations are per-

formed only on one side of the identity. In more complex identities sometimes both sides of the identity

are simplified or expanded. The thought process for establishing identities is to view each side of the iden-

tity separately, and at the end to show that both sides do in fact transform into identical mathematical

statements.

Option Two: Use the Trigonometric Pythagorean Theorem and other Fundamental Identities.

Example 2: Prove the identity: (1 − cos2 x)(1 + cot2 x) = 1

Solution: Use the Pythagorean Identity and its alternate form. Manipulate sin2 θ + cos2 θ = 1 to be

sin2 θ = 1 − cos2 θ. Also substitute csc2 x for 1 + cot2 x, then cross-cancel.

(1 − cos2 x)(1 + cot2 x) 1

sin2 x · csc2 x 1

sin2 x · 1sin2 x 1

1 1

Option Three: When working with identities where there are fractions- combine using algebraic techniques

for adding expressions with unlike denominators:

Example 3: Prove the identity: sin θ1+cos θ + 1+cos θsin θ = 2 csc θ.

Solution: Combine the two fractions on the left side of the equation by finding the common denominator:

(1 + cos θ) × sin θ, and the change the right side into terms of sine.

sin θ

1+cos θ +

1+cos θ

sin θ 2 csc θsin θ

sin θ · sin θ1+cos θ + 1+cos θsin θ · 1+cos θ1+cos θ 2 csc θ

sin2 θ+(1+cos θ)2

sin θ(1+cos θ) 2 csc θ

Now, we need to apply another algebraic technique, FOIL. Always leave the denominator factored, because

you might be able to cancel something out at the end.

sin2 θ+1+2 cos θ+cos2 θ

sin θ(1+cos θ) 2 csc θ

Using the second option, substitute sin2 θ + cos2 θ = 1 and simplify.

1+1+2 cos θ

sin θ(1+cos θ) 2 csc θ

2+2 cos θ

sin θ(1+cos θ) 2 csc θ

2(1+cos θ)

sin θ(1+cos θ) 2 csc θ

2

sin θ

2

sin θ

Option Four: If possible, factor trigonometric expressions. Actually procedure four was used in the above

example: 2+2 cos θsin θ(1+cos θ) = 2 csc θ can be factored to

2(1+cos θ)

sin θ(1+cos θ) = 2 csc θ and in this situation, the factors

cancel each other.

Example 4: Prove the identity: 1+tan θ

(1+cot θ) = tan θ.

Solution: Change cot θ to 1tan θ and find a common denominator.

1 + tan θ(

1 + 1tan θ

) = tan θ

1 + tan θ( tan θ

tan θ +

1

tan θ

) = tan θ or 1 + tan θtan θ+1

tan θ

= tan θ

171 www.ck12.org

Now invert the denominator and multiply.

tan θ(1 + tan θ)

tan θ + 1 = tan θ

tan θ = tan θ

Technology Note

A graphing calculator can help provide the correctness of an identity. For example looking at: csc x×tan x =

sec x, first graph y = csc x × tan x, and then graph y = sec x. Examining the viewing screen for each

demonstrates that the results produce the same graph.

To summarize, when verifying a trigonometric identity, use the following tips:

1. Work on one side of the identity- usually the more complicated looking side.

2. Try rewriting all given expressions in terms of sine and cosine.

3. If there are fractions involved, combine them.

4. After combining fractions, if the resulting fraction can be reduced, reduce it.

5. The goal is to make one side look exactly like the other—so as you change one side of the identity,

look at the other side for a potential hint to what to do next. If you are stumped, work with the

other side. Don’t limit yourself to working only on the left side, a problem might require you to work

on the right.

Points to Consider

• Are there other techniques that you could use to prove identities?

• What else, besides what is listed in this section, do you think would be useful in proving identities?

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Review Questions

Prove the following identities true:

1. sin x tan x+ cos x = sec x

2. cos x − cos x sin2 x = cos3 x

3. sin x1+cos x + 1+cos xsin x = 2 csc x

4. sin x1+cos x = 1−cos xsin x

5. 11+cos a + 11−cos a = 2 + 2 cot2 a

6. cos4 b − sin4 b = 1 − 2 sin2 b

7. sin y+cos ysin y − cos y−sin ycos y = sec y csc y

8. (sec x − tan x)2 = 1−sin x1+sin x

9. Show that 2 sin x cos x = sin 2x is true using 5pi6 .

10. Use the trig identities to prove sec x cot x = csc x

Review Answers

1. Step 1: Change everything into sine and cosine

sin x tan x+ cos x = sec x

sin x · sin xcos x + cos x =

1

cos x

Step 2: Give everything a common denominator, cos x.

sin2 x

cos x +

cos2 x

cos x =

1

cos x

Step 3: Because the denominators are all the same, we can eliminate them.

sin2 x+ cos2 x = 1

We know this is true because it is the Trig Pythagorean Theorem

2. Step 1: Pull out a cos x

cos x − cos x sin2 x = cos3 x

cos x(1 − sin2 x) = cos3 x

Step 2: We know sin2 x+ cos2 x = 1, so cos2 x = 1− sin2 x is also true, therefore cos x(cos2 x) = cos3 x.

This, of course, is true, we are done!

3. Step 1: Change everything in to sine and cosine and find a common denominator for left hand side.

sin x

1 + cos x +

1 + cos x

sin x = 2 csc x

sin x

1 + cos x +

1 + cos x

sin x =

2

sin x ← LCD : sin x(1 + cos x)

sin2 x+ (1 + cos x)2

sin x(1 + cos x)

Step 2: Working with the left side, FOIL and simplify.

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sin2 x+ 1 + 2 cos x+ cos2 x

sin x(1 + cos x) → FOIL (1 + cos x)

2

sin2 x+ cos2 x+ 1 + 2 cos x

sin x(1 + cos x) → move cos

2 x

1 + 1 + 2 cos x

sin x(1 + cos x) → sin

2 x+ cos2 x = 1

2 + 2 cos x

sin x(1 + cos x) → add

2(1 + cos x)

sin x(1 + cos x) → fator out 2

2

sin x → cancel (1 + cos x)

4. Step 1: Cross-multiply sin x

1 + cos x =

1 − cos x

sin x

sin2 x = (1 + cos x)(1 − cos x)

Step 2: Factor and simplify

sin2 x = 1 − cos2 x

sin2 x+ cos2 x = 1

5. Step 1: Work with left hand side, find common denominator, FOIL and simplify, using sin2 x+cos2 x =

1. 1

1 + cos x +

1

1 − cos x = 2 + 2 cot

2 x

1 − cos x+ 1 + cos x

(1 + cos x)(1 − cos x)

2

1 − cos2 x

2

sin2 x

Step 2: Work with the right hand side, to hopefully end up with 2sin2 x .= 2 + 2 cot2 x

= 2 + 2

cos2 x

sin2 x

= 2

(

1 +

cos2 x

sin2 x

)

→ factor out the 2

= 2

(sin2 x+ cos2 x

sin2 x

)

→ common denominator

= 2

( 1

sin2 x

)

→ trig pythagorean theorem

=

2

sin2 x → simply/multiply

Both sides match up, the identity is true.

6. Step 1: Factor left hand side cos4 b − sin4 b 1 − 2 sin2 b

(cos2 b+ sin2 b)(cos2 b − sin2 b) 1 − 2 sin2 b

cos2 b − sin2 b 1 − 2 sin2 b

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Step 2: Substitute 1 − sin2 b for cos2 b because sin2 x+ cos2 x = 1.

(1 − sin2 b) − sin2 b 1 − 2 sin2 b

1 − sin2 b − sin2 b 1 − 2 sin2 b

1 − 2 sin2 b 1 − 2 sin2 b

7. Step 1: Find a common denominator for the left hand side and change right side in terms of sine and

cosine. sin y+ cos y

sin y −

cos y − sin y

cos y = sec y csc y

cos y(sin y+ cos y) − sin y(cos y − sin y)

sin y cos y =

1

sin y cos y

Step 2: Work with left side, simplify and distribute.

sin y cos y+ cos2 y − sin y cos y+ sin2 y

sin y cos y

cos2 y+ sin2 y

sin y cos y

1

sin y cos y

8. Step 1: Work with left side, change everything into terms of sine and cosine.

(sec x − tan x)2 = 1 − sin x

1 + sin x(

1

cos x −

sin x

cos x

)2

(

1 − sin x

cos x

)2

(1 − sin x)2

cos2 x

Step 2: Substitute 1 − sin2 b for cos2 b because sin2 x+ cos2 x = 1

(1 − sin x)2

1 − sin2 x → be careful, these are NOT the same!

Step 3: Factor the denominator and cancel out like terms.

(1 − sin x)2

(1 + sin x)(1 − sin x)

1 − sin x

1 + sin x

9. Plug in 5pi6 for x into the formula and simplify.2 sin x cos x = sin 2x

2 sin 5pi

6

cos 5pi

6

= sin 2 · 5pi

6

2

√32

(−12

)

= sin 5pi

3

This is true because sin 300◦ is −

√

3

2

10. Change everything into terms of sine and cosine and simplify.sec x cot x = csc x

1

cos x ·

cos x

sin x =

1

sin x

1

sin x =

1

sin x

175 www.ck12.org

3.3 Solving Trigonometric Equations

Learning Objectives

• Use the fundamental identities to solve trigonometric equations.

• Express trigonometric expressions in simplest form.

• Solve trigonometric equations by factoring.

• Solve trigonometric equations by using the Quadratic Formula.

By now we have seen trigonometric functions represented in many ways: Ratios between the side lengths

of right triangles, as functions of coordinates as one travels along the unit circle and as abstract functions

with graphs. Now it is time to make use of the properties of the trigonometric functions to gain knowledge

of the connections between the functions themselves. The patterns of these connections can be applied to

simplify trigonometric expressions and to solve trigonometric equations.

Simplifying Trigonometric Expressions

Example 1: Simplify the following expressions using the basic trigonometric identities:

a. 1+tan2 xcsc2 x

b. sin2 x+tan2 x+cos2 xsec x

c. cos x − cos3 x

Solution:

a.

1 + tan2 x

csc2 x . . . (1 + tan

2 x = sec2 x)Pythagorean Identity

sec2 x

csc2 x . . . (sec

2 x =

1

cos2 x and csc

2 x =

1

sin2 x)Reciprocal Identity

1

cos2 x

1

sin2 x

=

( 1

cos2 x

)

÷

( 1

sin2 x

)

( 1

cos2 x

)

·

(sin2 x

1

)

=

sin2 x

cos2 x

= tan2 x→ Quotient Identity

b.

sin2 x+ tan2 x+ cos2 x

sec x . . . (sin

2 x+ cos2 x = 1)Pythagorean Identity

1 + tan2 x

sec x . . . (1 + tan

2 x = sec2 x)Pythagorean Identity

sec2 x

sec x = sec x

c.

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cos x − cos3 x

cos x(1 − cos2 x) . . .Factor out cos x and sin2 x = 1 − cos2 x

cos x(sin2 x)

In the above examples, the given expressions were simplified by applying the patterns of the basic trigono-

metric identities. We can also apply the fundamental identities to trigonometric equations to solve for x.

When solving trig equations, restrictions on x (or θ) must be provided, or else there would be infinitely

many possible answers (because of the periodicity of trig functions).

Solving Trigonometric Equations

Example 2: Without the use of technology, find all solutions tan2(x) = 3, such that 0 ≤ x ≤ 2pi.

Solution:

tan2 x = 3√

tan2 x =

√

3

tan x = ±√3

This means that there are four answers for x, because tangent is positive in the first and third quadrants and

negative in the second and fourth. Combine that with the values that we know would generate tan x =

√

3

or tan x = −√3, x = pi3 , 2pi3 , 4pi3 , and 5pi3 .

Example 3: Solve 2 cos x sin x − cos x = 0 for all values of x between [0, 2pi].

Solution:

cos x (2 sin x − 1) = 0→ set each factor equal to zero and solve them separately

↓ ↘

cos x = 0 2 sin x = 1

x =

pi

2

and x = 3pi

2

sin x = 1

2

x =

pi

6

and x = 5pi

6

In the above examples, exact values were obtained for the solutions of the equations. These solutions were

within the domain that was specified.

Example 4: Solve 2 sin2 x − cos x − 1 = 0 for all values of x.

Solution: The equation now has two functions – sine and cosine. Study the equation carefully and decide

in which function to rewrite the equation. sin2 x can be expressed in terms of cosine by manipulating the

Pythagorean Identity, sin2 x+ cos2 x = 1.

177 www.ck12.org

2 sin2 x − cos x − 1 = 0

2(1 − cos2 x) − cos x − 1 = 0

2 − 2 cos2 x − cos x − 1 = 0

− 2 cos2 x − cos x+ 1 = 0

2 cos2 x+ cos x − 1 = 0

(2 cos x − 1)(cos x+ 1) = 0

↙ ↘

2 cos x − 1 = 0 or cos x+ 1 = 0

cos x = 1

2

cos x = −1

x =

pi

3

+ 2pik, kZ x = pi+ 2pik, kZ

x =

5pi

3

+ 2pik, kZ

Solving Trigonometric Equations Using Factoring

Algebraic skills like factoring and substitution that are used to solve various equations are very useful when

solving trigonometric equations. As with algebraic expressions, one must be careful to avoid dividing by

zero during these maneuvers.

Example 5: Solve 2 sin2 x − 3 sin x+ 1 = 0 for 0 < x ≤ 2pi.

Solution:

2 sin2 x − 3 sin x+ 1 = 0 Factor this like a quadratic equation

(2 sin x − 1)(sin x − 1) = 0

↓ ↘

2 sin x − 1 = 0 or sin x − 1 = 0

2 sin x = 1 sin x = 1

sin x = 1

2

x =

pi

2

x =

pi

6

and x = 5pi

6

Example 6: Solve 2 tan x sin x+ 2 sin x = tan x+ 1 for all values of x.

Solution:

www.ck12.org 178

Pull out sin x

There is a common factor of (tan x+ 1)

Think of the −(tan x+ 1) as (−1)(tan x+ 1), which is why there is a −1 behind the 2 sin x.

Example 7: Solve 2 sin2 x+ 3 sin x − 2 = 0 for all x, [0, pi].

Solution:

2 sin2 x+ 3 sin x − 2 = 0→ Factor like a quadratic

(2 sin x − 1)(sin x+ 2) = 0

↙ ↘

2 sin x − 1 = 0 sin x+ 2 = 0

sin x = 1

2

sin x = −2

x =

pi

6

and x = 5pi

6

There is no solution because the range of sin x is [−1, 1].

Some trigonometric equations have no solutions. This means that there is no replacement for the variable

that will result in a true expression.

Example 8: Solve 4 sin3 x+ 2 sin2 x − 2 sin x − 1 = 0 for x in the interval [0, 2pi].

Solution: Even though this does not look like a factoring problem, it is. We are going to use factoring by

grouping, from Algebra II. First group together the first two terms and the last two terms. Then find the

greatest common factor for each pair.

4 sin3 x+ 2 sin2 x︸ ︷︷ ︸ −2 sin x − 1︸ ︷︷ ︸ = 0

2 sin2 x(2 sin x+ 1) − 1(2 sin x+ 1)

Notice we have gone from four terms to two. These new two terms have a common factor of 2 sin x+ 1.

We can pull this common factor out and reduce our number of terms from two to one, comprised of two

factors.

2 sin2 x(2 sin x+ 1) − 1(2 sin x+ 1) = 0

↘ ↙

(2 sin x+ 1)(2 sin2 x − 1) = 0

We can take this one step further because 2 sin2 x − 1 can factor again.

179 www.ck12.org

(2 sin x+ 1)

(√

2 sin x − 1

) (√

2 sin x+ 1

)

= 0

Set each factor equal to zero and solve.

2 sin x+ 1 = 0 or

√

2 sin x+ 1 = 0 or

√

2 sin x − 1 = 0

2 sin x = −1 √2 sin x = −1 √2 sin x = 1

sin x = −1

2

sin x = − 1√

2

= −

√

2

2

sin x = 1√

2

=

√

2

2

x =

7pi

6

,

11pi

6

x =

5pi

4

,

7pi

4

x =

pi

4

,

3pi

4

Notice there are six solutions for x. Graphing the original function would show that the equation crosses

the x−axis six times in the interval [0, 2pi].

Solving Trigonometric Equations Using the Quadratic Formula

When solving quadratic equations that do not factor, the quadratic formula is often used. The same can

be applied when solving trigonometric equations that do not factor. The values for a is the numerical

coefficient of the function’s squared term, b is the numerical coefficient of the function term that is to the

first power and c is a constant. The formula will result in two answers and both will have to be evaluated

within the designated interval.

Example 8: Solve 3 cot2 x − 3 cot x = 1 for exact values of x over the interval [0, 2pi].

Solution:

3 cot2 x − 3 cot x = 1

3 cot2 x − 3 cot x − 1 = 0

The equation will not factor. Use the quadratic formula for cot x, a = 3, b = −3, c = −1.

www.ck12.org 180

cot x = −b ±

√

b2 − 4ac

2a

cot x = −(−3) ±

√

(−3)2 − 4(3)(−1)

2(3)

cot x = 3 ±

√

9 + 12

6

cot x = 3 +

√

21

6

or cot x = 3 −

√

21

6

cot x = 3 + 4.5826

6

cot x = 3 − 4.5826

6

cot x = 1.2638 cot x = −0.2638

tan x = 1

1.2638

tan x = 1−0.2638

x = 0.6694, 3.81099 x = 1.8287, 4.9703

Example 2: Solve −5 cos2 x+ 9 sin x+ 3 = 0 for values of x over the interval [0, 2pi].

Solution: Change cos2 x to 1 − sin2 x from the Pythagorean Identity.

−5 cos2 x+ 9 sin x+ 3 = 0

−5(1 − sin2 x) + 9 sin x+ 3 = 0

−5 + 5 sin2 x+ 9 sin x+ 3 = 0

5 sin2 x+ 9 sin x − 2 = 0

sin x = −9 ±

√

92 − 4(5)(−2)

2(5)

sin x = −9 ±

√

81 + 40

10

sin x = −9 ±

√

121

10

sin x = −9 + 11

10

and sin x = −9 − 11

10

sin x = 1

5

and − 2

sin−1(0.2) and sin−1(−2)

x ≈ .201 rad and pi − .201 ≈ 2.941

This is the only solutions for x since −2 is not in the range of values.

To summarize, to solve a trigonometric equation, you can use the following techniques:

1. Simplify expressions with the fundamental identities.

2. Factor, pull out common factors, use factoring by grouping.

3. The Quadratic Formula.

4. Be aware of the intervals for x. Make sure your final answer is in the specified domain.

181 www.ck12.org

Points to Consider

• Are there other methods for solving equations that can be adapted to solving trigonometric equations?

• Will any of the trigonometric equations involve solving quadratic equations?

• Is there a way to solve a trigonometric equation that will not factor?

• Is substitution of a function with an identity a feasible approach to solving a trigonometric equation?

Review Questions

1. Solve the equation sin 2θ = 0.6 for 0 ≤ θ < 2pi.

2. Solve the equation cos2 x = 116 over the interval [0, 2pi]

3. Solve the trigonometric equation tan2 x = 1 for all values of θ such that 0 ≤ θ ≤ 2pi

4. Solve the trigonometric equation 4 sin x cos x+ 2 cos x − 2 sin x − 1 = 0 such that 0 ≤ x < pi.

5. Solve sin2 x − 2 sin x − 3 = 0 for x over [0, pi].

6. Solve tan2 x = 3 tan x for x over [0, pi].

7. Find all the solutions for the trigonometric equation 2 sin2 x4 − 3 cos x4 = 0 over the interval [0, 2pi).

8. Solve the trigonometric equation 3 − 3 sin2 x = 8 sin x over the interval [0, 2pi].

9. Solve 2 sin x tan x = tan x+ sec x for all values of x [0, 2pi].

10. Solve the trigonometric equation 2 cos2 x+ 3 sin x − 3 = 0 over the interval [0, 2pi].

11. Solve tan2 x+ tan x+ 2 = 0 for values of x over the interval

[

−pi2 , pi2

]

.

12. Solve the trigonometric equation such that 5 cos2 θ − 6 sin θ = 0 over the interval [0, 2pi].

Review Answers

1. Because the problem deals with 2θ, the domain values must be doubled, making the domain 0 ≤ 2θ <

4pi

The reference angle is α = sin−1 0.6 = 0.6435

2θ = 0.6435, pi − 0.6435, 2pi+ 0.6435, 3pi − 0.6435

2θ = 0.6435, 2.2980, 6.9266, 8.7812

The values for θ are needed so the above values must be divided by 2.

θ = 0.3218, 1.1490, 3.4633, 4.3906

The results can readily be checked by graphing the function. The four results are reasonable since

they are the only results indicated on the graph that satisfy sin 2θ = 0.6.

www.ck12.org 182

2.

cos2 x = 1

16

√

cos2 x =

√

1

16

cos x = ±1

4

Then cos x = 1

4

or cos x = −1

4

cos−1 1

4

= x cos−1 −1

4

= x

x = 1.3181 radians x = 1.8235 radians

However, cos x is also positive in the fourth quadrant, so the other possible solution for cos x = 14 is

2pi − 1.3181 = 4.9651 radians and cos x is also negative in the third quadrant, so the other possible

solution for cos x = −14 is 2pi − 1.8235 = 4.4597 radians3.

tan2 x = 1

tan x = ±√1

tan x = ±1

so, tan x = 1 or tan x = −1. Therefore, x is all critical values corresponding with pi4 within the interval.

x = pi4 ,

3pi

4 ,

5pi

4 ,

7pi

4

4. Use factoring by grouping.

2 sin x+ 1 = 0 or 2 cos x − 1 = 0

2 sin x = −1 2 cos x = 1

sin x = −1

2

cos x = 1

2

x =

7pi

6

,

11pi

6

x =

pi

x

,

2pi

3

5. You can factor this one like a quadratic.sin2 x − 2 sin x − 3 = 0

(sin x − 3)(sin x+ 1) = 0

sin x − 3 = 0 sin x+ 1 = 0

sin x = 3 or sin x = −1

x = sin−1(3) x = 3pi

2

For this problem the only solution is 3pi2 because sine cannot be 3 (it is not in the range).6.

tan2 x = 3 tan x

tan2 x − 3 tan x = 0

tan x(tan x − 3) = 0

tan x = 0 or tan x = 3

x = 0, pi x = 1.25, 4.39

183 www.ck12.org

7. 2 sin2 x4 − 3 cos x4 = 0

2

(

1 − cos2 x

4

)

− 3 cos x

4

= 0

2 − 2 cos2 x

4

− 3 cos x

4

= 0

2 cos2 x

4

+ 3 cos x

4

− 2 = 0(

2 cos x

4

− 1

) (

cos x

4

+ 2

)

= 0

↙ ↘

2 cos x

4

− 1 = 0 or cos x

4

+ 2 = 0

2 cos x

4

= 1 cos x

4

= −2

cos x

4

=

1

2

x

4

=

pi

3

or 5pi

3

x =

4pi

3

or 20pi

3

20pi

3 is eliminated as a solution because it is outside of the range and cos x4 = −2 will not generate any

solutions because −2 is outside of the range of cosine. Therefore, the only solution is 4pi3 .8.

3 − 3 sin2 x = 8 sin x

3 − 3 sin2 x − 8 sin x = 0

3 sin2 x+ 8 sin x − 3 = 0

(3 sin x − 1)(sin x+ 3) = 0

3 sin x − 1 = 0 or sin x+ 3 = 0

3 sin x = 1

sin x = 1

3

sin x = −3

x = 0.3398 radians No solution exists

x = pi − 0.3398 = 2.8018 radians

9. 2 sin x tan x = tan x+ sec x

2 sin x · sin xcos x =

sin x

cos x +

1

cos x

2 sin2 x

cos x =

sin x+ 1

cos

2 sin2 x = sin x+ 1

2 sin2 x − sin x − 1 = 0

(2 sin x+ 1)(sin x − 1) = 0

2 sin x+ 1 = 0 or sin x − 1 = 0

2 sin x = −1 sin x = 1

sin x = −1

2

x =

7pi

6

,

11pi

6

,

pi

2

www.ck12.org 184

10.

2 cos2 x+ 3 sin x − 3 = 0

2(1 − sin2 x) + 3 sin x − 3 = 0 Pythagorean Identity

2 − 2 sin2 x+ 3 sin x − 3 = 0

− 2 sin2 x+ 3 sin x − 1 = 0 Multiply by − 1

2 sin2 x − 3 sin x+ 1 = 0

(2 sin x − 1)(sin x − 1) = 0

2 sin x − 1 = 0 or sin x − 1 = 0

2 sin x = 1

sin x = 1

2

sin x = 1

x =

pi

6

,

5pi

6

x =

pi

2

11. tan2 x+ tan x − 2 = 0 −1 ± √12 − 4(1)(−2)

2

= tan x

−1 ± √1 + 8

2

= tan x

−1 ± 3

2

= tan x

tan x = −2 or 1

tan x = 1 when x = −3pi4 , in the interval

[

−pi2 , pi2

]

tan x = −2 when x = −4.249 rad

12. 5 cos2 θ − 6 sin θ = 0 over the interval [0, 2pi].5

(

1 − sin2 x

)

− 6 sin x = 0

−5 sin2 x − 6 sin x+ 5 = 0

5 sin2 x+ 6 sin x − 5 = 0

−6 ± √62 − 4(5)(−5)

2(5)

= sin x

−6 ± √36 + 100

10

= sin x

−6 ± √136

10

= sin x

−6 ± 2√34

10

= sin x

−3 ± √34

5

= sin x

x = sin−1

(

−3+√34

5

)

or sin−1

(

−3−√34

5

)

x = 0.6018 rad or 2.5398 rad from the first expression, the second

expression will not yield any answers because it is out the the range of sine.

3.4 Sum and Difference Identities

Learning Objectives

• Use and identify the sum and difference identities.

185 www.ck12.org

• Apply the sum and difference identities to solve trigonometric equations.

• Find the exact value of a trigonometric function for certain angles.

In this section we are going to explore cos(a ± b), sin(a ± b), and tan(a ± b). These identities have very

useful expansions and can help to solve identities and equations.

Sum and Difference Formulas: cosine

Is cos 15◦ = cos(45◦ − 30◦)? Upon appearance, yes, it is. This section explores how to find an expression

that would equal cos(45◦ − 30◦). To simplify this, let the two given angles be a and b where 0 < b < a < 2pi.

Begin with the unit circle and place the angles a and b in standard position as shown in Figure A. Point

Pt1 lies on the terminal side of b, so its coordinates are (cos b, sin b) and Point Pt2 lies on the terminal

side of a so its coordinates are (cos a, sin a). Place the a − b in standard position, as shown in Figure B.

The point A has coordinates (1, 0) and the Pt3 is on the terminal side of the angle a− b, so its coordinates

are (cos[a − b], sin[a − b]).

www.ck12.org 186

Triangles OP1P2 in figure A and Triangle OAP3 in figure B are congruent. (Two sides and the included

angle, a − b, are equal). Therefore the unknown side of each triangle must also be equal. That is:

d (A, P3) = d (P1, P2)

Applying the distance formula to the triangles in Figures A and B and setting them equal to each other:√

[cos(a − b) − 1]2 + [sin(a − b) − 0]2 =

√

(cos a − cos b)2 + (sin a − sin b)2

Square both sides to eliminate the square root.

[cos(a − b) − 1]2 + [sin(a − b) − 0]2 = (cos a − cos b)2 + (sin a − sin b)2

FOIL all four squared expressions and simplify.

cos2(a − b) − 2 cos(a − b) + 1 + sin2(a − b) = cos2 a − 2 cos a cos b+ cos2 b+ sin2 a − 2 sin a sin b+ sin2 b

sin2(a − b) + cos2(a − b)︸ ︷︷ ︸−2 cos(a − b) + 1 = sin2 a+ cos2 a︸ ︷︷ ︸−2 cos a cos b+ sin2 b+ cos2 b︸ ︷︷ ︸−2 sin a sin b

1 − 2 cos(a − b) + 1 = 1 − 2 cos a cos b+ 1 − 2 sin a sin b

2 − 2 cos(a − b) = 2 − 2 cos a cos b − 2 sin a sin b

−2 cos(a − b) = −2 cos a cos b − 2 sin a sin b

cos(a − b) = cos a cos b+ sin a sin b

In cos(a−b) = cos a cos b+sin a sin b, the difference formula for cosine, you can substitute a−(−b) = a+b to

obtain: cos(a+b) = cos[a−(−b)] or cos a cos(−b)+sin a sin(−b). since cos(−b) = cos b and sin(−b) = − sin b,

then cos(a+ b) = cos a cos b − sin a sin b, which is the sum formula for cosine.

Using the Sum and Difference Identities of cosine

The sum/difference formulas for cosine can be used to establish other identities:

Example 1: Find an equivalent form of cos

(

pi

2 − θ

)

using the cosine difference formula.

Solution:

cos

(

pi

2

− θ

)

= cos pi

2

cos θ + sin pi

2

sin θ

cos

(

pi

2

− θ

)

= 0 × cos θ + 1 × sin θ, substitute cos pi

2

= 0 and sin pi

2

= 1

cos

(

pi

2

− θ

)

= sin θ

We know that is a true identity because of our understanding of the sine and cosine curves, which are a

phase shift of pi2 off from each other.

The cosine formulas can also be used to find exact values of cosine that we weren’t able to find before,

such as 15◦ = (45◦ − 30◦), 75◦ = (45◦ + 30◦), among others.

Example 2: Find the exact value of cos 15◦

Solution: Use the difference formula where a = 45◦ and b = 30◦.

cos(45◦ − 30◦) = cos 45◦ cos 30◦ + sin 45◦ sin 30◦

cos 15◦ =

√

2

2

×

√

3

2

+

√

2

2

× 1

2

cos 15◦ =

√

6 +

√

2

4

187 www.ck12.org

Example 3: Find the exact value of cos 105◦.

Solution: There may be more than one pair of key angles that can add up (or subtract to) 105◦. Both

pairs, 45◦ + 60◦ and 150◦ − 45◦, will yield the correct answer.

1.

cos 105◦ = cos(45◦ + 60◦)

= cos 45◦ cos 60◦ − sin 45◦ sin 60◦, substitute in the known values

=

√

2

2

× 1

2

−

√

2

2

×

√

3

2

=

√

2 − √6

4

2.

cos 105◦ = cos(150◦ − 45◦)

= cos 150◦ cos 45◦ + sin 150◦ sin 45◦

= −

√

3

2

·

√

2

2

+

1

2

·

√

2

2

= −

√

6

4

+

√

2

4

=

√

2 − √6

4

You do not need to do the problem multiple ways, just the one that seems easiest to you.

Example 4: Find the exact value of cos 5pi12 , in radians.

Solution: cos 5pi12 = cos

(

pi

4 +

pi

6

)

, notice that pi4 = 3pi12 and pi6 = 2pi12

cos

(

pi

4

+

pi

6

)

= cos pi

4

cos pi

6

− sin pi

4

sin pi

6

cos pi

4

cos pi

6

− sin pi

4

sin pi

6

=

√

2

2

×

√

3

2

−

√

2

2

× 1

2

=

√

6 − √2

4

Sum and Difference Identities: sine

To find sin(a+ b), use Example 1, from above:

sin(a+ b) = cos

[

pi

2

− (a+ b)

]

Set θ = a+ b

= cos

[(

pi

2

− a

)

− b

]

Distribute the negative

= cos

(

pi

2

− a

)

cos b+ sin

(

pi

2

− a

)

sin b Difference Formula for cosines

= sin a cos b+ cos a sin b Co-function Identities

In conclusion, sin(a+ b) = sin a cos b+ cos a sin b, which is the sum formula for sine.

To obtain the identity for sin(a − b):

www.ck12.org 188

sin(a − b) = sin[a+ (−b)]

= sin a cos(−b) + cos a sin(−b) Use the sine sum formula

sin(a − b) = sin a cos b − cos a sin b Use cos(−b) = cos b, and sin(−b) = − sin b

In conclusion, sin(a − b) = sin a cos b − cos a sin b, so, this is the difference formula for sine.

Example 5: Find the exact value of sin 5pi12

Solution: Recall that there are multiple angles that add or subtract to equal any angle. Choose whichever

formula that you feel more comfortable with.

sin 5pi

12

= sin

(3pi

12

+

2pi

12

)

= sin 3pi

12

cos 2pi

12

+ cos 3pi

12

sin 2pi

12

sin 5pi

12

=

√

2

2

×

√

3

2

+

√

2

2

× 1

2

=

√

6 +

√

2

4

Example 6: Given sinα = 1213 , where α is in Quadrant II, and sin β = 35 , where β is in Quadrant I, find

the exact value of sin(α+ β).

Solution: To find the exact value of sin(α+β), here we use sin(α+β) = sinα cos β+cosα sin β. The values

of sinα and sin β are known, however the values of cosα and cos β need to be found.

Use sin2 α+ cos2 α = 1, to find the values of each of the missing cosine values.

For cos a : sin2 α + cos2 α = 1, substituting sinα = 1213 transforms to

(

12

13

)2

+ cos2 α = 144169 + cos2 α = 1 or

cos2 α = 25169 cosα = ± 513 , however, since α is in Quadrant II, the cosine is negative, cosα = − 513 .

For cos β use sin2 β+ cos2 β = 1 and substitute sin β = 35 ,

(

3

5

)2

+ cos2 β = 925 + cos2 β = 1 or cos2 β = 1625 and

cos β = ±45 and since β is in Quadrant I, cos β = 45

Now the sum formula for the sine of two angles can be found:

sin(α+ β) = 12

13

× 4

5

+

(

− 5

13

)

× 3

5

or 48

65

− 15

65

sin(α+ β) = 33

65

Sum and Difference Identities: Tangent

To find the sum formula for tangent:

189 www.ck12.org

tan(a+ b) = sin(a+ b)cos(a+ b) Using tan θ =

sin θ

cos θ

=

sin a cos b+ sin b cos a

cos a cos b − sin a sin b Substituting the sum formulas for sine and cosine

=

sin a cos b+sin b cos a

cos a cos b

cos a cos b−sin a sin b

cos a cos b

Divide both the numerator and the denominator by cos a cos b

=

sin a cos b

cos a cos b +

sin b cos a

cos a cos b

cos a cos b

cos a cos b − sin a sin bcos a cos b

Reduce each of the fractions

=

sin a

cos a +

sin b

cos b

1 − sin a sin bcos a cos b

Substitute sin θcos θ = tan θ

tan(a+ b) = tan a+ tan b

1 − tan a tan b Sum formula for tangent

In conclusion, tan(a+ b) = tan a+tan b1−tan a tan b . Substituting −b for b in the above results in the difference formula

for tangent:

tan(a − b) = tan a − tan b

1 + tan a tan b

Example 7: Find the exact value of tan 285◦.

Solution: Use the difference formula for tangent, with 285◦ = 330◦ − 45◦

tan(330◦ − 45◦) = tan 330

◦ − tan 45◦

1 + tan 330◦ tan 45◦

=

−

√

3

3 − 1

1 −

√

3

3 · 1

=

−3 − √3

3 − √3

=

−3 − √3

3 − √3 ·

3 +

√

3

3 +

√

3

=

−9 − 6√3 − 3

9 − 3

=

−12 − 6√3

6

= −2 − √3

To verify this on the calculator, tan 285◦ = −3.732 and −2 − √3 = −3.732.

Using the Sum and Difference Identities to Verify Other Identi-

ties

Example 8: Verify the identity cos(x−y)sin x sin y = cot x cot y+ 1

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cot x cot y+ 1 = cos(x − y)sin x sin y

=

cos x cos y

sin x sin y +

sin x sin y

sin x sin y Expand using the cosine difference formula.

=

cos x cos y

sin x sin y + 1

cot x cot y+ 1 = cot x cot y+ 1 cotangent equals cosine over sine

Example 9: Show cos(a+ b) cos(a − b) = cos2 a − sin2 b

Solution: First, expand left hand side using the sum and difference formulas:

cos(a+ b) cos(a − b) = (cos a cos b − sin a sin b)(cos a cos b+ sin a sin b)

= cos2 a cos2 b − sin2 a sin2 b→ FOIL, middle terms cancel out

Substitute(1 − sin2 b)for cos2 b and(1 − cos2 a)for sin2 a and simplify.

cos2 a(1 − sin2 b) − sin2 b(1 − cos2 a)

cos2 a − cos2 a sin2 b − sin2 b+ cos2 a sin2 b

cos2 a − sin2 b

Solving Equations with the Sum and Difference Formulas

Just like the section before, we can incorporate all of the sum and difference formulas into equations and

solve for values of x. In general, you will apply the formula before solving for the variable. Typically, the

goal will be to isolate sin x, cos x, or tan x and then apply the inverse. Remember, that you may have to

use the identities in addition to the formulas seen in this section to solve an equation.

Example 10: Solve 3 sin(x − pi) = 3 in the interval [0, 2pi).

Solution: First, get sin(x − pi) by itself, by dividing both sides by 3.

3 sin(x − pi)

3

=

3

3

sin(x − pi) = 1

Now, expand the left side using the sine difference formula.

sin x cos pi − cos x sin pi = 1

sin x(−1) − cos x(0) = 1

− sin x = 1

sin x = −1

The sin x = −1 when x is 3pi2 .

Example 11: Find all the solutions for 2 cos2

(

x+ pi2

)

= 1 in the interval [0, 2pi).

Solution: Get the cos2

(

x+ pi2

)

by itself and then take the square root.

191 www.ck12.org

2 cos2

(

x+

pi

2

)

= 1

cos2

(

x+

pi

2

)

=

1

2

cos

(

x+

pi

2

)

=

√

1

2

=

1√

2

=

√

2

2

Now, use the cosine sum formula to expand and solve.

cos x cos pi

2

− sin x sin pi

2

=

√

2

2

cos x(0) − sin x(1) =

√

2

2

− sin x =

√

2

2

sin x = −

√

2

2

The sin x = −

√

2

2 is in Quadrants III and IV, so x = 5pi4 and 7pi4 .

Points to Consider

• What are the angles that have 15◦ and 75◦ as reference angles?

• Are the only angles that we can find the exact sine, cosine, or tangent values for, multiples of pi12?

(Recall that pi2 would be 6 · pi12 , making it a multiple of pi12)

Review Questions

1. Find the exact value for:

(a) cos 5pi12

(b) cos 7pi12

(c) sin 345◦

(d) tan 75◦

(e) cos 345◦

(f) sin 17pi12

2. If sin y = 1213 , y is in quad II, and sin z = 35 , z is in quad I find cos(y − z)

3. If sin y = − 513 , y is in quad III, and sin z = 45 , z is in quad II find sin(y+ z)

4. Simplify:

(a) cos 80◦ cos 20◦ + sin 80◦ sin 20◦

(b) sin 25◦ cos 5◦ + cos 25◦ sin 5◦

5. Prove the identity: cos(m−n)sinm cos n = cotm+ tan n

6. Simplify cos(pi+ θ) = − cos θ

7. Verify the identity: sin(a+ b) sin(a − b) = cos2 b − cos2 a

8. Simplify tan(pi+ θ)

9. Verify that sin pi2 = 1, using the sine sum formula.

10. Reduce the following to a single term: cos(x+ y) cos y+ sin(x+ y) sin y.

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11. Prove cos(c+d)cos(c−d) = 1−tan c tan d1+tan c tan d

12. Find all solutions to 2 cos2

(

x+ pi2

)

= 1, when x is between [0, 2pi).

13. Solve for all values of x between [0, 2pi) for 2 tan2

(

x+ pi6

)

− 1 = 7.

14. Find all solutions to sin

(

x+ pi6

)

= sin

(

x − pi4

)

, when x is between [0, 2pi).

Review Answers

1. (a)

cos 5pi

12

= cos

(2pi

12

+

3pi

12

)

= cos

(

pi

6

+

pi

4

)

= cos pi

6

cos pi

4

− sin pi

6

sin pi

4

=

√

3

2

·

√

2

2

− 1

2

·

√

2

2

=

√

6

4

−

√

2

4

=

√

6 − √2

4(b)

cos 7pi

12

= cos

(4pi

12

+

3pi

12

)

= cos

(

pi

3

+

pi

4

)

= cos pi

3

cos pi

4

− sin pi

3

sin pi

4

=

1

2

·

√

2

2

−

√

3

2

·

√

2

2

=

√

2

4

−

√

6

4

=

√

2 − √6

4(c)

sin 345◦ = sin(300◦ + 45◦) = sin 300◦ cos 45◦ + cos 300◦ sin 45◦

= −

√

3

2

·

√

2

2

+

1

2

·

√

2

2

= −

√

6

4

+

√

2

4

=

√

6 +

√

2

4(d)

tan 75◦ = tan(45◦ + 30◦) = tan 45

◦ + tan 30◦

1 − tan 45◦ tan 30◦

=

1 +

√

3

3

1 − 1 ·

√

3

3

=

3+

√

3

3

3−√3

3

=

3 +

√

3

3 − √3 ·

3 +

√

3

3 +

√

3

=

9 + 6

√

3 + 3

9 − 3 =

12 + 6

√

3

6

= 2 +

√

3

(e)

cos 345◦ = cos(315◦ + 30◦) = cos 315◦ cos 30◦ − sin 315◦ sin 30◦

=

√

2

2

·

√

3

2

−

√

2

2

· 1

2

=

√

6 − √2

4(f)

sin 17pi

12

= sin

(9pi

12

+

8pi

12

)

= sin

(3pi

4

+

2pi

3

)

= sin 3pi

4

cos 2pi

3

+ cos 3pi

4

sin 2pi

3

=

√

2

2

· 1

2

+ −

√

2

2

·

√

3

2

=

√

2

4

−

√

6

4

=

√

2 − √6

4

2. If sin y = 1213 and in Quadrant II, then by the Pythagorean Theorem cos y = − 513(122 + b2 = 132).

And, if sin z = 35 and in Quadrant I, then by the Pythagorean Theorem cos z = 45(a2 + 32 = 52). So,

to find cos(y − z) = cos y cos z+ sin y sin z and = − 513 · 45 + 1213 · 35 = −2065 + 3665 = 1665

3. If sin y = − 513 and in Quadrant III, then cosine is also negative. By the Pythagorean Theorem, the

second leg is 12(52 + b2 = 132), so cos y = −1213 . If the sin z = 45 and in Quadrant II, then the cosine

is also negative. By the Pythagorean Theorem, the second leg is 3(42 + b2 = 52), so cos = −35 . To

find sin(y+ z), plug this information into the sine sum formula.sin(y+ z) = sin y cos z+ cos y sin z

= − 5

13

· −3

5

+ −12

13

· 4

5

=

15

65

− 48

65

= −33

65

4. (a) This is the cosine difference formula, so: cos 80◦ cos 20◦+sin 80◦20◦ = cos(80◦−20◦) = cos 60◦ = 12

193 www.ck12.org

(b) This is the expanded sine sum formula, so: sin 25◦ cos 5◦ + cos 25◦ sin 5◦ = sin(25◦ + 5◦) =

sin 30◦ = 12

5. Step 1: Expand using the cosine sum formula and change everything into sine and cosinecos(m − n)

sinm cos n = cotm+ tan n

cosm cos n+ sinm sin n

sinm cos n =

cosm

sinm +

sin n

cos n

Step 2: Find a common denominator for the right hand side.

=

cosm cos n+ sinm sin n

sinm cos n

The two sides are the same, thus they are equal to each other and the identity is true.

6. cos(pi+ θ) = cos pi cos θ − sin pi sin θ = − cos θ

7. Step 1: Expand sin(a+b) and sin(a−b) using the sine sum and difference formulas. sin(a+b) sin(a−

b) = cos2 b − cos2 a (sin a cos b+ cos a sin b)(sin a cos b − cos a sin b)

Step 2: FOIL and simplify.

sin2 a cos2 b − sin a cos a sin b cos b+ sin a sin b cos a cos b − cos2 a sin2 b sin2 a cos2 b − cos a2 sin2 b

Step 3: Substitute (1 − cos2 a) for sin2 a and (1 − cos2 b) for sin2 b, distribute and simplify.

(1 − cos2 a) cos2 b − cos a2(1 − cos2 b)

cos2 b − cos2 a cos2 b − cos2 a+ cos2 a cos2 b

cos2 b − cos2 a

8. tan(pi+ θ) = tan pi+tan θ1−tan pi tan θ = tan θ1 = tan θ

9. sin pi2 = sin

(

pi

4 +

pi

4

)

= sin pi4 cos pi4 −cos pi4 sin pi4 =

√

2

2 ·

√

2

2 −

√

2

2 ·

√

2

2 =

2

4 − 24 = 0 This could also be verified

by using 60◦ + 30◦

10. Step 1: Expand using the cosine and sine sum formulas.

cos(x+ y) cos y+ sin(x+ y) sin y = (cos x cos y − sin x sin y) cos y+ (sin x cos y+ cos x sin y) sin y

Step 2: Distribute cos y and sin y and simplify.

= cos x cos2 y − sin x sin y cos y+ sin x sin y cos y+ cos x sin2 y

= cos x cos2 y+ cos x sin2 y

= cos x (cos2 y+ sin2 y)︸ ︷︷ ︸

1

= cos x

11. Step 1: Expand left hand side using the sum and difference formulascos(c+ d)

cos(c − d) =

1 − tan c tan d

1 + tan c tan d

cos c cos d − sin c sin d

cos c cos d + sin c sin d =

1 − tan c tan d

1 + tan c tan d

Step 2: Divide each term on the left side by cos c cos d and simplify

cos c cos d

cos c cos d − sin c sin dcos c cos d

cos c cos d

cos c cos d − sin c sin dcos c cos d

=

1 − tan c tan d

1 + tan c tan d

1 − tan c tan d

1 + tan c tan d =

1 − tan c tan d

1 + tan c tan d

www.ck12.org 194

12. To find all the solutions, between [0, 2pi), we need to expand using the sum formula and isolate the

cos x.

2 cos2

(

x+

pi

2

)

= 1

cos2

(

x+

pi

2

)

=

1

2

cos

(

x+

pi

2

)

=

√

1

2

=

√

2

2

cos x cos pi

2

− sin x sin pi

2

=

√

2

2

cos x · 0 − sin x · 1 =

√

2

2

− sin x =

√

2

2

sin x = −

√

2

2

This is true when x = 5pi4 or 7pi4

13. First, solve for tan().

2 tan2

(

x+

pi

6

)

− 1 = 7

2 tan2

(

x+

pi

6

)

= 6

tan2

(

x+

pi

6

)

= 3

tan

(

x+

pi

6

)

=

√

3

Now, use the tangent sum formula to expand.

tan x+ tan pi6

1 − tan x tan pi6

=

√

3

tan x+ tan pi

6

=

√

3

(

1 − tan x tan pi

6

)

tan x+

√

3

3

=

√

3 − √3 tan x ·

√

3

3

tan x+

√

3

3

=

√

3 − tan x

2 tan x = 2

√

3

3

tan x =

√

3

3

This is true when x = pi6 or 7pi6 .

14. To solve, expand each side:

sin

(

x+

pi

6

)

= sin x cos pi

6

+ cos x sin pi

6

=

√

3

2

sin x+ 1

2

cos x

sin

(

x − pi

4

)

= sin x cos pi

4

− cos x sin pi

4

=

√

2

2

sin x −

√

2

2

cos x

Set the two sides equal to each other:

195 www.ck12.org

√

3

2

sin x+ 1

2

cos x =

√

2

2

sin x −

√

2

2

cos x

√

3 sin x+ cos x =

√

2 sin x − √2 cos x√

3 sin x − √2 sin x = − cos x − √2 cos x

sin x

(√

3 − √2

)

= cos x

(

−1 − √2

)

sin x

cos x =

−1 − √2√

3 − √2

tan x = −1 −

√

2√

3 − √2 ·

√

3 +

√

2√

3 +

√

2

=

−√3 − √2 + √6 − 2

3 − 2

= −2 + √6 − √3 − √2

As a decimal, this is −2.69677, so tan−1(−2.69677) = x, x = 290.35◦ and 110.35◦.

3.5 Double Angle Identities

Learning Objectives

• Use the double angle identities to solve other identities.

• Use the double angle identities to solve equations.

Deriving the Double Angle Identities

One of the formulas for calculating the sum of two angles is:

sin(α+ β) = sinα cos β+ cosα sin β

If α and β are both the same angle in the above formula, then

sin(α+ α) = sinα cosα+ cosα sinα

sin 2α = 2 sinα cosα

This is the double angle formula for the sine function. The same procedure can be used in the sum formula

for cosine, start with the sum angle formula:

cos(α+ β) = cosα cos β − sinα sin β

If α and β are both the same angle in the above formula, then

cos(α+ α) = cosα cosα − sinα sinα

cos 2α = cos2 α − sin2 α

This is one of the double angle formulas for the cosine function. Two more formulas can be derived by

using the Pythagorean Identity, sin2 α+ cos2 α = 1.

www.ck12.org 196

sin2 α = 1 − cos2 α and likewise cos2 α = 1 − sin2 α

Using sin2 α = 1 − cos2 α : Using cos2 α = 1 − sin2 α :

cos 2α = cos2 α − sin2 α cos 2α = cos2 α − sin2 α

= cos2 α − (1 − cos2 α) = (1 − sin2 α) − sin2 α

= cos2 α − 1 + cos2 α = 1 − sin2 α − sin2 α

= 2 cos2 α − 1 = 1 − 2 sin2 α

Therefore, the double angle formulas for cos 2a are:

cos 2α = cos2 α − sin2 α

cos 2α = 2 cos2 α − 1

cos 2α = 1 − 2 sin2 α

Finally, we can calculate the double angle formula for tangent, using the tangent sum formula:

tan(α+ β) = tanα+ tan β

1 − tanα tan β

If α and β are both the same angle in the above formula, then

tan(α+ α) = tanα+ tanα

1 − tanα tanα

tan 2α = 2 tanα

1 − tan2 α

Applying the Double Angle Identities

Example 1: If sin a = 513 and a is in Quadrant II, find sin 2a cos 2a, and tan 2a.

Solution: To use sin 2a = 2 sin a cos a, the value of cos a must be found first.

= cos2 a+ sin2 a = 1

= cos2 a+

( 5

13

)2

= 1

= cos2 a+ 25

169

= 1

= cos2 a = 144

169

, cos a = ±12

13

.

However since a is in Quadrant II, cos a is negative or cos a = −1213 .

sin 2a = 2 sin a cos a = 2

( 5

13

)

×

(

−12

13

)

= sin 2a = −120

169

For cos 2a, use cos(2a) = cos2 a − sin2 a

cos(2a) =

(

−12

13

)2

−

( 5

13

)2

or 144 − 25

169

cos(2a) = 119

169

197 www.ck12.org

For tan 2a, use tan 2a = 2 tan a

1−tan2 a . From above, tan a =

5

13

− 1213

= − 512 .

tan(2a) = 2 ·

−5

12

1 −

(−5

12

)2 = −561 − 25144 =

−5

6

119

144

= −5

6

· 144

119

= −120

119

Example 2: Find cos 4θ.

Solution: Think of cos 4θ as cos(2θ + 2θ).

cos 4θ = cos(2θ + 2θ) = cos 2θ cos 2θ − sin 2θ sin 2θ = cos2 2θ − sin2 2θ

Now, use the double angle formulas for both sine and cosine. For cosine, you can pick which formula you

would like to use. In general, because we are proving a cosine identity, stay with cosine.

= (2 cos2 θ − 1)2 − (2 sin θ cos θ)2

= 4 cos4 θ − 4 cos2 θ + 1 − 4 sin2 θ cos2 θ

= 4 cos4 θ − 4 cos2 θ + 1 − 4(1 − cos2 θ) cos2 θ

= 4 cos4 θ − 4 cos2 θ + 1 − 4 cos2 θ + 4 cos4 θ

= 8 cos4 θ − 8 cos2 θ + 1

Example 3: If cot x = 43 and x is an acute angle, find the exact value of tan 2x .

Solution: Cotangent and tangent are reciprocal functions, tan x = 1cot x and tan x = 34 .

tan 2x = 2 tan x

1 − tan2 x

=

2 · 34

1 −

(

3

4

)2

=

3

2

1 − 916

=

3

2

7

16

=

3

2

· 16

7

=

24

7

Example 4: Given sin(2x) = 23 and x is in Quadrant I, find the value of sin x.

Solution: Using the double angle formula, sin 2x = 2 sin x cos x. Because we do not know cos x, we need

to solve for cos x in the Pythagorean Identity, cos x =

√

1 − sin2 x. Substitute this into our formula and

solve for sin x.

sin 2x = 2 sin x cos x

2

3

= 2 sin x

√

1 − sin2 x(2

3

)2

=

(

2 sin x

√

1 − sin2 x

)2

4

9

= 4 sin2 x(1 − sin2 x)

4

9

= 4 sin2 x − 4 sin4 x

At this point we need to get rid of the fraction, so multiply both sides by the reciprocal.

www.ck12.org 198

9

4

(4

9

= 4 sin2 x − 4 sin4 x

)

1 = 9 sin2 x − 9 sin4 x

0 = 9 sin4 x − 9 sin2 x+ 1

Now, this is in the form of a quadratic equation, even though it is a quartic. Set a = sin2 x, making the

equation 9a2 − 9a+ 1 = 0. Once we have solved for a, then we can substitute sin2 x back in and solve for

x. In the Quadratic Formula, a = 9, b = −9, c = 1.

9 ± √(−9)2 − 4(9)(1)

2(9)

=

9 ± √81 − 36

18

=

9 ± √45

18

=

9 ± 3√5

18

=

3 ± √5

6

So, a = 3+

√

5

6 ≈ 0.873 or 3−

√

5

6 ≈ .1273. This means that sin2 x ≈ 0.873 or .1273 so sin x ≈ 0.934 or

sin x ≈ .357.

Example 5: Prove tan θ = 1−cos 2θsin 2θ

Solution: Substitute in the double angle formulas. Use cos 2θ = 1− 2 sin2 θ, since it will produce only one

term in the numerator.

tan θ = 1 − (1 − 2 sin

2 θ)

2 sin θ cos θ

=

2 sin2 θ

2 sin θ cos θ

=

sin θ

cos θ

= tan θ

Solving Equations with Double Angle Identities

Much like the previous sections, these problems all involve similar steps to solve for the variable. Isolate

the trigonometric function, using any of the identities and formulas you have accumulated thus far.

Example 6: Find all solutions to the equation sin 2x = cos x in the interval [0, 2pi]

Solution: Apply the double angle formula sin 2x = 2 sin x cos x

2 sin x cos x = cos x

2 sin x cos x − cos x = cos x − cos x

2 sin x cos x − cos x = 0

cos x(2 sin x − 1) = 0 Factor out cos x

Then cos x = 0 or 2 sin x − 1 = 0

cos x = 0 or 2 sin x − 1 + 1 = 0 + 1

2

2

sin x = 1

2

sin x = 1

2

The values for cos x = 0 in the interval [0, 2pi] are x = pi2 and x = 3pi2 and the values for sin x = 12 in the

interval [0, 2pi] are x = pi6 and x = 5pi6 . Thus, there are four solutions.

Example 7: Solve the trigonometric equation sin 2x = sin x such that (−pi ≤ x < pi)

199 www.ck12.org

Solution: Using the sine double angle formula:

sin 2x = sin x

2 sin x cos x = sin x

2 sin x cos x − sin x = 0

sin x(2 cos x − 1) = 0y ↘

2 cos x − 1 = 0

2 cos x = 1

sin x = 0

x = 0,−pi cos x = 1

2

x =

pi

3

,−pi

3

Example 8: Find the exact value of cos 2x given cos x = −1314 if x is in the second quadrant.

Solution: Use the double-angle formula with cosine only.

cos 2x = 2 cos2 x − 1

cos 2x = 2

(

−13

14

)2

− 1

cos 2x = 2

(169

196

)

− 1

cos 2x =

(338

196

)

− 1

cos 2x = 338

196

− 196

196

cos 2x = 142

196

=

71

98

Example 9: Solve the trigonometric equation 4 sin θ cos θ =

√

3 over the interval [0, 2pi).

Solution: Pull out a 2 from the left-hand side and this is the formula for sin 2x.

4 sin θ cos θ =

√

3

2(2 sin θ cos θ) =

√

3

2(2 sin θ cos θ) = 2 sin 2θ

2 sin 2θ =

√

3

sin 2θ =

√

3

2

The solutions for 2θ are pi3 , 2pi3 , 7pi3 , 8pi3 , dividing each of these by 2, we get the solutions for θ, which are

pi

6 ,

pi

3 ,

7pi

6 ,

8pi

6 .

Points to Consider

• Are there similar formulas that can be derived for other angles?

• Can technology be used to either solve these trigonometric equations or to confirm the solutions?

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Review Questions

1. If sin x = 45 and x is in Quad II, find the exact values of cos 2x, sin 2x and tan 2x

2. Find the exact value of cos2 15◦ − sin2 15◦

3. Verify the identity: cos 3θ = 4 cos3 θ − 3 cos θ

4. Verify the identity: sin 2t − tan t = tan t cos 2t

5. If sin x = − 941 and x is in Quad III, find the exact values of cos 2x, sin 2x and tan 2x

6. Find all solutions to sin 2x+ sin x = 0 if 0 ≤ x < 2pi

7. Find all solutions to cos2 x − cos 2x = 0 if 0 ≤ x < 2pi

8. If tan x = 34 and 0◦ < x < 90◦, use the double angle formulas to determine each of the following:

(a) tan 2x

(b) sin 2x

(c) cos 2x

9. Use the double angle formulas to prove that the following equations are identities.

(a) 2 csc 2x = csc2 x tan x

(b) cos4 θ − sin4 θ = cos 2θ

(c) sin 2x1+cos 2x = tan x

10. Solve the trigonometric equation cos 2x − 1 = sin2 x such that [0, 2pi)

11. Solve the trigonometric equation cos 2x = cos x such that 0 ≤ x < pi

12. Prove 2 csc 2x tan x = sec2 x.

13. Solve sin 2x − cos 2x = 1 for x in the interval [0, 2pi).

14. Solve the trigonometric equation sin2 x − 2 = cos 2x such that 0 ≤ x < 2pi

Review Answers

1. If sin x = 45 and in Quadrant II, then cosine and tangent are negative. Also, by the Pythagorean

Theorem, the third side is 3(b =

√

52 − 42). So, cos x = −35 and tan x = −43 . Using this, we can find

sin 2x, cos 2x, and tan 2x.

cos 2x = 1 − sin2 x tan 2x = 2 tan x

1 − tan2 x

= 1 − 2 ·

(4

5

)2

=

2 · −43

1 −

(

−43

)2

sin 2x = 2 sin x cos x = 1 − 2 · 16

25

=

−83

1 − 169

= −8

3

÷ −7

9

= 2 · 4

5

· −3

5

= 1 − 32

25

= −8

3

· −9

7

= −24

25

= − 7

25

=

24

7

2. This is one of the forms for cos 2x.cos2 15◦ − sin2 15◦ = cos(15◦ · 2)

= cos 30◦

=

√

3

2

3. Step 1: Use the cosine sum formula cos 3θ = 4 cos3 θ − 3 cos θ

cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ

201 www.ck12.org

Step 2: Use double angle formulas for cos 2θ and sin 2θ

= (2 cos2 θ − 1) cos θ − (2 sin θ cos θ) sin θ

Step 3: Distribute and simplify.

= 2 cos3 θ − cos θ − 2 sin2 θ cos θ

= − cos θ(−2 cos2 θ + 2 sin2 θ + 1)

= − cos θ[−2 cos2 θ + 2(1 − cos2 θ) + 1] → Substiture 1 − cos2 θ for sin2 θ

= − cos θ[−2 cos2 θ + 2 − 2 cos2 θ + 1]

= − cos θ(−4 cos2 θ + 3)

= 4 cos3 θ − 3 cos θ

4. Step 1: Expand sin 2t using the double angle formula.

sin 2t − tan t = tan t cos 2t

2 sin t cos t − tan t = tan t cos 2t

Step 2: change tan t and find a common denominator.

2 sin t cos t − sin tcos t

2 sin t cos2 t − sin t

cos t

sin t(2 cos2 t − 1)

cos t

sin t

cos t · (2 cos

2 t − 1)

tan t cos 2t

5. If sin x = − 941 and in Quadrant III, then cos x = −4041 and tan x = 940 (Pythagorean Theorem,

b =

√

412 − (−9)2). So,

cos 2x = 2 cos2 x − 1

sin 2x = 2 sin x cos x = 2

(

−40

41

)2

− 1 tan 2x = sin 2xcos 2x

= 2 · − 9

41

· −40

41

=

3200

1681

− 1681

1681

=

720

1681

1519

1681

=

720

1681

=

1519

1681

=

720

1519

6. Step 1: Expand sin 2x

sin 2x+ sin x = 0

2 sin x cos x+ sin x = 0

sin x(2 cos x+ 1) = 0

Step 2: Separate and solve each for x.

2 cos x+ 1 = 0

sin x = 0 cos x = −1

2

x = 0, pi or x = 2pi

3

,

4pi

3

www.ck12.org 202

7. Expand cos 2x and simplify

cos2 x − cos 2x = 0

cos2 x − (2 cos2 x − 1) = 0

− cos2 x+ 1 = 0

cos2 x = 1

cos x = 1

cos x = 1 when x = 0, 2pi

8. (a) 3.429

(b) 0.960

(c) 0.2809. (a)

2 csc x 2x = 2sin 2x

2 csc x 2x = 2

2 sin x cos x

2 csc x 2x = 1sin x cos x

2 csc x 2x =

(sin x

sin x

) ( 1

sin x cos x

)

2 csc x 2x = sin xsin2 x cos x

2 csc x 2x = 1sin2 x ·

sin x

cos x

2 csc x 2x = csc2 x tan x

(b)

cos4 θ − sin4 θ = (cos2 θ + sin2 θ)(cos2 θ − sin2 θ)

cos4 − sin4 θ = 1(cos2 θ − sin2 θ)

cos 2θ = cos2 θ − sin2 θ

∴ cos4 θ − sin4 θ = cos 2θ

(c)

sin 2x

1 + cos 2x =

2 sin x cos x

1 + (1 − 2 sin2 x)

sin 2x

1 + cos 2x =

2 sin x cos x

2 − 2 sin2 x

sin 2x

1 + cos 2x =

2 sin x cos x

2(1 − sin2 x)

sin 2x

1 + cos 2x =

2 sin x cos x

2 cos2 x

sin 2x

1 + cos 2x =

sin x

cos x

sin 2x

1 + cos 2x = tan x

10. cos 2x − 1 = sin2 x

203 www.ck12.org

1 − 2 sin2 x = sin2 x

1 = 3 sin2 x

1

3

= sin2 x

√

3

3

= sin x

x = 35.26◦, 144.74◦

11.

cos 2x = cos x

2 cos2 x − 1 = cos x

2 cos2 x − cos x − 1 = 0

(2 cos x+ 1)(cos x − 1) = 0

↘ ↘

2 cos x+ 1 = 0 or cos x − 1 = 0

2 cos x = −1 cos x = 1

cos x = −1

2

cos x = 1 when x = 0 and cos x = −12 when x = 5pi6 .12.

2 csc 2x tan x = sec2 x

2

sin 2x ·

sin x

cos x =

1

cos2 x

2

2 sin x cos x ·

sin x

cos x =

1

cos2 x

1

cos2 x =

1

cos2 x

13. sin 2x − cos 2x = 1

www.ck12.org 204

2 sin x cos x − (1 − 2 sin2 x) = 1

2 sin x cos x − 1 + 2 sin2 x = 1

2 sin x cos x+ 2 sin2 x = 2

sin x cos x+ sin2 x = 1

sin x cos x = 1 − sin2 x

sin x cos x = cos2 x(√

1 − cos2 x

)

cos x = cos2 x(

1 − cos2 x

)

cos2 x = cos4 x

cos2 x − cos4 x = cos4 x

cos2 x − 2 cos4 x = 0

cos2 x(1 − 2 cos2 x) = 0

↙ ↘

1 − 2 cos2 x = 0

cos2 x = 0 − 2 cos2 x = −1

cos x = 0 or cos2 x = 1

2

x =

pi

2

,

3pi

2

cos x =

√

2

2

x =

pi

4

,

7pi

4

14. Use the double angle identity for cos 2x.

sin2 x − 2 = cos 2x

sin2 x − 2 = cos 2x

sin2 x − 2 = 1 − 2 sin2 x

3 sin2 x = 3

sin2 x = 1

sin x = ±1

x =

pi

2

,

3pi

2

3.6 Half-Angle Identities

Learning Objectives

• Apply the half angle identities to expressions, equations and other identities.

• Use the half angle identities to find the exact value of trigonometric functions for certain angles.

Just as there are double angle identities, there are also half angle identities. For example: sin 12a can be

found in terms of the angle “a”. Recall that 12a and a2 are the same thing and will be used interchangeably

throughout this section.

205 www.ck12.org

Deriving the Half Angle Formulas

In the previous lesson, one of the formulas that was derived for the cosine of a double angle is: cos 2θ =

1 − 2 sin2 θ. Set θ = α2 , so the equation above becomes cos 2α2 = 1 − 2 sin2 α2 .

Solving this for sin α2 , we get:

cos 2α

2

= 1 − 2 sin2 α

2

cosα = 1 − 2 sin2 α

2

2 sin2 α

2

= 1 − cosα

sin2 α

2

=

1 − cosα

2

sin α

2

= ±

√

1 − cosα

2

sin α2 =

√

1−cosα

2 if α2 is located in either the first or second quadrant.

sin α2 = −

√

1−cosα

2 if α2 is located in the third or fourth quadrant.

Example 1: Determine the exact value of sin 15◦.

Solution: Using the half angle identity, α = 30◦, and 15◦ is located in the first quadrant. Therefore,

sin α2 =

√

1−cosα

2 .

sin 15◦ =

√

1 − cos 30◦

2

=

√

1 −

√

3

2

2

=

√

2−√3

2

2

=

√

2 − √3

4

Plugging this into a calculator,

√

2−√3

4 ≈ 0.2588. Using the sine function on your calculator will validate

that this answer is correct.

Example 2: Use the half angle identity to find exact value of sin 112.5◦

Solution: since sin 225◦2 = sin 112.5◦, use the half angle formula for sine, where α = 225◦. In this example,

the angle 112.5◦ is a second quadrant angle, and the sin of a second quadrant angle is positive.

sin 112.5◦ = sin 225

◦

2

= ±

√

1 − cos 225◦

2

= +

√√

1 −

(

−

√

2

2

)

2

=

√

2

2 +

√

2

2

2

=

√

2 +

√

2

4

www.ck12.org 206

One of the other formulas that was derived for the cosine of a double angle is:

cos 2θ = 2 cos2 θ − 1. Set θ = α2 , so the equation becomes cos 2α2 = −1 + 2 cos2 α2 . Solving this for cos α2 , we

get:

cos 2α

2

= 2 cos2 α

2

− 1

cosα = 2 cos2 α

2

− 1

2 cos2 α

2

= 1 + cosα

cos2 α

2

=

1 + cosα

2

cos α

2

= ±

√

1 + cosα

2

cos α2 =

√

1+cosα

2 if α2 is located in either the first or fourth quadrant.

cos α2 = −

√

1+cosα

2 if α2 is located in either the second or fourth quadrant.

Example 3: Given that the cos θ = 34 , and that θ is a fourth quadrant angle, find cos 12 θ

Solution: Because θ is in the fourth quadrant, the half angle would be in the second quadrant, making

the cosine of the half angle negative.

cos θ

2

= −

√

1 + cos θ

2

= −

√

1 + 34

2

= −

√

7

4

2

= −

√

7

8

= −

√

7

2

√

2

= −

√

14

4

Example 4: Use the half angle formula for the cosine function to prove that the following expression is

an identity: 2 cos2 x2 − cos x = 1

Solution: Use the formula cos α2 =

√

1+cosα

2 and substitute it on the left-hand side of the expression.

2

√1 + cos θ2

2 − cos θ = 1

2

(1 + cos θ

2

)

− cos θ = 1

1 + cos θ − cos θ = 1

1 = 1

The half angle identity for the tangent function begins with the reciprocal identity for tangent.

tanα = sinαcosα ⇒ tan

α

2

=

sin α2

cos α2

The half angle formulas for sine and cosine are then substituted into the identity.

207 www.ck12.org

tan α

2

=

√

1−cosα

2√

1+cosα

2

=

√

1 − cosα√

1 + cosα

At this point, you can multiply by either

√

1−cosα√

1−cosα or

√

1+cosα√

1+cosα . We will show both, because they produce

different answers.

=

√

1 − cosα√

1 + cosα

·

√

1 − cosα√

1 − cosα =

√

1 − cosα√

1 + cosα

·

√

1 + cosα√

1 + cosα

=

1 − cosα√

1 − cos2 α

or =

√

1 − cos2 α

1 + cosα

=

1 − cosα√

sin2 α

=

√

sin2 α

1 + cosα

=

1 − cosα

sinα =

sinα

1 + cosα

So, the two half angle identities for tangent are tan α2 = 1−cosαsinα and tan α2 = sinα1+cosα .

Example 5: Use the half-angle identity for tangent to determine an exact value for tan 7pi12 .

Solution:

tan α

2

=

1 − cosα

sinα

tan 7pi

12

=

1 − cos 7pi6

sin 7pi6

tan 7pi

12

=

1 +

√

3

2

−12

tan 7pi

12

= −2 − √3

Example 6: Prove the following identity: tan x = 1−cos 2xsin 2x

Solution: Substitute the double angle formulas for cos 2x and sin 2x.

tan x = 1 − cos 2xsin 2x

=

1 − (1 − 2 sin2 x)

2 sin x cos x

=

1 − 1 + 2 sin2 x

2 sin x cos x

=

2 sin2 x

2 sin x cos x

=

sin x

cos x

= tan x

www.ck12.org 208

Solving Trigonometric Equations Using Half Angle Formulas

Example 7: Solve the trigonometric equation sin2 θ = 2 sin2 θ2 over the interval [0, 2pi).

Solution:

sin2 θ = 2 sin2 θ

2

sin2 θ = 2

(1 − cos θ

2

)

Half angle identity

1 − cos2 θ = 1 − cos θ Pythagorean identity

cos θ − cos2 θ = 0

cos θ(1 − cos θ) = 0

Then cos θ = 0 or 1 − cos θ = 0, which is cos θ = 1.

θ = 2pi or θ = 0.

Points to Consider

• Can you derive a third or fourth angle formula?

• How do 12 sin x and sin 12 x differ? Is there a formula for 12 sin x?

Review Questions

1. Find the exact value of:

(a) cos 112.5◦

(b) sin 105◦

(c) tan 7pi8

(d) tan pi8

(e) sin 67.5◦

(f) tan 165◦

2. If sin θ = 725 and θ is in Quad II, find sin θ2 , cos θ2 , tan θ2

3. Prove the identity: tan b2 = sec bsec b csc b+csc b

4. Verify the identity: cot c2 = sin c1−cos c

5. Prove that sin x tan pi2 + 2 cos x = 2 cos2 pi2

6. If sin u = − 813 , find cos u2

7. Solve 2 cos2 x2 = 1 for 0 ≤ x < 2pi

8. Solve tan a2 = 4 for 0 ≤ x < 2pi

9. Solve the trigonometric equation cos x2 = 1 + cos x such that 0 ≤ x < 2pi.

10. Prove sin x1+cos x = 1−cos xsin x .

Review Answers

1. (a)

cos 112.5◦ = cos 225

◦

2

= −

√

1 + cos 225◦

2

=

√

1 −

√

2

2

2

= −

√

2−√2

2

2

= −

√

2 − √2

4

= −

√

2 − √2

2

209 www.ck12.org

(b)

sin 105◦ = sin 210

◦

2

=

√

1 − cos 210◦

2

=

√

1 −

√

3

2

2

=

√

2−√3

2

2

=

√

2 − √3

4

=

√

2 − √3

2(c)

tan 7pi

8

= tan 1

2

· 7pi

4

=

1 − cos 7pi4

sin 7pi4

=

1 −

√

2

2

−

√

2

2

=

2−√2

2

−

√

2

2

= −2 −

√

2√

2

=

−2√2 + 2

2

= −√2 + 1

(d) tan pi8 = tan 12 · pi4 =

1−cos pi4

sin pi4

= 1−

√

2

2√

2

2

=

2−√2

2√

2

2

= 2−

√

2√

2

= 2

√

2−2

2 =

√

2 − 1

(e) sin 67.5◦ = sin 135◦2 =

√

1−cos 135◦

2 =

√

1+

√

2

2

2 =

√

2+

√

2

2

2 =

√

2+

√

2

4 =

√

2+

√

2

2

(f) tan 165◦ = tan 330◦2 = 1−cos 330

◦

sin 330◦ =

1−

√

3

2

− 12

=

2−√3

2

− 12

= −

(

2 − √3

)

= −2 + √3

But, because 165◦ is in the second quadrant, tangent is negative, so the answer is −

(

−2 + √3

)

=

2 − √3.

2. If sin θ = 725 , then by the Pythagorean Theorem the third side is 24. Because θ is in the second

quadrant, cos θ = −2425 .

sin θ

2

=

√

1 − cos θ

2

cos θ

2

=

√

1 + cos θ

2

=

√

1 + 2425

2

=

√

1 − 2425

2

tan θ

2

=

√

1 − cos θ

1 + cos θ

=

√

49

50

=

√

1

50

=

√

1 + 2425

1 − 2425

=

7

5

√

2

√

2√

2

=

1

5

√

2

√

2√

2

=

√

49

50

· 50

1

=

7

√

2

10

=

√

2

10

=

√

49

= 7

3. Step 1: Change right side into sine and cosine.tan b

2

=

sec b

sec b csc b+ csc b

=

1

cos b ÷ csc b(sec b+ 1)

=

1

cos b ÷

1

sin b

( 1

cos b + 1

)

=

1

cos b ÷

1

sin b

(

1 + cos b

cos b

)

=

1

cos b ÷

1 + cos b

sin b cos b

=

1

cos b ·

sin b cos b

1 + cos b

=

sin b

1 + cos b

www.ck12.org 210

Step 2: At the last step above, we have simplified the right side as much as possible, now we simplify

the left side, using the half angle formula.√

1 − cos b

1 + cos b =

sin b

1 + cos b

1 − cos b

1 + cos b =

sin2 b

(1 + cos b)2

(1 − cos b)(1 + cos b)2 = sin2 b(1 + cos b)

(1 − cos b)(1 + cos b) = sin2 b

1 − cos2 b = sin2 b

4. Step 1: change cotangent to cosine over sine, then cross-multiply.cot c

2

=

sin c

1 − cos c

=

cos c2

sin c2

=

√

1 + cos c

1 − cos c√

1 + cos c

1 − cos c =

sin c

1 − cos c

1 + cos c

1 − cos c =

sin2 c

(1 − cos c)2

(1 + cos c)(1 − cos c)2 = sin2 c(1 − cos c)

(1 + cos c)(1 − cos c) = sin2 c

1 − cos2 c = sin2 c5.

sin x tan x

2

+ 2 cos x = sin x

(1 − cos x

sin x

)

+ 2 cos x

sin x tan x

2

+ 2 cos x = 1 − cos x+ 2 cos x

sin x tan x

2

+ 2 cos x = 1 + cos x

sin x tan x

2

+ 2 cos x = 2 cos2 x

2

6. First, we need to find the third side. Using the Pythagorean Theorem, we find that the final side is√

105

(

b =

√

132 − (−8)2

)

. Using this information, we find that cos u =

√

105

13 . Plugging this into the

half angle formula, we get:

cos u

2

= −

√

1 +

√

105

13

2

= −

√

13+

√

105

13

2

= −

√

13 +

√

105

26

7. To solve 2 cos2 x2 = 1, first we need to isolate cosine, then use the half angle formula.2 cos2 x

2

= 1

cos2 x

2

=

1

2

1 + cos x

2

=

1

2

1 + cos x = 1

cos x = 0

211 www.ck12.org

cos x = 0 when x = pi2 , 3pi2

8. To solve tan a2 = 4, first isolate tangent, then use the half angle formula.

tan a

2

= 4√

1 − cos a

1 + cos a = 4

1 − cos a

1 + cos a = 16

16 + 16 cos a = 1 − cos a

17 cos a = −15

cos a = −15

17

Using your graphing calculator, cos a = −1517 when x = 152◦, 208◦9.

cos x

2

= 1 + cos x

±

√

1 + cos x

2

= 1 + cos x Half angle identity±√1 + cos x2

2 = (1 + cos x)2 square both sides

1 + cos x

2

= 1 + 2 cos x+ cos2 x

2

(1 + cos x

2

)

= 2(1 + 2 cos x+ cos2 x)

1 + cos x = 2 + 4 cos x+ 2 cos2 x

2 cos2 x+ 3 cos x+ 1 = 0

(2 cos x+ 1)(cos x+ 1) = 0

Then 2 cos x+ 1 = 0

2 cos x

2

=

−1

2

x =

2pi

3

,

4pi

3

Or cos x+ 1 = 0

cos x = −1

x = pi

10. sin x1+cos x = 1−cos xsin x This is the two formulas for tan x2 . Cross-multiply.

sin x

1 + cos x =

1 − cos x

sin x

(1 − cos x)(1 + cos x) = sin2 x

1 + cos x − cos x − cos2 x = sin2 x

1 − cos2 x = sin2 x

1 = sin2 x+ cos2 x

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3.7 Products, Sums, Linear Combinations, and

Applications

Learning Objectives

• Use the transformation formulas to go from product to sum and sum to product.

• Derive multiple angle formulas.

• Use linear combinations to solve trigonometric equations.

• Apply trigonometric equations to real-life situations.

Sum to Product Formulas for Sine and Cosine

In some problems, the product of two trigonometric functions is more conveniently found by the sum of

two trigonometric functions by use of identities such as this one:

sinα+ sin β = 2 sin α+ β

2

× cos α − β

2

This can be verified by using the sum and difference formulas:

2 sin α+ β

2

cos α − β

2

= 2

[

sin

(

α

2 +

β

2

)

cos

(

α

2 − β2

)]

= 2

[(

sin α2 cos

β

2 + cos α2 sin

β

2 )( cos α2 cos

β

2 + sin α2 sin

β

2

)]

= 2

[

sin α2 cos α2 cos2

β

2 + sin2 α2 sin

β

2 cos

β

2 + sin

β

2 cos2 α2 cos

β

2 + sin α2 sin2

β

2 cos α2

]

= 2

[

sin α2 cos α2

(

sin2 β2 + cos2

β

2

)

+ sin β2 cos

β

2

(

sin2 α2 + cos2 α2

)]

= 2

[

sin α2 cos α2 + sin

β

2 cos

β

2

]

= 2 sin α

2

cos α

2

+ 2 sin β

2

cos β

2

= sin

(

2 · α

2

)

+ sin

(

2 · β

2

)

= sinα+ sin β

The following variations can be derived similarly:

sinα − sin β = 2 sin α − β

2

× cos α+ β

2

cosα+ cos β = 2 cos α+ β

2

× cos α − β

2

cosα − cos β = −2 sin α+ β

2

× sin α − β

2

Example 1: Change sin 5x − sin 9y into a product.

Solution: Use the formula sinα − sin β = 2 sin α−β2 × cos α+β2 .

sin 5x − sin 9x = 2 sin 5x − 9x

2

cos 5x+ 9x

2

= 2 sin(−2x) cos 7x

= −2 sin 2x cos 7x

213 www.ck12.org

Example 2: Change cos(−3x) + cos 8x into a product.

Solution: Use the formula cosα+ cos β = 2 cos α+β2 × cos α−β2 .

cos(−3x) + cos(8x) = 2 cos −3x+ 8x

2

cos −3x − 8x

2

= 2 cos(2.5x) cos(−5.5x)

= 2 cos(2.5) cos(5.5x)

Example 3: Change 2 sin 7x cos 4x to a sum.

Solution: This is the reverse of what was done in the previous two examples. Looking at the four formulas

above, take the one that has sine and cosine as a product, sinα − sin β = 2 sin α−β2 × cos α+β2 . Therefore,

7x = α−β2 and 4x =

α+β

2 .

7x = α − β

24x= α+β2 and14x=α−β8x=α+βα=14x+β8x=[14x+β]+βso−6x=2β−3x=βα=14x+(−3x)α=11x

So, this translates to sin(11x) + sin(−3x) or sin(11x) − sin(3x). A shortcut for this problem, would be to

notice that the sum of 7x and 4x is 11x and the difference is 3x.

Product to Sum Formulas for Sine and Cosine

There are two formulas for transforming a product of sine or cosine into a sum or difference. First, let’s

look at the product of the sine of two angles. To do this, start with cosine.

cos(a − b) = cos a cos b+ sin a sin b and cos(a+ b) = cos a cos b − sin a sin b

cos(a − b) − cos(a+ b) = cos a cos b+ sin a sin b − (cos a cos b − sin a sin b)

cos(a − b) − cos(a+ b) = cos a cos b+ sin a sin b − cos a cos b+ sin a sin b

cos(a − b) − cos(a+ b) = 2 sin a sin b

1

2

[cos(a − b) − cos(a+ b)] = sin a sin b

The following product to sum formulas can be derived using the same method:

cosα cos β = 1

2

[cos(α − β) + cos(α+ β)]

sinα cos β = 1

2

[sin(α+ β) + sin(α − β)]

cosα sin β = 1

2

[sin(α+ β) − sin(α − β)]

Example 4: Change cos 2x cos 5y to a sum.

Solution: Use the formula cosα cos β = 12 [cos(α − β) + cos(α+ β)]. Set α = 2x and β = 5y.

cos 2x cos 5y = 1

2

[cos(2x − 5y) + cos(2x+ 5y)]

Example 5: Change sin 11z+sin z2 to a product.

www.ck12.org 214

Solution: Use the formula sinα cos β = 12 [sin(α+ β) + sin(α − β)]. Therefore, α+ β = 11z and α − β = z.

Solve the second equation for α and plug that into the first.

α = z+ β→ (z+ β) + β = 11z and α = z+ 5z = 6z

z+ 2β = 11z

2β = 10z

β = 5z

sin 11z+sin z

2 = sin 6z sin 5z. Again, the sum of 6z and 5z is 11z and the difference is z.

Solving Equations with Product and Sum Formulas

Example 6: Solve sin 4x+ sin 2x = 0.

Solution: Use the formula sinα+ sin β = 2 sin α+β2 × cos α−β2 .

sin 4x+ sin 2x = 0 So, sin 3x = 0 and cos x = 0→ x = pi

2

,

3pi

2

2 sin 3x cos x = 0 3x = 0, pi, 2pi, 3pi, 4pi, 5pi

sin 3x cos x = 0 x = 0, pi

3

,

2pi

3

, pi,

4pi

3

,

5pi

3

Example 7: Solve cos 5x+ cos x = cos 2x.

Solution: Use the formula cosα+ cos β = 2 cos α+β2 × cos α−β2 .

cos 5x+ cos x = cos 2x

2 cos 3x cos 2x = cos 2x

2 cos 3x cos 2x − cos 2x = 0

cos 2x(2 cos 3x − 1) = 0

↙ ↘

cos 2x = 0 2 cos 3x − 1 = 0

2 cos 3x = 1

2x =

pi

2

,

3pi

2

and cos 3x = 1

2

x =

pi

4

,

3pi

4

3x =

pi

3

,

5pi

3

,

7pi

3

,

11pi

3

,

13pi

3

,

17pi

3

x =

pi

9

,

5pi

9

,

7pi

9

,

11pi

9

,

13pi

9

,

17pi

9

Triple-Angle Formulas and Beyond

By combining the sum formula and the double angle formula, formulas for triple angles and more can be

found.

Example 8: Find the formula for sin 3x

Solution: Use both the double angle formula and the sum formula.

215 www.ck12.org

sin 3x = sin(2x+ x)

= sin(2x) cos x+ cos(2x) sin x

= (2 sin x cos x) cos x+ (cos2 x − sin2 x) sin x

= 2 sin x cos2 x+ cos2 x sin x − sin3 x

= 3 sin x cos2 x − sin3 x

= 3 sin x(1 − sin2 x) − sin3 x

= 3 sin x − 4 sin3 x

Example 9: Find the formula for cos 4x

Solution: Using the same method from the previous example, you can obtain this formula.

cos 4x = cos(2x+ 2x)

= cos2 2x − sin2 2x

= (cos2 x − sin2 x)2 − (2 sin x cos x)2

= cos4 −2 sin2 x cos2 x+ sin4 x − 4 sin2 x cos2 x

= cos4 −6 sin2 x cos2 x+ sin4 x

= cos4 −6(1 − cos2 x) cos2 x+ (1 − cos2 x)2

= 1 − 8 cos2 x+ 8 cos4 x

Linear Combinations

Here, we take an equation which takes a linear combination of sine and cosine and converts it into a simpler

cosine function.

A cos x+ B sin x = C cos(x − D), where C = √A2 + B2, cosD = AC and sinD = BC .

Example 10: Transform 3 cos 2x − 4 sin 2x into the form C cos(2x − D)

Solution: A = 3 and B = −4, so C = √32 + (−4)2 = 5. Therefore cosD = 35 and sinD = −45 which

makes the reference angle is −53.1◦ or -0.927 radians. since cosine is positive and sine is negative, the

angle must be a fourth quadrant angle. D must therefore be 306.9◦ or 5.36 radians.The final answer is

3 cos 2x − 4 sin 2x = 5 cos(2x − 5.36).

Example 11: Solve 5 cos x+ 12 sin x = 6.

Solution: First, transform the left-hand side into the form C cos(x − D). A = 5 and B = 12, so C =√

52 + 122 = 13. From this cosD = 513 and sinD = 1213 , which makes the angle in the first quadrant and

1.176 radians. Now, our equation looks like this: 13 cos(x − 1.176) = 6 and we can solve for x.

cos(x − 1.176) = 6

13

x − 1.176 = cos−1

( 6

13

)

x − 1.176 = 1.09

x = 2.267 radians

www.ck12.org 216

Applications & Technology

Example 12: The range of a small rocket that is launched with an initial velocity v at an angle with θ

the horizontal is given by R(range) = v2(velocity)g(9.8m/s2) sin 2θ. If the rocket is launched with an initial velocity of

15 m/s, what angle is needed to reach a range of 20 m?

Solution: Plug in 15 m/s for v and 20 m for the range to solve for the angle.

20 =

152

9.8

sin 2θ

20 = 22.96 sin 2θ

0.871̄ = sin 2θ

sin−1(0.871̄) = 2θ

60.59◦, 119.41◦ = 2θ

30.3◦, 59.7◦ = θ

You can also use the TI-83 to solve trigonometric equations. It is sometimes easier than solving the

equation algebraically. Just be careful with the directions and make sure your final answer is in the form

that is called for. You calculator cannot put radians in terms of pi.

Example 13: Solve sin x = 2 cos x such that 0 ≤ x ≤ 2pi using a graphing calculator.

Solution: In y =, graph y1 = sin x and y2 = 2 cos x.

Next, use CALC to find the intersection points of the graphs.

Review Questions

1. Express the sum as a product: sin 9x+ sin 5x

2. Express the difference as a product: cos 4y − cos 3y

3. Verify the identity (using sum-to-product formula): cos 3a−cos 5asin 3a+sin 5a = − tan 4a

4. Express the product as a sum: sin(6θ) sin(4θ)

5. Transform to the form C cos(x − D)

(a) 5 cos x − 5 sin x

(b) −15 cos 3x − 8 sin 3x

6. Solve sin 11x − sin 5x = 0 for all solutions 0 ≤ x < 2pi.

7. Solve cos 4x+ cos 2x = 0 for all solutions 0 ≤ x < 2pi.

8. Solve sin 5x+ sin x = sin 3x for all solutions 0 ≤ x < 2pi.

9. In the study of electronics, the function f (t) = sin(200t+pi)+sin(200t−pi) is used to analyze frequency.

Simplify this function using the sum-to-product formula.

217 www.ck12.org

10. Derive a formula for tan 4x.

11. A spring is being moved up and down. Attached to the end of the spring is an object that undergoes

a vertical displacement. The displacement is given by the equation y = 3.50 sin t + 1.20 sin 2t. Find

the first two values of t (in seconds) for which y = 0.

Review Answers

1. Using the sum-to-product formula:

sin 9x+ sin 5x

1

2

(

sin

(9x+ 5x

2

)

cos

(9x − 5x

2

))

1

2

sin 7x cos 2x

2. Using the difference-to-product formula:

cos 4y − cos 3y

− 2 sin

(4y+ 3y

2

)

sin

(4y − 3y

2

)

− 2 sin 7y

2

sin y

2

3. Using the difference-to-product formulas: cos 3a − cos 5a

sin 3a − sin 5a = − tan 4a

−2 sin

(

3a+5a

2

)

sin

(

3a−5a

2

)

2 sin

(

3a−5a

2

)

cos

(

3a+5a

2

)

− sin 4acos 4a

− tan 4a

4. Using the product-to-sum formula:

sin 6θ sin 4θ

1

2

(cos(6θ − 4θ − cos(6θ + 4θ))

1

2

(cos 2θ − cos 10θ)

5. (a) If 5 cos x − 5 sin x, then A = 5 and B = −5. By the Pythagorean Theorem, C = 5√2 and

cosD = 5

5

√

2

= 1√

2

=

√

2

2 . So, because B is negative, D is in Quadrant IV. Therefore, D = 7pi4 .

Our final answer is 5

√

2 cos

(

x − 7pi4

)

.

(b) If −15 cos 3x − 8 sin 3x, then A = −15 and B = −8. By the Pythagorean Theorem, C = 17.

Because A and B are both negative, D is in Quadrant III, which means D = cos−1

(

−1517

)

= 2.65

rad. Our final answer is 17 cos 3(x − 2.65).

6. Using the sum-to-product formula:

sin 11x − sin 5x = 0 sin 3x = 0 or cos 8x = 0

2 sin 11x − 5x

2

cos 11x+ 5x

2

= 0 So, 3x = 0, pi 8x = pi

2

,

3pi

2

2 sin 3x cos 8x = 0

sin 3x cos 8x = 0 x = 0, pi

3

x =

pi

16

,

3pi

16

www.ck12.org 218

7. Using the sum-to-product formula:

cos 4x+ cos 2x = 0

2 cos 4x+ 2x

2

cos 4x − 2x

2

= 0

2 cos 3x cos x = 0

cos 3x cos x = 0

So, either cos 3x = 0 or cos x = 0

3x =

pi

2

,

3pi

2

,

5pi

2

,

7pi

2

,

9pi

2

,

11pi

2

x =

pi

6

,

pi

2

,

5pi

6

,

7pi

6

,

3pi

2

,

11pi

6

8. Move sin 3x over to the other side and use the sum-to-product formula:

sin 5x+ sin x = sin 3x

sin 5x − sin 3x+ sin x = 0

2 cos

(5x+ 3x

2

)

sin

(5x − 3x

2

)

+ sin x = 0

2 cos 4x sin x+ sin x = 0

sin x(2 cos 4x+ 1) = 0

So sin x = 0

x = 0, pi or 2 cos 4x = −1

cos 4x = −1

2

4x =

2pi

3

,

4pi

3

,

8pi

3

,

10pi

3

,

14pi

3

,

16pi

3

,

20pi

3

,

22pi

3

=

pi

6

,

pi

3

,

2pi

3

,

5pi

6

,

7pi

6

,

4pi

3

,

5pi

3

,

11pi

6

x = 0, =

pi

6

,

pi

3

,

2pi

3

,

5pi

6

, pi,

7pi

6

,

4pi

3

,

5pi

3

,

11pi

6

9. Using the sum-to-product formula:

f (x) = sin(200x+ pi) + sin(200x − pi)

= 2 sin

(

(200x+ pi) + (200t − pi)

2

)

cos

(

(200x+ pi) − (200x − pi)

2

)

= 2 sin

(400x

2

)

cos

(2pi

2

)

= 2 sin 200x cos pi

= 2 sin 200x(−1)

= −2 sin 200x

10. Derive a formula for tan 4x.

219 www.ck12.org

tan 4x = tan(2x+ 2x)

=

tan 2x+ tan 2x

1 − tan 2x tan 2x

=

2 tan 2x

1 − tan2 2x

=

2 · 2 tan x

1−tan2 x

1 −

(

2 tan x

1−tan2 x

)2

=

4 tan x

1 − tan2 x ÷

(1 − tan2 x)2 − 4 tan2 x

(1 − tan2 x)2

=

4 tan x

1 − tan2 x ÷

1 − 2 tan2 x+ tan4 x − 4 tan2 x

(1 − tan2 x)2

=

4 tan x

1 − tan2 x ·

(1 − tan2 x)2

1 − 6 tan2 x+ tan4 x

=

4 tan x − 4 tan3 x

1 − 6 tan2 x+ tan4 x

11. Let y = 0.

3.50 sin t + 1.20 sin 2t = 0

3.50 sin t + 2.40 sin t cos t = 0, Double-Angle Identity

sin t(3.50 + 2.40 cos t) = 0

sin t = 0 or 3.50 + 2.40 cos t = 0

2.40 cos t = −3.50

cos t = −1.46→ no solution because − 1 ≤ cos t ≤ 1.

t = 0, pi

3.8 Chapter Review

Chapter Summary

Here are the identities studied in this chapter:

Quotient & Reciprocal Identities

tan θ = sin θcos θ cot θ =

cos θ

sin θ

csc θ = 1sin θ sec θ =

1

cos θ cot θ =

1

tan θ

Pythagorean Identities

sin2 θ + cos2 θ = 1 1 + cot2 θ = csc2 θ tan2 θ + 1 = sec2 θ

Even & Odd Identities

sin(−x) = − sin x cos(−x) = cos x tan(−x) = − tan x

csc(−x) = − csc x sec(−x) = sec x cot(−x) = − cot x

Co-Function Identities

www.ck12.org 220

sin

(

pi

2

− θ

)

= cos θ cos

(

pi

2

− θ

)

= sin θ tan

(

pi

2

− θ

)

= cot θ

csc

(

pi

2

− θ

)

= sec θ sec

(

pi

2

− θ

)

= csc θ cot

(

pi

2

− θ

)

= tan θ

Sum and Difference Identities

cos(α+ β) = cosα cos β − sinα sin β cos(α − β) = cosα cos β+ sinα sin β

sin(α+ β) = sinα cos β+ cosα sin β sin(α − β) = sinα cos β − cosα sin β

tan(α+ β) = tanα+ tan β

1 − tanα tan β tan(α − β) =

tanα − tan β

1 + tanα tan β

Double Angle Identities

cos(2α) = cos2 α − sin2 α = 2 cos2 α − 1 = 1 − 2 sin2 α

sin(2α) = 2 sinα cos β

tan(2α) = 2 tanα

1 − tan2 α

Half Angle Identities

cos α

2

= ±

√

1 + cosα

2

sin α

2

= ±

√

1 − cosα

2

tan α

2

=

1 − cosα

sinα =

sinα

1 + cosα

Product to Sum & Sum to Product Identities

sin a+ sin b = 2 sin a+ b

2

cos a − b

2

sin a sin b = 1

2

[cos(a − b) − cos(a+ b)]

sin a − sin b = 2 sin a − b

2

cos a+ b

2

cos a cos b = 1

2

[cos(a − b) + cos(a+ b)]

cos a+ cos b = 2 cos a+ b

2

cos a − b

2

sin a cos b = 1

2

[sin(a+ b) + sin(a − b)]

cos a − cos b = 2 − 2 sin a+ b

2

sin a − b

2

cos a sin b = 1

2

[sin(a+ b) − sin(a − b)]

Linear Combination Formula

A cos x+ B sin x = C cos(x − D), where C = √A2 + B2, cosD = AC and sinD = BC

Review Questions

1. Find the sine, cosine, and tangent of an angle with terminal side on (−8, 15).

2. If sin a =

√

5

3 and tan a < 0, find sec a.

3. Simplify: cos4 x−sin4 xcos2 − sin2 x .

4. Verify the identity: 1+sin xcos x sin x = sec x(csc x+ 1)

For problems 5-8, find all the solutions in the interval [0, 2pi).

5. sec

(

x+ pi2

)

+ 2 = 0

221 www.ck12.org

6. 8 sin

(

x

2

)

− 8 = 0

7. 2 sin2 x+ sin 2x = 0

8. 3 tan2 2x = 1

9. Solve the trigonometric equation 1 − sin x = √3 sin x over the interval [0, pi].

10. Solve the trigonometric equation 2 cos 3x − 1 = 0 over the interval [0, 2pi].

11. Solve the trigonometric equation 2 sec2 x − tan4 x = −1 for all real values of x.

Find the exact value of:

12. cos 157.5◦

13. sin 13pi12

14. Write as a product: 4(cos 5x+ cos 9x)

15. Simplify: cos(x − y) cos y − sin(x − y) sin y

16. Simplify: sin

(

4pi

3 − x

)

+ cos

(

x+ 5pi6

)

17. Derive a formula for sin 6x.

18. If you solve cos 2x = 2 cos2 x − 1 for cos2 x, you would get cos2 x = 12(cos 2x + 1). This new formula

is used to reduce powers of cosine by substituting in the right part of the equation for cos2 x. Try

writing cos4 x in terms of the first power of cosine.

19. If you solve cos 2x = 1 − 2 sin2 x for sin2 x, you would get sin2 x = 12(1 − cos 2x). Similar to the new

formula above, this one is used to reduce powers of sine. Try writing sin4 x in terms of the first power

of cosine.

20. Rewrite in terms of the first power of cosine:

(a) sin2 x cos2 2x

(b) tan4 2x

Review Answers

1. If the terminal side is on (−8, 15), then the hypotenuse of this triangle would be 17 (by the Pythagorean

Theorem, c =

√

(−8)2 + 152). Therefore, sin x = 1517 , cos x = − 817 , and tan x = −158 .

2. If sin a =

√

5

3 and tan a < 0, then a is in Quadrant II. Therefore sec a is negative. To find the third

side, we need to do the Pythagorean Theorem.(√

5

)2

+ b2 = 32

5 + b2 = 9 So, sec a = 3

2

,

b2 = 4

b = 2

3. Factor top, cancel like terms, and use the Pythagorean Theorem Identity.

cos4 x − sin4 x

cos2 x − sin2 x

(cos2 x+ sin2 x)(cos2 x − sin2 x)

cos2 x − sin2 x

cos2 x+ sin2 x

1

4. Change secant and cosecant into terms of sine and cosine, then find a common denominator.

www.ck12.org 222

1 + sin x

cos x sin x = sec x(csc x+ 1)

=

1

cos x

( 1

sin x + 1

)

=

1

cos x

(

1 + sin x

sin x

)

=

1 + sin x

cos x sin x5.

sec

(

x+

pi

2

)

+ 2 = 0

sec

(

x+

pi

2

)

= −2

cos

(

x+

pi

2

)

= −1

2

x+

pi

2

=

2pi

3

,

4pi

3

x =

2pi

3

− pi

2

,

4pi

3

− pi

2

x =

pi

6

,

5pi

66.

8 sin

( x

2

)

− 8 = 0

8 sin x

2

= 8

sin x

2

= 1

x

2

=

x

2

x = pi

7.

2 sin2 x+ sin 2x = 0

2 sin2 x+ 2 sin x cos x = 0

2 sin x(sin x+ cos x) = 0

So, 2 sin x = 0 or sin x+ cos x = 0

2 sin x = 0 sin x+ cos x = 0

sin x = 0 sin x = − cos x

x = 0, pi x =

3pi

4

,

7pi

48.

tan2 2x = 1

3

tan 2x =

√

3

3

2x =

pi

6

,

7pi

6

x =

pi

12

,

7pi

12

223 www.ck12.org

9.

1 − sin x = √3 sin x

1 = sin x+

√

3 sin x

1 = sin x

(

1 +

√

3

)

1

1 +

√

3

= sin x

sin−1

(

1

1+

√

3

)

= x or x = .3747 radians and x = 2.7669 radians

10. Because this is cos 3x, you will need to divide by 3 at the very end to get the final answer. This is

why we went beyond the limit of 2pi when finding 3x.

2 cos 3x − 1 = 0

2 cos 3x = 1

cos 3x = 1

2

3x = cos−1

(1

2

)

=

pi

3

,

5pi

3

,

7pi

3

,

11pi

3

,

13pi

3

,

17pi

3

x =

pi

9

,

5pi

9

,

7pi

9

,

11pi

9

,

13pi

9

,

17pi

9

11. Rewrite the equation in terms of tan by using the Pythagorean identity, 1 + tan2 θ = sec2 θ.

2 sec2 x − tan4 x = −1

2(1 + tan2 x) − tan4 x = −1

2 + 2 tan2 x − tan4 x = −1

tan4 x − 2 tan2 x+ 1 = 0

(tan2 x − 1)(tan2 x − 1) = 0

Because these factors are the same, we only need to solve one for x.

tan2 x − 1 = 0

tan2 x = 1

tan x = ±1

x =

pi

4

+ pik and 3pi

4

+ pik

Where k is any integer.

12. Use the half angle formula with 315◦.

cos 157.5◦ = cos 315

◦

2

= −

√

1 + cos 315◦

2

= −

√

1 +

√

2

2

2

= −

√

2 +

√

2

4

= −

√

2 +

√

2

2

www.ck12.org 224

13. Use the sine sum formula.

sin 13pi

12

= sin

(10pi

12

+

3pi

12

)

= sin

(5pi

6

+

pi

4

)

= sin 5pi

6

cos pi

4

+ cos 5pi

6

sin pi

4

=

√

3

2

·

√

2

2

− 1

2

·

√

2

2

=

√

6 − √2

414.

4(cos 5x+ cos 9x) = 4

[

2 cos

(5x+ 9x

2

)

cos

(5x − 9x

2

)]

= 8 cos 7x cos(−2x)

= 8 cos 7x cos 2x

15.

cos(x − y) cos y − sin(x − y) sin y

cos y(cos x cos y+ sin x sin y) − sin y (sin x cos y − cos x sin y)

cos x cos2 y+ sin x sin y cos y − sin x sin y cos y+ cos x sin2 y

cos x cos2 y+ cos x sin2 y

cos x(cos2 y+ sin2 y)

cos x

16. Use the sine and cosine sum formulas.

sin

(4pi

3

− x

)

+ cos

(

x+

5pi

6

)

sin 4pi

3

cos x − cos 4pi

3

sin x+ cos x cos 5pi

6

− sin x sin 5pi

6

−

√

3

2

cos x+ 1

2

sin x −

√

3

2

cos x − 1

2

sin x

− √3 cos x

17. Use the sine sum formula as well as the double angle formula.

sin 6x = sin(4x+ 2x)

= sin 4x cos 2x+ cos 4x sin 2x

= sin(2x+ 2x) cos 2x+ cos(2x+ 2x) sin 2x

= cos 2x (sin 2x cos 2x+ cos 2x sin 2x) + sin 2x(cos 2x cos 2x − sin 2x sin 2x)

= 2 sin 2x cos2 2x+ sin 2x cos2 2x − sin3 2x

= 3 sin 2x cos2 2x − sin3 2x

= sin 2x(3 cos2 2x − sin2 2x)

= 2 sin x cos x[3(cos2 x − sin2 x)2 − (2 sin x cos x)2

= 2 sin x cos x[3(cos4 x − 2 sin2 x cos2 x+ sin4 x) − 4 sin2 x cos2 x]

= 2 sin x cos x[3 cos4 x − 6 sin2 x cos2 x+ 3 sin4 x − 4 sin2 x cos2 x]

= 2 sin x cos x[3 cos4 x+ 3 sin4 x − 10 sin2 x cos2 x]

= 6 sin x cos5 x+ 6 sin5 x cos x − 20 sin3 x cos3 x

225 www.ck12.org

18. Using our new formula, cos4 x =

[

1

2(cos 2x+ 1)

]2 Now, our final answer needs to be in the first power

of cosine, so we need to find a formula for cos2 2x. For this, we substitute 2x everywhere there is an

x and the formula translates to cos2 2x = 12(cos 4x+ 1).

19. Using our new formula, sin4 x =

[

1

2(1 − cos 2x)

]2 Now, our final answer needs to be in the first power

of cosine, so we need to find a formula for cos2 2x. For this, we substitute 2x everywhere there is an

x and the formula translates to cos2 2x = 12(cos 4x+ 1).

20. (a) First, we use both of our new formulas, then simplify:

sin2 x cos2 2x = 1

2

(1 − cos 2x)1

2

(cos 4x+ 1)

=

(1

2

− 1

2

cos 2x

) (1

2

cos 4x+ 1

2

)

=

1

4

cos 4x+ 1

4

− 1

4

cos 2x cos 4x − 1

4

cos 2x

=

1

4

(1 − cos 2x+ cos 4x − cos 2x cos 4x)

(b) For tangent, we using the identity tan x = sin xcos x and then substitute in our new formulas. tan4 2x =

sin4 2x

cos4 2x → now, use the formulas we derived in #8 and 9.

Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator

activities designed to supplement the objectives for some of the lessons in this chapter. See

http://www.ck12.org/flexr/chapter/9701.

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Chapter 4

Inverse Trigonometric

Functions - 2nd edition

4.1 Basic Inverse Trigonometric Functions

Introduction

Recall that an inverse function is a reflection of the function over the line y = x. In order to find the inverse

of a function, you must switch the x and y values and then solve for y. A function has an inverse if and

only if it has exactly one output for every input and exactly one input for every output. All of the trig

functions fit these criteria over a specific range. In this chapter, we will explore inverse trig functions and

equations.

Learning Objectives

• Understand and evaluate inverse trigonometric functions.

• Extend the inverse trigonometric functions to include the csc−1, sec−1 and cot−1 functions.

• Apply inverse trigonometric functions to the critical values on the unit circle.

Defining the Inverse of the Trigonometric Ratios

Recall from Chapter 1, the ratios of the six trig functions and their inverses, with regard to the unit circle.

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sin θ = y

r

→ sin−1 y

r

= θ cos θ = x

r

→ cos−1 x

r

= θ

tan θ = y

x

→ tan−1 y

x

= θ cot θ = x

y

→ cot−1 x

y

= θ

csc θ = r

y

→ csc−1 r

y

= θ sec θ = r

x

→ sec−1 r

x

= θ

These ratios can be used to find any θ in standard position or in a triangle.

Example 1: Find the measure of the angles below.

a.

b.

Solution: For part a, you need to use the sine function and part b utilizes the tangent function. Because

both problems require you to solve for an angle, the inverse of each must be used.

a. sin x = 725 → sin−1 725 = x→ x = 16.26◦

b. tan x = 409 → tan−1 409 = x→ x = 77.32◦

The trigonometric value tan θ = 409 of the angle is known, but not the angle. In this case the inverse of the

trigonometric function must be used to determine the measure of the angle. (Directions for how to find

inverse function values in the graphing calculator are in Chapter 1). The inverse of the tangent function

is read “tangent inverse” and is also called the arctangent relation. The inverse of the cosine function is

read “cosine inverse” and is also called the arccosine relation. The inverse of the sine function is read “sine

inverse” and is also called the arcsine relation.

Example 2: Find the angle, θ, in standard position.

Solution: The tan θ = yx or, in this case, tan θ = 8−11 . Using the inverse tangent, you get tan−1 − 811 =−36.03◦. This is the reference angle and in the 4th quadrant. This value of −36.03◦ is the angle you also

see if you move clockwise from the -x axis. To find the corresponding angle in the second quadrant (which

is the same as though you started at the +x axis and moved counterclockwise), subtract 36.03◦ from 180◦,

www.ck12.org 228

yielding 143.97◦.

Recall that inverse trigonometric functions are also used to find the angle of depression or elevation.

Example 3: A new outdoor skating rink has just been installed outside a local community center. A

light is mounted on a pole 25 feet above the ground. The light must be placed at an angle so that it will

illuminate the end of the skating rink. If the end of the rink is 60 feet from the pole, at what angle of

depression should the light be installed?

Solution: In this diagram, the angle of depression, which is located outside of the triangle, is not known.

Recall, the angle of depression equals the angle of elevation. For the angle of elevation, the pole where the

light is located is the opposite and is 25 feet high. The length of the rink is the adjacent side and is 60

feet in length. To calculate the measure of the angle of elevation the trigonometric ratio for tangent can

be applied.

tan θ = 25

60

tan θ = 0.4166

tan−1(tan θ) = tan−1(0.4166)

θ = 22.6◦

The angle of depression at which the light must be placed to light the rink is 22.6◦

Exact Values for Inverse Sine, Cosine, and Tangent

Recall the unit circle and the critical values. With the inverse trigonometric functions, you can find the

angle value (in either radians or degrees) when given the ratio and function. Make sure that you find all

solutions within the given interval.

Example 4: Find the exact value of each expression without a calculator, in [0, 2pi).

a. sin−1

(

−

√

3

2

)

b. cos−1

(

−

√

2

2

)

c. tan−1

√

3

Solution: These are all values from the special right triangles and the unit circle.

a. Recall that −

√

3

2 is from the 30 − 60 − 90 triangle. The reference angle for sin and

√

3

2 would be 60◦.

Because this is sine and it is negative, it must be in the third or fourth quadrant. The answer is either 4pi3

or 5pi3 .

b. −

√

2

2 is from an isosceles right triangle. The reference angle is then 45◦. Because this is cosine and

negative, the angle must be in either the second or third quadrant. The answer is either 3pi4 or 5pi4 .

c.

√

3 is also from a 30 − 60 − 90 triangle. Tangent is √3 for the reference angle 60◦. Tangent is positive

229 www.ck12.org

in the first and third quadrants, so the answer would be pi3 or 4pi3 .

Notice how each one of these examples yield two answers. This poses a problem when finding a singular

inverse for each of the trig functions. Therefore, we need to restrict the domain in which the inverses can

be found, which will be addressed in the next section. Unless otherwise stated, all angles are in radians.

Finding Inverses Algebraically

In the Prerequisite Chapter, you learned that each function has an inverse relation and that this inverse

relation is a function only if the original function is one-to-one. A function is one-to-one when its graph

passes both the vertical and the horizontal line test. This means that every vertical and horizontal line

will intersect the graph in exactly one place.

This is the graph of f (x) = xx+1 . The graph suggests that f is one-to-one because it passes both the vertical

and the horizontal line tests. To find the inverse of f , switch the x and y and solve for y.

First, switch x and y.

x =

y

y+ 1

Next, multiply both sides by (y+ 1).

(y+ 1)x =

y

y+ 1

(y+ 1)

x(y+ 1) = y

Then, apply the distributive property and put all the y terms on one side so you can pull out the y.

xy+ x = y

xy − y = −x

y(x − 1) = −x

Divide by (x − 1) to get y by itself.

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y =

−x

x − 1

Finally, multiply the right side by −1−1 .

y =

x

1 − x

Therefore the inverse of f is f −1(x) = x1−x .

The symbol f −1 is read “ f inverse” and is not the reciprocal of f .

Example 5: Find the inverse of f (x) = 1x−5 algebraically.

Solution: To find the inverse algebraically, switch f (x) to y and then switch x and y.

y =

1

x − 5

x =

1

y − 5

x(y − 5) = 1

xy − 5x = 1

xy = 5x+ 1

y =

5x+ 1

x

Example 6: Find the inverse of f (x) = 5 sin−1

(

2

x−3

)

Solution:

a.

f (x) = 5 sin−1

( 2

x − 3

)

x = 5 sin−1

(

2

y − 3

)

x

5

= sin−1

(

2

y − 3

)

sin x

5

=

(

2

y − 3

)

(y − 3) sin x

5

= 2

(y − 3) = 2sin x5

y =

2

sin x5

+ 3

Example 7: Find the inverse of the trigonometric function f (x) = 4 tan−1(3x+ 4)

Solution:

231 www.ck12.org

x = 4 tan−1(3y+ 4)

x

4

= tan−1(3y+ 4)

tan x

4

= 3y+ 4

tan x

4

− 4 = 3y

tan x4 − 4

3

= y

f −1(x) =

tan x4 − 4

3

Points to Consider

• What is the difference between an inverse and a reciprocal?

• Considering that most graphing calculators do not have csc, sec or cot buttons, how would you find

the inverse of each of these?

• Besides algebraically, is there another way to find the inverse?

Review Questions

1. Use the special triangles or the unit circle to evaluate each of the following:

(a) cos 120◦

(b) csc 3pi4

(c) tan 5pi3

2. Find the exact value of each inverse function, without a calculator in [0, 2pi):

(a) cos−1(0)

(b) tan−1

(

−√3

)

(c) sin−1

(

−12

)

Find the value of the missing angle.

3.

4.

5. What is the value of the angle with its terminal side passing through (-14, -23)?

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6. A 9-foot ladder is leaning against a wall. If the foot of the ladder is 4 feet from the base of the wall,

what angle does the ladder make with the floor?

Find the inverse of the following functions.

7. f (x) = 2x3 − 5

8. y = 13 tan−1

(

3

4 x − 5

)

9. g(x) = 2 sin(x − 1) + 4

10. h(x) = 5 − cos−1(2x+ 3)

Review Answers

1. (a) −12

(b)

√

2

(c) −√3

2. (a) pi2 , 3pi2

(b) 2pi3 , 5pi3

(c) 11pi6 , 7pi6

3. cos θ = 1217 → cos−1 1217 = 45.1◦

4. sin θ = 2536 → sin−1 3136 = 59.44◦

5. This problem uses tangent inverse. tan x = −14−23 → x = tan−1 1423 = 31.33◦ (value graphing calculator

will produce). However, this is the reference angle. Our angle is in the third quadrant because both

the x and y values are negative. The angle is 180◦ + 31.33◦ = 211.33◦.

6.

cos A = 4

9

cos−1 4

9

= A

∠A = 63.6◦

7.

f (x) = 2x3 − 5

y = 2x3 − 5

x = 2y3 − 5

x+ 5 = 2y3

x+ 5

2

= y3

3

√

x+ 5

2

= y

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8.

y =

1

3

tan−1

(3

4

x − 5

)

x =

1

3

tan−1

(3

4

y − 5

)

3x = tan−1

(3

4

y − 5

)

tan(3x) = 3

4

y − 5

tan(3x) + 5 = 3

4

y

4(tan(3x) + 5)

3

= y9.

g(x) = 2 sin(x − 1) + 4

y = 2 sin(x − 1) + 4

x = 2 sin(y − 1) + 4

x − 4 = 2 sin(y − 1)

x − 4

2

= sin(y − 1)

sin−1

( x − 4

2

)

= y − 1

1 + sin−1

( x − 4

2

)

= y10.

h(x) = 5 − cos−1(2x+ 3)

y = 5 − cos−1(2x+ 3)

x = 5 − cos−1(2y+ 3)

x − 5 = − cos−1(2y+ 3)

5 − x = cos−1(2y+ 3)

cos(5 − x) = 2y+ 3

cos(5 − x) − 3 = 2y

cos(5 − x) − 3

2

= y

4.2 Graphing Inverse Trigonometric Functions

Learning Objectives

• Understand the meaning of restricted domain as it applies to the inverses of the six trigonometric

functions.

• Apply the domain, range and quadrants of the six inverse trigonometric functions to evaluate ex-

pressions.

Finding the Inverse by Mapping

Determining an inverse function algebraically can be both involved and difficult, so it is useful to know

how to map f to f −1. The graph of f can be used to produce the graph of f −1 by applying the inverse

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reflection principle:

The points (a, b) and (b, a) in the coordinate plane are symmetric with respect to the line y = x.

The points (a, b) and (b, a) are reflections of each other across the line y = x.

Example 1: Find the inverse of f (x) = 1x−5 mapping.

Solution: From the last section, we know that the inverse of this function is y = 5x+1x . To find the inverse

by mapping, pick several points on f (x), reflect them using the reflection principle and plot.

A: (4, -1)

B: (4.8, -5)

C: (2, -0.3)

D: (0, -0.2)

E: (5.3, 4)

F: (6, 1)

G: (8, 0.3)

H: (11, 0.2)

Now, take these eight points, switch the x and y and plot (y, x). Connect them to make the inverse function.

A−1 : (−1, 4)

B−1 : (−5, 4.8)

C−1 : (−0.3, 2)

D−1 : (−0.2, 0)

E−1 : (4, 5.3)

F−1 : (1, 6)

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G−1 : (0.3, 8)

H−1 : (0.2, 11)

Not all functions have inverses that are one-to-one. However, the inverse can be modified to a one-to-one

function if a “restricted domain” is applied to the inverse function.

Example 2: Find the inverse of f (x) = x2 − 4.

Solution: Let’s use the graphic approach for this one. The function is graphed in blue and its inverse is

red.

Clearly, the inverse relation is not a function because it does not pass the vertical line test. This is because

all parabolas fail the horizontal line test. To “make” the inverse a function, we restrict the domain of the

original function. For parabolas, this is fairly simple. To find the inverse of this function algebraically, we

get f −1(x) = √x+ 4. Technically, however, the inverse is ±√x+ 4 because the square root of any number

could be positive or negative. So, the inverse of f (x) = x2 − 4 is both parts of the square root equation,√

x+ 4 and −√x+ 4. √x+ 4 will yield the top portion of the horizontal parabola and −√x+ 4 will yield

the bottom half. Be careful, because if you just graph f −1(x) = √x+ 4 in your graphing calculator, it will

only graph the top portion of the inverse.

This technique of sectioning the inverse is applied to finding the inverse of trigonometric functions because

it is periodic.

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Finding the Inverse of the Trigonometric Functions

In order to consider the inverse of this function, we need to restrict the domain so that we have a section

of the graph that is one-to-one. If the domain of f is restricted to −pi2 ≤ x ≤ pi2 a new function f (x) =

sin x,−pi2 ≤ x ≤ pi2 . is defined. This new function is one-to-one and takes on all the values that the function

f (x) = sin x takes on. Since the restricted domain is smaller, f (x) = sin x,−pi2 ≤ x ≤ pi2 takes on all values

once and only once.

In the previous lesson the inverse of f (x) was represented by the symbol f −1(x), and y = f −1(x)⇔ f (y) = x.

The inverse of sin x,−pi2 ≤ x ≤ pi2 will be written as sin−1 x. or arcsin x.

y = sin−1 x

or

y = arcsin x

⇔ sin y = x

In this lesson we will use both sin−1 x and arcsin x and both are read as “the inverse sine of x” or “the

number between −pi2 and pi2 whose sine is x.”

The graph of y = sin−1 x is obtained by applying the inverse reflection principle and reflecting the graph

of y = sin x,−pi2 ≤ x ≤ pi2 in the line y = x. The domain of y = sin x becomes the range of y = sin−1 x, and

hence the range of y = sin x becomes the domain of y = sin−1 x.

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Another way to view these graphs is to construct them on separate grids. If the domain of y = sin x is

restricted to the interval

[

−pi2 , pi2

]

, the result is a restricted one-to one function. The inverse sine function

y = sin−1 x is the inverse of the restricted section of the sine function.

The domain of y = sin x is

[

−pi2 , pi2

]

and the range is [-1, 1].

The restriction of y = sin x is a one-to-one function and it has an inverse that is shown below.

The domain of y = sin−1 is [-1, 1] and the range is

[

−pi2 , pi2

]

.

The inverse functions for cosine and tangent are defined by following the same process as was applied for

the inverse sine function. However, in order to create one-to-one functions, different intervals are used. The

cosine function is restricted to the interval 0 ≤ x ≤ pi and the new function becomes y = cos x, 0 ≤ x ≤ pi.

The inverse reflection principle is then applied to this graph as it is reflected in the line y = x The result

is the graph of y = cos−1 x (also expressed as y = arccos x).

Again, construct these graphs on separate grids to determine the domain and range. If the domain of

y = cos x is restricted to the interval [0, pi], the result is a restricted one-to one function. The inverse cosine

function y = cos−1 x is the inverse of the restricted section of the cosine function.

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The domain of y = cos x is [0, pi] and the range is [-1, 1].

The restriction of y = cos x is a one-to-one function and it has an inverse that is shown below.

The statements y = cos x and x = cos y are equivalent for y−values in the restricted domain [0, pi]

and x−values between -1 and 1.

The domain of y = cos−1 x is [-1, 1] and the range is [0, pi].

The tangent function is restricted to the interval −pi2 < x < pi2 and the new function becomes y = tan x,−pi2 <

x < pi2 . The inverse reflection principle is then applied to this graph as it is reflected in the line y = x. The

result is the graph of y = tan−1 x (also expressed as y = arctan x).

Graphing the two functions separately will help us to determine the domain and range. If the domain of

y = tan x is restricted to the interval

[

−pi2 , pi2

]

, the result is a restricted one-to one function. The inverse

tangent function y = tan−1 x is the inverse of the restricted section of the tangent function.

239 www.ck12.org

The domain of y = tan x is

[

−pi2 , pi2

]

and the range is [−∞,∞].

The restriction of y = tan x is a one-to-one function and it has an inverse that is shown below.

The statements y = tan x and x = tan y are equivalent for y−values in the restricted domain[

−pi2 , pi2

]

and x−values between -4 and +4.

The domain of y = tan−1 x is [−∞,∞] and the range is

[

−pi2 , pi2

]

.

The above information can be readily used to evaluate inverse trigonometric functions without the use of

a calculator. These calculations are done by applying the restricted domain functions to the unit circle.

To summarize:

Table 4.1:

Restricted Do-

main Function

Inverse Trigono-

metric Function

Domain Range Quadrants

y = sin x

[

−pi2 , pi2

]

[-1, 1] 1 AND 4

y = arcsin x

y = sin−1 x

[-1, 1]

[

−pi2 , pi2

]

y = cos x [0, pi] [-1, 1] 1 AND 2

y = arccos x

y = cos−1 x

[-1, 1] [0, pi]

y = tan x

(

−pi2 , pi2

)

(−∞,∞) 1 AND 4

y = arctan x

y = tan−1 x

(−∞,∞)

(

−pi2 , pi2

)

Now that the three trigonometric functions and their inverses have been summarized, let’s take a look at

the graphs of these inverse trigonometric functions.

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241 www.ck12.org

Points to Consider

• What are the restricted domains for the inverse relations of the trigonometric functions?

• Can the values of the special angles of the unit circle be applied to the inverse trigonometric functions?

Review Questions

Study each of the following graphs and answer these questions:

(a) Is the graphed relation a function?

(b) Does the relation have an inverse that is a function?

1.

2.

www.ck12.org 242

3.

Find the inverse of the following functions using the mapping principle.

4. f (x) = x2 + 2x − 15

5. y = 1 + 2 sin x

6. Sketch a graph of y = 12 cos−1(3x+ 1). Sketch y = cos−1 x on the same set of axes and compare how

the two differ.

7. Sketch a graph of y = 3 − tan−1(x − 2). Sketch y = tan−1 x on the same set of axes and compare how

the two differ.

8. Graph y = 2 sin−1(2x − 1) + 1

9. Graph y = 4 + cos−1 13 x

10. Remember that sine and cosine are out of phase with each other, sin x = cos

(

x − pi2

)

. Find the inverse

of y = cos

(

x − pi2

)

. Is the inverse of y =

(

cos x − pi2

)

the same as y = sin−1 x? Why or why not?

Review Answers

1. The graph represents a one-to-one function. It passes both a vertical and a horizontal line test. The

inverse would be a function.

2. The graph represents a function, but is not one-to-one because it does not pass the horizontal line

test. Therefore, it does not have an inverse that is a function.

3. The graph does not represent a one-to-one function. It fails a vertical line test. However, its inverse

would be a function.

4. By selecting 4-5 points and switching the x and y values, you will get the red graph below.

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5. By selecting 4-5 points and switching the x and y values, you will get the red graph below.

6. y = 12 cos−1(3x+ 1) is in blue and y = cos−1(x) is in red. Notice that y = 12 cos−1(3x+ 1) has half the

amplitude and is shifted over -1. The 3 seems to narrow the graph.

7. y = 3 − tan−1(x − 2) is in blue and y = tan−1 x is in red. y = 3 − tan−1(x − 2) is shifted up 3 and to

the right 2 (as indicated by point C, the “center”) and is flipped because of the − tan−1.

www.ck12.org 244

8.

9.

10.

y = cos

(

x − pi

2

)

x = cos

(

y − pi

2

)

cos−1 x = y − pi

2

pi

2

+ cos−1 x = y

sin−1 x , pi2 + cos−1 x, graphing the two equations will illustrate that the two are not the same.

This is because of the restricted domain on the inverses. Since the functions are periodic, there is a

phase shift of cosine that, when the inverse is found, is equal to sine inverse.

4.3 Inverse Trigonometric Properties

Learning Objectives

• Relate the concept of inverse functions to trigonometric functions.

• Reduce the composite function to an algebraic expression involving no trigonometric functions.

• Use the inverse reciprocal properties.

• Compose each of the six basic trigonometric functions with tan−1 x.

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Composing Trig Functions and their Inverses

In the Prerequisite Chapter, you learned that for a function f ( f −1(x)) = x for all values of x for which

f −1(x) is defined. If this property is applied to the trigonometric functions, the following equations will be

true whenever they are defined:

sin(sin−1(x)) = x cos(cos−1(x)) = x tan(tan−1(x)) = x

As well, you learned that f −1( f (x)) = x for all values of x for which f (x) is defined. If this property

is applied to the trigonometric functions, the following equations that deal with finding an inverse trig.

function of a trig. function, will only be true for values of x within the restricted domains.

sin−1(sin(x)) = x cos−1(cos(x)) = x tan−1(tan(x)) = x

These equations are better known as composite functions and are composed of one trigonometric function in

conjunction with another different trigonometric function. The composite functions will become algebraic

functions and will not display any trigonometry. Let’s investigate this phenomenon.

Example 1: Find sin

(

sin−1

√

2

2

)

.

Solution: We know that sin−1

√

2

2 =

pi

4 , within the defined restricted domain. Then, we need to find sin pi4 ,

which is

√

2

2 . So, the above properties allow for a short cut. sin

(

sin−1

√

2

2

)

=

√

2

2 , think of it like the sine

and sine inverse cancel each other out and all that is left is the

√

2

2 .

Composing Trigonometric Functions

Besides composing trig functions with their own inverses, you can also compose any trig functions with

any inverse. When solving these types of problems, start with the function that is composed inside of the

other and work your way out. Use the following examples as a guideline.

Example 2: Without using technology, find the exact value of each of the following:

a. cos

(

tan−1

√

3

)

b. tan

(

sin−1

(

−12

))

c. cos(tan−1(−1))

d. sin

(

cos−1

√

2

2

)

Solution: For all of these types of problems, the answer is restricted to the inverse functions’ ranges.

a. cos

(

tan−1

√

3

)

: First find tan−1

√

3, which is pi3 . Then find cos pi3 . Your final answer is 12 . Therefore,

cos

(

tan−1

√

3

)

= 12 .

b. tan

(

sin−1

(

−12

))

= tan

(

−pi6

)

= −

√

3

3

c. cos(tan−1(−1)) = cos−1

(

−pi4

)

=

√

2

2 .

d. sin

(

cos−1

√

2

2

)

= sin pi4 =

√

2

2

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Inverse Reciprocal Functions

We already know that the cosecant function is the reciprocal of the sine function. This will be used to

derive the reciprocal of the inverse sine function.

y = sin−1 x

x = sin y

1

x

= csc y

csc−1 1

x

= y

csc−1 1

x

= sin−1 x

Because cosecant and sect are inverses, sin−1 1x = csc−1 x is also true.

The inverse reciprocal identity for cosine and secant can be proven by using the same process as above.

However, remember that these inverse functions are defined by using restricted domains and the reciprocals

of these inverses must be defined with the intervals of domain and range on which the definitions are valid.

sec−1 1

x

= cos−1 x↔ cos−1 1

x

= sec−1 x

Tangent and cotangent have a slightly different relationship. Recall that the graph of cotangent differs

from tangent by a reflection over the y−axis and a shift of pi2 . As an equation, this can be written as

cot x = − tan

(

x − pi2

)

. Taking the inverse of this function will show the inverse reciprocal relationship

between arccotangent and arctangent.

y = − tan

(

x − pi

2

)

x = − tan

(

y − pi

2

)

−x = tan

(

y − pi

2

)

tan−1(−x) = y − pi

2

pi

2

+ tan−1(−x) = y

pi

2

− tan−1 x = y

Remember that tangent is an odd function, so that tan(−x) = − tan(x). Because tangent is odd, its inverse

is also odd. So, this tells us that cot−1 x = pi2 − tan−1 x and tan−1 x = pi2 − cot−1 x. You will determine

the domain and range of all of these functions when you graph them in the exercises for this section. To

graph arcsecant, arccosecant, and arccotangent in your calculator you will use these conversion identities:

sec−1 x = cos−1 1x , csc−1 x = sin−1 1x , cot−1 x = pi2 − tan−1 x.

Now, let’s apply these identities to some problems that will give us an insight into how they work.

Example 3: Evaluate sec−1

√

2

Solution: Use the inverse reciprocal property. sec−1 x = cos−1 1x → sec−1

√

2 = cos−1 1√

2

. Recall that

1√

2

= 1√

2

·

√

2√

2

=

√

2

2 . So, sec−1

√

2 = cos−1

√

2

2 , and we know that cos−1

√

2

2 =

pi

4 . Therefore, sec−1

√

2 = pi4 .

Example 4: Find the exact value of each expression within the restricted domain, without a calculator.

247 www.ck12.org

a. sec−1

√

2

b. cot−1

(

−√3

)

c. csc−1 2

√

3

3

Solution: For each of these problems, first find the reciprocal and then determine the angle from that.

a. sec−1

√

2 = cos−1

√

2

2 From the unit circle, we know that the answer is pi4 .

b. cot−1

(

−√3

)

= pi2 − tan−1

(

−√3

)

From the unit circle, the answer is 5pi6 .

c. csc−1 2

√

3

3 = sin−1

√

3

2 Within our interval, there are is one answer, pi3 .

Example 5: Using technology, find the value in radian measure, of each of the following:

a. arcsin 0.6384

b. arccos(−0.8126)

c. arctan(−1.9249)

Solution:

a.

b.

c.

Make sure that your calculator’s MODE is RAD (radians).

Composing Inverse Reciprocal Trig Functions

In this subsection, we will combine what was learned in the previous two sections. Here are a few examples:

Example 6: Without a calculator, find cos

(

cot−1

√

3

)

.

Solution: First, find cot−1

√

3, which is also tan−1

√

3

3 . This is pi6 . Now, find cos pi6 , which is

√

3

2 . So, our

answer is

√

3

2 .

Example 7: Without a calculator, find sec−1

(

csc pi3

)

.

Solution: First, csc pi3 = 1sin pi3 =

1√

3

2

= 2√

3

= 2

√

3

3 . Then sec−1 2

√

3

3 = cos−1

√

3

2 =

pi

6 .

Example 8: Evaluate cos

(

sin−1 35

)

.

Solution: Even though this problem is not a critical value, it can still be done without a calculator. Recall

that sine is the opposite side over the hypotenuse of a triangle. So, 3 is the opposite side and 5 is the

hypotenuse. This is a Pythagorean Triple, and thus, the adjacent side is 4. To continue, let θ = sin−1 35 or

sin θ = 35 , which means θ is in the Quadrant 1 (from our restricted domain, it cannot also be in Quadrant

II). Substituting in θ we get cos

(

sin−1 35

)

= cos θ and cos θ = 45 .

Example 9: Evaluate tan

(

sin−1

(

−34

))

Solution: Even though 34 does not represent two lengths from a Pythagorean Triple, you can still use the

Pythagorean Theorem to find the missing side. (−3)2+b2 = 42, so b = √16 − 9 = √7. From the restricted

domain, sine inverse is negative in the 4th Quadrant. To illustrate:

www.ck12.org 248

Let

θ = sin−1

(

−3

4

)

sin θ = −3

4

tan

(

sin−1

(

−3

4

))

= tan θ

tan θ = −3√

7

or −3

√

7

7

Trigonometry in Terms of Algebra

All of the trigonometric functions can be rewritten in terms of only x, when using one of the inverse

trigonometric functions. Starting with tangent, we draw a triangle where the opposite side (from θ)

is defined as x and the adjacent side is 1. The hypotenuse, from the Pythagorean Theorem would be√

x2 + 1. Substituting tan−1 x for θ, we get:

tan θ = x

1

tan θ = x hypotenuse =

√

x2 + 1

θ = tan−1 x

sin(tan−1 x) = sin θ = x√

x2 + 1

csc(tan−1 x) = csc θ =

√

x2 + 1

x

cos(tan−1 x) = cos θ = 1√

x2 + 1

sec(tan−1 x) = sec θ =

√

x2 + 1

tan(tan−1 x) = tan θ = x cot(tan−1 x) = cot θ = 1

x

Example 10: Find sin(tan−1 3x).

Solution: Instead of using x in the ratios above, use 3x.

249 www.ck12.org

sin(tan−1 3x) = sin θ = 3x√

(3x)2 + 1

=

3x√

9x2 + 1

Example 11: Find sec2(tan−1 x).

Solution: This problem might be better written as [sec(tan−1 x)]2. Therefore, all you need to do is square

the ratio above.

[sec(tan−1 x)]2 =

( √

x2 + 1

)2

= x2 + 1

You can also write the all of the trig functions in terms of arcsine and arccosine. However, for each inverse

function, there is a different triangle. You will derive these formulas in the exercise for this section.

Points to Consider

• Is it possible to graph these composite functions? What happens when you graph y = sin(cos−1 x) in

your calculator?

• Do exact values of functions of inverse functions exist if any value is used?

Review Questions

1. Evaluate each of the following:

(a) cos−1

√

3

2

(b) sec−1

√

2

(c) sec−1

(

−√2

)

(d) sec−1(−2)

(e) cot−1(−1)

(f) csc−1

(√

2

)

2. Use your calculator to find:

(a) arccos(−0.923)

(b) arcsin 0.368

(c) arctan 5.698

3. Find the exact value of the functions, without a calculator, over their restricted domains.

(a) csc

(

cos−1

√

3

2

)

(b) sec−1(tan(cot−1 1))

(c) tan−1

(

cos pi2

)

(d) cot

(

sec−1 2

√

3

3

)

4. Using your graphing calculator, graph y = sec−1 x. Sketch this graph, determine the domain and

range, x− and/or y−intercepts. (Your calculator knows the restriction on this function, there is no

need to input it into Y =.)

5. Using your graphing calculator, graph y = csc−1 x. Sketch this graph, determine the domain and

range, x− and/or y−intercepts. (Your calculator knows the restriction on this function, there is no

need to input it into Y =.)

www.ck12.org 250

6. Using your graphing calculator, graph y = cot−1 x. Sketch this graph, determine the domain and

range, x− and/or y−intercepts. (Your calculator knows the restriction on this function, there is no

need to input it into Y =.)

7. Evaluate:

(a) sin

(

cos−1 513

)

(b) tan

(

sin−1

(

− 611

))

(c) cos

(

csc−1 257

)

8. Express each of the following functions as an algebraic expression involving no trigonometric func-

tions.

(a) cos2(tan−1 x)

(b) cot(tan−1 x2)

9. To find trigonometric functions in terms of sine inverse, use the following triangle.

(a) Determine the sine, cosine and tangent in terms of arcsine.

(b) Find tan(sin−1 2x3).

10. To find the trigonometric functions in terms of cosine inverse, use the following triangle.

(a) Determine the sine, cosine and tangent in terms of arccosine.

(b) Find sin2

(

cos−1 12 x

)

.

Review Answers

1. (a) pi6

(b) pi4

(c) 5pi4

(d) 2pi3 , 4pi3

(e) 3pi4 , 7pi4

(f) pi4 , 3pi4

2. (a) 2.747

(b) 0.377

(c) 1.397

3. (a) csc

(

cos−1

√

3

2

)

= csc pi6 = 2

(b) sec−1(tan(cot−1 1)) = sec−1

(

tan pi4

)

= sec−1 1 = 0

251 www.ck12.org

(c) tan−1

(

cos pi2

)

= tan−1 0 = 0

(d) cot

(

sec−1 2

√

3

3

)

= cot

(

cos−1

√

3

2

)

= cot pi6 = 1tan pi6 =

1√

3

3

=

√

3

4. y = sec−1 x when plugged into your graphing calculator is y = cos−1 1x .

The domain is all reals, excluding the interval (-1, 1). The range is all reals in the interval [0, pi], y , pi2 .

There are no y intercepts and the only x intercept is at 1.

5. y = csc−1 x when plugged into your graphing calculator is y = sin−1 1x .

The domain is all reals, excluding the interval (-1, 1). The range is all reals in the interval [−pi2 , pi2 ], y , 0.

There are no x or y intercepts.

6. The domain is all real numbers and the range is from (0, pi). There is an x−intercept at pi2 .

www.ck12.org 252

7. (a)

cos θ = 5

13

sin

(

cos−1

( 5

13

))

= sin θ

sin θ = 12

13

(b) tan

(

sin−1

(

− 611

))

→ sin θ = − 611 . The third side is b =

√

121 − 36 = √85. tan θ = − 6√

85

= −6

√

85

85

(c) cos

(

csc−1

(

25

7

))

→ csc θ = 257 → sin θ = 725 . This two lengths of a Pythagorean Triple, with the

third side being 24. cos θ = 2425

8. (a) 1x2+1

(b) 1x2

9. The adjacent side to θ is

√

1 − x2, so the three trig functions are:(a)

sin(sin−1 x) = sin θ = x

cos(sin−1 x) = cos θ =

√

1 − x2

tan(sin−1 x) = tan θ = x√

1 − x2(b)

tan(sin−1(2x3)) = 2x

3√

1 − (2x3)2 =

2x3√

1 − 8x6

10. The opposite side to θ is

√

1 − x2, so the three trig functions are:(a)

sin(cos−1 x) = sin θ =

√

1 − x2

cos(cos−1 x) = cos θ = x

tan(cos−1 x) = tan θ =

√

1 − x2

x(b)

sin2

(

cos−1

(1

2

x

))

=

√

1 −

(1

2

x

)2

2

= 1 − 1

4

x2

253 www.ck12.org

4.4 Applications & Models

Learning Objectives

• Apply inverse trigonometric functions to real life situations.

The following problems are real-world problems that can be solved using the trigonometric functions. In

everyday life, indirect measurement is used to obtain answers to problems that are impossible to solve

using measurement tools. However, mathematics will come to the rescue in the form of trigonometry to

calculate these unknown measurements.

Example 1: On a cold winter day the sun streams through your living room window and causes a warm,

toasty atmosphere. This is due to the angle of inclination of the sun which directly affects the heating and

the cooling of buildings. Noon is when the sun is at its maximum height in the sky and at this time, the

angle is greater in the summer than in the winter. Because of this, buildings are constructed such that the

overhang of the roof can act as an awning to shade the windows for cooling in the summer and yet allow

the sun’s rays to provide heat in the winter. In addition to the construction of the building, the angle of

inclination of the sun varies according to the latitude of the building’s location.

If the latitude of the location is known, then the following formula can be used to calculate the angle of

inclination of the sun on any given date of the year:

Angle of sun = 90◦ − latitude+ −23.5◦ · cos

[

(N + 10)360365

]

where N represents the number of the day of the

year that corresponds to the date of the year. Note: This formula is accurate to ±1◦2

a. Determine the measurement of the sun’s angle of inclination for a building located at a latitude of 42◦,

March 10th, the 69th day of the year.

Solution:

Angle of sun = 90◦ − 42◦ + −23.5◦ · cos

[

(69 + 10)

360

365

]

Angle of sun = 48◦ + −23.5◦(0.2093)

Angle of sun = 48◦ − 4.92◦

Angle of sun = 43.08◦

b. Determine the measurement of the sun’s angle of inclination for a building located at a latitude of 20◦,

September 21st.

Solution:

Angle of sun = 90◦ − 20◦ + −23.5◦ · cos

[

(264 + 10)

360

365

]

Angle of sun = 70◦ + −23.5◦(0.0043)

Angle of sun = 70.10◦

Example 2: A tower, 28.4 feet high, must be secured with a guy wire anchored 5 feet from the base of

the tower. What angle will the guy wire make with the ground?

Solution: Draw a picture.

www.ck12.org 254

tan θ = opp.

ad j.

tan θ = 28.4

5

tan θ = 5.68

tan−1(tan θ) = tan−1(5.68)

θ = 80.02◦

The following problem that involves functions and their inverses will be solved using the property f ( f −1(x)) =

f −1( f (x)). In addition, technology will also be used to complete the solution.

Example 3: In the main concourse of the local arena, there are several viewing screens that are available

to watch so that you do not miss any of the action on the ice. The bottom of one screen is 3 feet above

eye level and the screen itself is 7 feet high. The angle of vision (inclination) is formed by looking at both

the bottom and top of the screen.

a. Sketch a picture to represent this problem.

b. Calculate the measure of the angle of vision that results from looking at the bottom and then the top

of the screen. At what distance from the screen does this value of the angle occur?

Solution:

a.

b.

255 www.ck12.org

θ2 = tan θ − tan θ1

tan θ = 10

x

and tan θ1 =

3

x

θ2 = tan−1

(10

x

)

− tan−1

(3

x

)

To determine these values, use a graphing calculator and the trace function to determine when the actual

maximum occurs.

From the graph, it can be seen that the maximum occurs when x ≈ 5.59 f t. and θ ≈ 32.57◦.

Example 4: A silo is 40 feet high and 12 feet across. Find the angle of depression from the top edge of

the silo to the floor of the opposite edge.

Solution: tan θ = 4012 → θ = tan−1 4012 = 73.3◦

Example 5: The pilot of an airplane flying at an elevation of 5000 feet sights two towers that are 300 feet

apart. If the angle of depression to the tower closer to him is 30◦, determine the angle of depression to the

second tower.

Solution: Draw a picture. First we need to find x in order to find y.

www.ck12.org 256

tan 30◦ = x

5000

→ x = 5000 tan 30◦, x = 2886.75

tan y = 3186.75

5000

y = tan−1 3186.75

5000

y = 32.51◦

Which means that the two towers are 2.51◦ apart.

Review Questions

1. The intensity of a certain type of polarized light is given by the equation I = I0 sin 2θ cos 2θ. Solve

for θ.

2. The following diagram represents the ends of a water-trough. The ends are actually isosceles trape-

zoids. Determine the maximum value of the trough and the value of θ that maximizes the volume.

3. A boat is docked at the end of a 10 foot pier. The boat leaves the pier and drops anchor 230 feet

away 3 feet straight out from shore (which is perpendicular to the pier). What was the bearing of

the boat?

4. The electric current in a certain circuit is given by i = Im[sin(wt + α) cosϕ+ cos(wt + α) sinϕ] Solve

for t.

5. Using the formula from Example 1, determine the measurement of the sun’s angle of inclination for

a building located at a latitude of:

(a) 64◦ on the 16th of November

(b) 15◦ on the 8th of August

6. A ship leaves port and travels due east 15 nautical miles, then changes course to N 20◦ W and travels

40 more nautical miles. Find the bearing to the port of departure.

7. Find the maximum displacement for the simple harmonic motion described by d = 4 cos pit.

8. The pilot of an airplane flying at an elelvation of 10,000 feet sights two towers that are 500 feet apart.

If the angle of depression to the tower closer to him is 18◦, determine the anglf of depression to the

second tower.

257 www.ck12.org

Review Answers

1.

I = I0 sin 2θ cos 2θ

I

I0

=

I0

I0

sin 2θ cos 2θ

I

I0

= sin 2θ cos 2θ

2I

I0

= 2 sin 2θ cos 2θ

2I

I0

= sin 4θ

sin−1 2I

I0

= 4θ

1

4

sin−1 2I

I0

= θ

2. The volume is 10 feet times the area of the end. The end consists of two congruent right triangles and

one rectangle. The area of each right triangle is 12(sin θ)(cos θ) and that of the rectangle is (1)(cos θ).

The maximum value is approximately 13 cubic feet and occurs when θ = pi6 .

3.

cos x = 7

230

→ x = cos−1 7

230

x = 88.26◦

4.

i = Im[sin(wt + α) cosϕ+ cos(wt + α) sinϕ]

i

Im

= sin(wt + α) cosϕ+ cos(wt + α) sinϕ︸ ︷︷ ︸

sin(wt+α+ϕ)

i

Im

= sin(wt + α+ ϕ)

sin−1 i

Im

= wt + α+ ϕ

sin−1 i

Im

− α − ϕ = wt

1

w

(

sin−1 i

Im

− α − ϕ

)

= t

5. (a) 64◦ on the 16th of November = 90◦ − 64◦ − 23.5◦ cos

[

(320 + 10)360365

]

= 6.64◦

(b) 15◦ on the 8th of August = 90◦ − 15◦ − 23.5◦ cos

[

(220 + 10)360365

]

= 91.07◦

6. We need to find y and z before we can find x◦.

www.ck12.org 258

sin 70◦ = y

40

→ y = 40 sin 70◦ = 37.59

cos 70◦ = z

40

→ z = 40 cos 70◦ = 13.68

Using 15-13.68 as the adjacent side for x, we can now find the missing angle. tan x◦ = 37.591.32 = 28.48→

x◦ = tan−1(28.48) = 87.99◦.

7. The maximum displacement for this equation is simply the amplitude, 4.

8. You can use the same picture from Example 5 for this problem.

tan 18◦ = x

10, 000

→ x = 10, 000 tan 18◦x = 3249.2

tan y = 3749.2

10, 000

→ y = tan−1 3749.2

10, 000

y = 20.6◦

So, the towers are 2.6◦ apart.

4.5 Chapter Review

Chapter Summary

In this chapter, we studied all aspects of inverse trigonometric functions. First, we defined the function by

finding inverses algebraically. Second, we analyzed the graphs of inverse functions. We needed to restrict

the domain of the trigonometric functions in order to take the inverse of each of them. This is because

they are periodic and did not pass the horizontal line test. Then, we learned about the properties of the

inverse functions, mostly composing a trig function and an inverse. Finally, we applied the principles of

inverse trig functions to real-life situations.

Chapter Vocabulary

Arccosecant Read “cosecant inverse” and also written csc−1. The domain of this function is all reals,

excluding the interval (-1, 1). The range is all reals in the interval [−pi2 , pi2 ], y , 0.

Arccosine Read “cosine inverse” and also written cos−1. The domain of this function is [-1, 1]. The

range is [0, pi].

Arccotangent Read “cotangent inverse” and also written cot−1. The domain of this function is all reals.

The range is (0, pi).

259 www.ck12.org

Arcsecant Read “secant inverse” and also written sec−1. The domain of this function is all reals, exclud-

ing the interval (-1, 1). The range is all reals in the interval [0, pi], y , pi2 .

Arcsine Read “sine inverse” and also written sin−1. The domain of this function is [-1, 1]. The range is

[−pi2 , pi2 ].

Arctangent Read “tangent inverse” and also written tan−1. The domain of this function is all reals. The

range is (−pi2 , pi2).

Composite Function The final result from when one function is plugged into another, f (g(x)).

Harmonic Motion A motion that is consistent and periodic, in a sinusoidal pattern. The general

equation is x(t) = A cos(2pi f t + ϕ) where A is the amplitude, f is the frequency, and ϕ is the phase

shift.

Horizontal Line Test The test applied to a function to see if it has an inverse. Continually draw

horizontal lines across the function and if a horizontal line touches the function more than once, it

does not have an inverse.

Inverse Function Two functions that are symmetric over the line y = x.

Inverse Reflection Principle The points (a, b) and (b, a) in the coordinate plane are symmetric with

respect to the line y = x. The points (a, b) and (b, a) are reflections of each other across the line

y = x.

Invertible If a function has an inverse, it is invertible.

One-to-One Function A function, where, for every x value, there is EXACTLY one y−value. These are

the only invertible functions.

Review Questions

1. Find the exact value of the following expressions:

(a) csc−1(−2)

(b) cos−1

√

3

2

(c) cot−1

(

−

√

3

3

)

(d) sec−1

(

−√2

)

(e) arcsin 0

(f) arctan 1

2. Use your calculator to find the value of each of the following expressions:

(a) arccos 35

(b) csc−1 2.25

(c) tan−1 8

(d) arcsin(−0.98)

(e) cot−1

(

− 940

)

(f) sec−1 65

www.ck12.org 260

3. Find the exact value of the following expressions:

(a) cos

(

sin−1

√

2

2

)

(b) tan(cot−1 1)

(c) csc

(

sec−1 2

√

3

3

)

(d) sin

(

arccos 1213

)

(e) tan

(

arcsin 57

)

(f) sec−1

(

csc pi6

)

4. Find the inverse of each of the following:

(a) f (x) = 5 + cos(2x − 1)

(b) g(x) = −4 sin−1(x+ 3)

5. Sketch a graph of each of the following:

(a) y = 3 − arcsin

(

1

2 x+ 1

)

(b) f (x) = 2 tan−1(3x − 4)

(c) h(x) = sec−1(x − 1) + 2

(d) y = 1 + 2 arccos 2x

6. Using the triangles from Section 4.3, find the following:

(a) sin(cos−1 x3)

(b) tan2

(

sin−1 x23

)

(c) cos4(arctan(2x)2)

7. A ship leaves port and travels due west 20 nautical miles, then changes course to S 40◦E and travels

65 more nautical miles. Find the bearing to the port of departure.

8. Using the formula from Example 1 in Section 4.4, determine the measurement of the sun’s angle of

inclination for a building located at a latitude of 36◦ on the 12th of May.

9. Find the inverse of sin(x ± y) = sin x cos y ± cos x sin y. HINT: Set a = sin x and b = sin y and rewrite

cos x and cos y in terms of sine.

10. Find the inverse of cos(x ± y) = cos x cos y ∓ sin x sin y. HINT: Set a = cos x and b = cos y and rewrite

sin x and sin y in terms of sine.

Review Answers

1. (a) −pi6

(b) pi6

(c) −pi3

(d) 3pi4

(e) 0

(f) pi4

2. (a) 0.927

(b) 0.461

(c) 1.446

(d) -1.37

(e) 1.792

(f) 0.586

3. (a)

√

2

2

(b) 1

(c) 2

261 www.ck12.org

(d) 513

(e) 5

2

√

6

or 5

√

6

12

(f) pi34. (a)

f (x) = 5 + cos(2x − 1)

y = 5 + cos(2x − 1)

x = 5 + cos(2y − 1)

x − 5 = cos(2y − 1)

cos−1(x − 5) = 2y − 1

1 + cos−1(x − 5) = 2y

1 + cos−1(x − 5)

2

= y

(b)

g(x) = −4 sin−1(x+ 3)

y = −4 sin−1(x+ 3)

x = −4 sin−1(y+ 3)

− x

4

= sin−1(y+ 3)

sin

(

− x

4

)

= y+ 3

sin

(

− x

4

)

− 3 = y

5. (a)

(b)

www.ck12.org 262

(c)

(d)

6. (a) sin(cos−1 x3) =

√

1 − (x3)2 = √1 − x6

(b) tan2

(

sin−1 x23

)

=

x23√

1−

(

x2

3

)2

2

=

x4

9

1−

(

x4

9

) = x4

9

(

1− x49

) = x4

9−x4

(c) cos4(arctan(2x)2) = cos4(tan−1 4x2) =

(

1√

(4x2)2+1

)4

= 1√

16x4+14

= 1

(16x4+1)2

7. x◦ is our final answer, but we need to find y and z first.

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sin 40◦ = y

65

→ y = 65 sin 40◦ = 41.78

cos 40◦ = 20 + z

65

→ 20 + z = 65 cos 40◦

20 + z = 49.79→ z = 29.79

tan x = 41.78

29.79

→ x = tan−1 41.78

29.79

x = 54.51◦

8. 36◦ on the 12th of May = 90◦ − 36◦ − 23.5◦ cos

[

(132 + 10)360365

]

= 72.02◦

9. sin(x ± y) = sin x cos y ± cos x sin y, a = sin x and b = sin y→ x = sin−1 a and y = sin−1 b

sin(x ± y) = a

√

1 − sin2 y ± b

√

1 − sin2 x

sin(x ± y) = a

√

1 − b2 ± b

√

1 − a2

x ± y = sin−1

(

a

√

1 − b2 ± b

√

1 − a2

)

sin−1 a ± sin−1 b = sin−1

(

a

√

1 − b2 ± b

√

1 − a2

)

10. cos(x ± y) = cos x cos y ∓ sin x sin y, a = cos x and b = cos y→ x = cos−1 a and y = cos−1 b

cos(x ± y) = ab ∓ b

√

(1 − cos2 x)(1 − cos2 y)

cos(x ± y) = ab ∓

√

(1 − a2)(1 − b2)

x ± y = cos−1

(

ab ∓

√

(1 − a2)(1 − b2)

)

cos−1 a ± cos−1 b = cos−1

(

ab ∓

√

(1 − a2)(1 − b2)

)

Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator

activities designed to supplement the objectives for some of the lessons in this chapter. See

http://www.ck12.org/flexr/chapter/9702.

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Chapter 5

Triangles and Vectors

5.1 The Law of Cosines

Introduction

This chapter takes concepts that had only been applied to right triangles and interprets them so that they

can be used for any type of triangle. First, the laws of sines and cosines take the Pythagorean Theorem

and ratios and apply them to any triangle. The second half of the chapter introduces and manipulates

vectors. Vectors can be added, subtracted, multiplied and divided.

Learning Objectives

• Understand how the Law of Cosines is derived.

• Apply the Law of Cosines when you know two sides and the included angle of an oblique (non-right)

triangle (SAS).

• Apply the Law of Cosines when you know all three sides of an oblique triangle.

• Identify accurate drawings of oblique triangles.

• Use the Law of Cosines in real-world and applied problems.

Derive the Law of Cosines

4ABC contains an altitude BD that extends from B and intersects AC. We will refer to the length of BD

as y. The sides of 4ABC measure a units, b units, and c units. If DC is x units long, then AD measures

(b − x) units.

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Using the Pythagorean Theorem we know that:

c2 = y2 + (b − x)2 Pythagorean Theorem

c2 = y2 + b2 − 2bx+ x2 Expand (b − x)2

c2 = a2 + b2 − 2bx a2 = y2 + x2 by Pythagorean Theorem

c2 = a2 + b2 − 2b(a cosC) cosC = x

a

, so a cosC = x (cross multiply)

c2 = a2 + b2 − 2ab cosC Simplify

We can use a similar process to derive all three forms of the Law of Cosines:

a2 = b2 + c2 − 2bc cos A

b2 = a2 + c2 − 2ac cos B

c2 = a2 + b2 − 2ab cosC

Note that if either ∠A, ∠B or ∠C is 90◦ then cos 90◦ = 0 and the Law of Cosines is identical to the Pythagorean

Theorem.

The Law of Cosines is one tool we use in certain situations involving all triangles: right, obtuse, and acute.

It is a general statement relating the lengths of the sides of any general triangle to the cosine of one of its

angles. There are two situations in which we can and want to use the Law of Cosines:

1. When we know two sides and the included angle in an oblique triangle and want to find the third

side (SAS).

2. When we know all three sides in an oblique triangle and want to find one of the angles (SSS).

Case #1: Finding the Side of an Oblique Triangle

One case where we can use the Law of Cosines is when we know two sides and the included angle in a

triangle (SAS) and want to find the third side.

Example 1: Using 4DEF, ∠E = 12◦, d = 18, and f = 16.8. Find e.

Solution: Since 4DEF isn’t a right triangle, we cannot use the Pythagorean Theorem or trigonometry

functions to find the third side. However, we can use our newly derived Law of Cosines.

e2 = 182 + 16.82 − 2(18)(16.8) cos 12 Law of Cosines

e2 = 324 + 282.24 − 2(18)(16.8) cos 12 Simplify squares

e2 = 324 + 282.24 − 591.5836689 Multiply

e2 = 14.6563311 Add and subtract from left to right

e ≈ 3.8 Square root

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∗Note that the negative answer is thrown out as having no geometric meaning in this case.

Example 2: An architect is designing a kitchen for a client. When designing a kitchen, the architect must

pay special attention to the placement of the stove, sink, and refrigerator. In order for a kitchen to be

utilized effectively, these three amenities must form a triangle with each other. This is known as the “work

triangle.” By design, the three parts of the work triangle must be no less than 3 feet apart and no more

than 7 feet apart. Based on the dimensions of the current kitchen, the architect has determined that the

sink will be 3.6 feet away from the stove and 5.7 feet away from the refrigerator. If the sink forms a 103◦

angle with the stove and the refrigerator, will the distance between the stove and the refrigerator remain

within the confines of the work triangle?

Solution: In order to find the distance from the sink to the refrigerator, we need to find side x. To find

side x, we will use the Law of Cosines because we are dealing with an obtuse triangle (and thus have

no right angles to work with). We know the length two sides: the sink to the stove and the sink to the

refrigerator. We also know the included angle (the angle between the two known lengths) is 103◦. This

means we have the SAS case and can apply the Law of Cosines.

x2 = 3.62 + 5.72 − 2(3.6)(5.7) cos 103 Law of Cosines

x2 = 12.96 + 32.49 − 2(3.6)(5.7) cos 103 Simplify squares

x2 = 12.96 + 32.49 + 9.23199127 Multiply

x2 = 54.68199127 Evaluate

x ≈ 7.4 Square root

No, this does not confirm the work triangle. The sink and the refrigerator are too far apart by 0.4 feet.

Case #2: Finding any Angle of a Triangle

Another situation where we can apply the Law of Cosines is when we know all three sides in a triangle

(SSS) and we need to find one of the angles. The Law of Cosines allows us to find any of the three angles

in the triangle. First, we will look at how to apply the Law of Cosines in this case, and then we will look

at a real-world application.

Example 3: Continuing on from Example 2, if the architect moves the stove so that it is 4.2 feet from

the sink and makes the fridge 6.8 feet from the stove, how does this affect the angle the sink forms with

the stove and the refrigerator?

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Solution: In order to find how the angle is affected, we will again need to utilize the Law of Cosines, but

because we do not know the measures of any of the angles, we solve for Y.

6.82 = 4.22 + 5.72 − 2(4.2)(5.7) cos Y Law of Cosines

46.24 = 17.64 + 32.49 − 2(4.2)(5.7) cosY Simplify squares

46.24 = 17.64 + 32.49 − 47.88 cosY Multiply

46.24 = 50.13 − 47.88 cosY Add

−3.89 = −47.88 cos Y Subtract

0.0812447786 = cos Y Divide

85.3◦ ≈ Y cos−1 (0.081244786)

The new angle would be 85.3◦, which means it would be 17.7◦ less than the original angle.

Example 4: In oblique 4MNO,m = 45, n = 28, and o = 49. Find ∠M.

Solution: Since we know all three sides of the triangle, we can use the Law of Cosines to find ∠M.

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452 = 282 + 492 − 2(28)(49) cosM Law of Cosines

2025 = 784 + 2401 − 2(28)(49) cosM Simplify squares

2025 = 784 + 2401 − 2744 cosM Multiply

2025 = 3185 − 2744 cosM Add

−1160 = −2744 cosM Subtract 3185

0.422740525 = cosM Divide by − 2744

65◦ ≈ M cos−1 (0.422740525)

It is important to note that we could use the Law of Cosines to find ∠N or ∠O also.

Example 5: Sam is building a retaining wall for a garden that he plans on putting in the back corner

of his yard. Due to the placement of some trees, the dimensions of his wall need to be as follows: side

1 = 12 f t, side 2 = 18 f t, and side 3 = 22 f t. At what angle do side 1 and side 2 need to be? Side 2 and side

3? Side 1 and side 3?

Solution: Since we know the measures of all three sides of the retaining wall, we can use the Law of

Cosines to find the measures of the angles formed by adjacent walls. We will refer to the angle formed by

side 1 and side 2 as ∠A, the angle formed by side 2 and side 3 as ∠B, and the angle formed by side 1 and

side 3 as ∠C. First, we will find ∠A.

222 = 122 + 182 − 2(12)(18) cos A Law of Cosines

484 = 144 + 324 − 2(12)(18) cos A Simplify squares

484 = 144 + 324 − 432 cos A Multiply

484 = 468 − 432 cos A Add

16 = −432 cos A Subtract 468

−0.037037037 ≈ cos A Divide by − 432

92.1◦ ≈ A cos−1 (−0.037037037)

Next we will find the measure of ∠B also by using the Law of Cosines.

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182 = 122 + 222 − 2(12)(22) cos B Law of Cosines

324 = 144 + 484 − 2(12)(22) cos B Simplify squares

324 = 144 + 484 − 528 cos B Multiply

324 = 628 − 528 cos B Add

−304 = −528 cos B Subtract 628

0.575757576 = cos B Divide by − 528

54.8◦ ≈ B cos−1 (0.575757576)

Now that we know two of the angles, we can find the third angle using the Triangle Sum Theorem,

∠C = 180 − (92.1 + 54.8) = 33.1◦.

Identify Accurate Drawings of General Triangles

The Law of Cosines can also be used to verify that drawings of oblique triangles are accurate. In a right

triangle, we might use the Pythagorean Theorem to verify that all three sides are the correct length, or we

might use trigonometric ratios to verify an angle measurement. However, when dealing with an obtuse or

acute triangle, we must rely on the Law of Cosines.

Example 6: In 4ABC at the right, a = 32, b = 20, And c = 16. Is the drawing accurate if it labels ∠C as

35.2◦? If not, what should ∠C measure?

Solution: We will use the Law of Cosines to check whether or not ∠C is 35.2◦.

162 = 202 + 322 − 2(20)(32) cos 35.2 Law of Cosines

256 = 400 + 1024 − 2(20)(32) cos 35.2 Simply squares

256 = 400 + 1024 − 1045.94547 Multiply

256 , 378.05453 Add and subtract

Since 256 , 378.05453, we know that ∠C is not 35.2◦. Using the Law of Cosines, we can figure out the

correct measurement of ∠C.

162 = 202 + 322 − 2(20)(32) cosC Law of Cosines

256 = 400 + 1024 − 2(20)(32) cosC Simplify Squares

256 = 400 + 1024 − 1280 cosC Multiply

256 = 1424 − 1280 cosC Add

−1168 = −1280 cosC Subtract 1424

0.9125 = cosC Divide

24.1◦ ≈ ∠C cos−1(0.9125)

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For some situations, it will be necessary to utilize not only the Law of Cosines, but also the Pythagorean

Theorem and trigonometric ratios to verify that a triangle or quadrilateral has been drawn accurately.

Example 7: A builder received plans for the construction of a second-story addition on a house. The

diagram shows how the architect wants the roof framed, while the length of the house is 20 ft. The builder

decides to add a perpendicular support beam from the peak of the roof to the base. He estimates that new

beam should be 8.3 feet high, but he wants to double-check before he begins construction. Is the builder’s

estimate of 8.3 feet for the new beam correct? If not, how far off is he?

Solution: If we knew either ∠A or ∠C, we could use trigonometric ratios to find the height of the support

beam. However, neither of these angle measures are given to us. Since we know all three sides of 4ABC,

we can use the Law of Cosines to find one of these angles. We will find ∠A.

142 = 122 + 202 − 2(12)(20) cos A Law of Cosines

196 = 144 + 400 − 480 cos A Simplify

196 = 544 − 480 cos A Add

−348 = −480 cos A Subtract

0.725 = cos A Divide

43.5◦ ≈ ∠A cos−1(0.725)

Now that we know ∠A, we can use it to find the length of BD.

sin 43.5 = x

12

12 sin 43.5 = x

8.3 ≈ x

Yes, the builder’s estimate of 8.3 feet for the support beam is accurate.

Points to Consider

• How is the Pythagorean Theorem a special case of the Law of Cosines?

• In the SAS case, is it possible to use the Law of Cosines to find all missing sides and angles?

• In which cases can we not use the Law of Cosines? Explain.

• Give an example of three side lengths that do not form a triangle.

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Review Questions

1. Using each figure and the given information below, decide which side(s) or angle(s) you could find

using the Law of Cosines.

Table 5.1:

Given Information Figure What can you find?

a. ∠A = 50◦, b = 8, c = 11

b. t = 6, r = 7, i = 8

c. ∠L = 79.5◦,m = 22.4, p = 13.7

d. q = 17, d = 12.8, r =

18.6, ∠Q = 62.4◦

e. ∠B = 67.2◦, d = 43, e = 39

f. c = 9, d = 11,m = 13

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2. Using the figures and information from the chart above, use the Law of Cosines to find the following:

(a) side a

(b) the largest angle

(c) side l

(d) the smallest angle

(e) side b

(f) the second largest angle

3. In 4CIR, c = 63, i = 52, and r = 41.9. Find the measure of all three angles.

4. Find AD using the Pythagorean Theorem, Law of Cosines, trig functions, or any combination of the

three.

5. Find HK using the Pythagorean Theorem, Law of Cosines, trig functions, or any combination of the

three if JK = 3.6,KI = 5.2, JI = 1.9,HI = 6.7, and ∠KJI = 96.3◦.

6. Use the Law of Cosines to determine whether or not the following triangle is drawn accurately. If

not, determine how much side d is off by.

7. A businessman is traveling down Interstate 43 and has intermittent cell phone service. There is a

transmission tower near Interstate 43. The range of service from the tower forms a 47◦ angle and the

range of service is 26 miles to one section of I-43 and 31 miles to another point on I-43.

(a) If the businessman is traveling at a speed of 45 miles per hour, how long will he have service

for?

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(b) If he slows down to 35mph, how much longer will he be able to have service?

8. A dock is being built so that it is 183 yards away from one buoy and 306 yards away from a second

buoy. The two buoys are 194.1 yards apart.

(a) What angle does the dock form with the two buoys?

(b) If the second buoy is moved so that it is 329 yards away from the dock and 207 yards away from

the first buoy, how does this affect the angle formed by the dock and the two buoys?

9. A golfer hits the ball from the 18th tee. His shot is a 235 yard hook (curves to the left) 9◦ from the

path straight to the flag on the green.

(a) If the tee is 329 yards from the flag, how far is the ball away from the flag?

(b) If the golfer’s next shot is 98 yards and is hooked 3◦ from the path straight to the flag, how far

is ball away now?

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10. Given the numbers 127, 210 and 17 degrees write a problem that uses the Law of Cosines.

11. The sides of a triangle are 15, 27 and 39. What is its area?

12. A person inherits a triangular piece of land with dimensions 300 ft, 600ft, and 850 ft. What is the

area of the piece of land? How much of an acre is it?

Review Answers

1. (a) side a

(b) ∠T, ∠R, and ∠I

(c) side l

(d) ∠R and ∠D

(e) side b

(f) ∠C, ∠D, ∠M

2. (a) a2 = 82 + 112 − 2 · 8 · 11 · cos 50◦, a ≈ 8.5

(b) 112 = 62 + 72 − 2 · 6 · 7 · cos I, ∠I ≈ 115.4◦

(c) l2 = 22.42 + 13.172 − 2 · 22.4 · 13.17 · cos 79.5◦, l ≈ 23.7

(d) 12.82 = 172 + 18.62 − 2 · 17 · 18.6 · cosD, ∠D ≈ 41.8◦

(e) b2 = 392 + 432 − 2 · 39 · 43 · cos 67.2◦, b ≈ 45.5

(f) 112 = 92 + 133 − 2 · 9 · 13 · cosD, ∠D ≈ 56.5◦

3. 632 = 522 + 41.92 − 2 · 52 · 41.9 · cosC, 522 = 632 + 41.92 − 2 · 63 · 41.9 · cos I, 180◦ − 83.5◦ − 55.1◦ =

41.4◦, ∠C ≈ 83.5◦, ∠I ≈ 55.1◦, ∠R ≈ 41.4◦

4. First, find AB. AB2 = 14.22 + 152 − 2 · 14.2 · 15 · cos 37.4◦, AB = 9.4. sin 23.3◦ = AD9.4 , AD = 3.7.

5. ∠HJI = 180◦ − 96.3◦ = 83.7◦ (these two angles are a linear pair). HJ2 = 6.72 + 1.92 − 2 · 6.7 · 1.9 ·

cos 83.7◦,HJ = 6.8. So, HJ + JK = HK, 6.8 + 3.6 = HK = 10.4

6. To determine this, use the Law of Cosines and solve for d to determine if the picture is accurate.

d2 = 122 + 242 − 2 · 12 · 24 · cos 30◦, d = 14.9, which means d in the picture is off by 1.9.

7. (a) First, find x: x2 = 312 + 262 − 2 · 31 · 26 · cos 47◦, x = 23.2 miles. Dividing the miles by his speed

will tell us how long he will have service. 23.245 = 0.52 hr or 30.9 min.

(b) 23.235 = 0.66 hr or 39.8 min, so he will have service for 8.9 minutes longer.

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8. (a) 194.12 = 1832 + 3062 − 2 · 183 · 306 · cos a. The angle formed, a, is 37◦.

(b) 2072 = 1832 + 3292 − 2 · 183 · 329 · cos b. The new angle, b, will need to be 34.8◦ rather than 37◦

or 2.2◦ less.

9. (a) x2 = 2352 + 3292 − 2 · 235 · 329 · cos 9◦, making the ball 103.6 yards away from the flag.

(b)

x2 = 982 + 103.62 − 2 · 98 · 103.6 · cos 3◦, making his second shot 7.8 yards away from the flag.

10. Students answers will vary. The goal is to have each student create their own word problem.

11. We need to find the height in order to get the area.

272 = 152 + 392 − 2 · 15 · 39 · cos x, x = 29.6◦

sin 29.6◦ = h

15

→ h = 7.4

A =

1

2

· 39 · 7.4 = 27.5

12. Recall that the area of a trapezoid is A = 12h(b1 + b2). We need to find the angle

x, in order to find y and then h.

21002 = 24002 + 22002 − 2 · 2400 · 2200 · cos x, x = 54.1◦.

90◦ − 54.1◦ = 35.9◦ = y. sin 35.9◦ = h

2200

→ h = 1782.1.

A =

1

2

1782.1(2400 + 3000) = 4, 811, 370 sq. f t. or 110.5 acres.

5.2 Area of a Triangle

Learning Objectives

• Apply the area formula to triangles where you know two sides and the included angle.

• Apply the area formula to triangles where you know all three sides, Heron’s Formula.

• Use the area formulas in real-world and applied problems.

In this section, we will look at how we can derive a new formula using the area formula that we already

know and the sine function. This new formula will allow us to find the area of a triangle when we don’t

know the height. We will also look at when we can use this formula and how to apply it to real-world

situations.

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Deriving an Alternate Formula to A = 12bh

We can use the area formula from Geometry, A = 12bh, as well as the sine function, to derive a new formula

that can be used when the height, or altitude, is unknown.

In 4ABC below, BD is altitude from B to AC. We will refer to the length of BD as h since it also represents

the height of the triangle. Also, we will refer to the area of the triangle as K to avoid confusing the area

with ∠A.

k =

1

2

bh Area of a triangle

k =

1

2

b(c sin A) sin A = h

c

therefore c sin A = h

k =

1

2

bc sin A Simplify

We can use a similar method to derive all three forms of the area formula, regardless of the angle:

K =

1

2

bc sin A

K =

1

2

ac sin B

K =

1

2

ab sinC

The formula K = 12 bc sin A requires us to know two sides and the included angle (SAS) in a triangle. Once

we know these three things, we can easily calculate the area of an oblique triangle.

Example 1: In 4ABC, ∠C = 62◦, b = 23.9, and a = 31.6. Find the area of the triangle.

Solution: Using our new formula, K = 12 ab sinC, plug in what is known and solve for the area.

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K =

1

2

(31.6)(23.9) sin 62

K ≈ 333.4

Example 2: The Pyramid Hotel recently installed a triangular pool. One side of the pool is 24 feet,

another side is 26 feet, and the angle in between the two sides is 87◦. If the hotel manager needs to order

a cover for the pool, and the cost is $35 per square foot, how much can he expect to spend?

Solution: In order to find the cost of the cover, we first need to know the area of the cover. Once we

know how many square feet the cover is, we can calculate the cost. In the illustration above, you can see

that we know two of the sides and the included angle. This means we can use the formula K = 12 bc sin A.

K =

1

2

(24)(26) sin 87

K ≈ 311.6

311.6 sq. f t. × $35/sq. f t. = $10, 905.03

The cost of the cover will be $10, 905.03.

Find the Area Using Three Sides: Heron’s Formula

In the last section, we learned how to find the area of an oblique triangle when we know two sides and

the included angle using the formula K = 12 bc sin A. We could also find the area of a triangle in which

we know all three sides by first using the Law of Cosines to find one of the angles and then using the

formula K = 12 bc sin A. While this process works, it is time-consuming and requires a lot of calculation.

Fortunately, we have another formula, called Heron’s Formula, which allows us to calculate the area of

a triangle when we know all three sides. It is derived from K = 12 bc sin A, the Law of Cosines and the

Pythagorean Identity.

Heron’s Formula:

K =

√

s(s − a)(s − b)(s − c) where s = 12(a+ b+ c) or half of the perimeter of the triangle.

Example 3: In 4ABC, a = 23, b = 46, and C = 41. Find the area of the triangle.

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Solution: First, you need to find s: s = 12(23+41+46) = 55. Now, plug s and the three sides into Heron’s

Formula and simplify.

K =

√

55(55 − 23)(55 − 46)(55 − 41)

K =

√

55(32)(9)(14)

k =

√

221760

K ≈ 470.9

Example 4: A handyman is installing a tile floor in a kitchen. Since the corners of the kitchen are not

exactly square, he needs to have special triangular shaped tile made for the corners. One side of the tile

needs to be 11.3”, the second side needs to be 11.9”,and the third side is 13.6”. If the tile costs $4.89 per

square foot, and he needs four of them, how much will it cost to have the tiles made?

Solution: In order to find the cost of the tiles, we first need to find the area of one tile. Since we know

the measurements of all three sides, we can use Heron’s Formula to calculate the area.

s =

1

2

(11.3 + 11.6 + 13.6) = 18.25

K =

√

18.25(18.25 − 11.3)(18.25 − 11.9)(18.25 − 13.6)

K =

√

18.25(6.95)(6.35)(4.65)

K =

√

3745.19

K ≈ 61.2 in2

The area of one tile is 61.2 square inches. The cost of the tile is given to us in square feet, while the area

of the tile is in square inches. In order to find the cost of one tile, we must first convert the area of the tile

into square feet.

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1 square foot = 12in × 12in = 144in2

61.2

144

= .425 f t2 Covert square inches into square feet

.425 f t2 × 4.89 = 2.08 Multiply by the cost of the tile.

2.08 × 4 = 8.32

The cost for four tiles would be $8.32.

Finding a Part of the Triangle, Given the Area

We have already looked at two examples of situations where we can apply the two new area formulas we

learned in this section. In this section, we will look at another real-world application where we know the

area but need to find another part of the triangle, as well as an application involving a quadrilateral.

Example 5: The jib sail on a sailboat came untied and the rope securing it was lost. If the area of the

jib sail is 56.1 square feet, use the figure and information belowto find the length of the rope.

Solution: Since we know the area, one of the sides, and one angle of the jib sail, we can use the formula

K = 12 bc sin A to find the side of the jib sail that is attached to the mast. We will call this side y.

56.1 =

1

2

28(y) sin 11

56.1 = 2.671325935 y

21.0 = y

Now that we know side y, we know two sides and the included angle in the triangle formed by the mast,

the rope, and the jib sail. We can now use the Law of Cosines to calculate the length of the rope.

x2 = 212 + 272 − 2(21)(27) cos 18

x2 = 91.50191052

x ≈ 9.6 f t

The length of the rope is approximately 9.6 feet.

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Example 6: In quadrilateral QUAD at the right, The area of 4QUA = 5.64, the area of 4UAD =

6.39, ∠QUD = 31◦, ∠UAD = 40◦, and UD = 7.8. Find the perimeter of QUAD.

Solution: In order to find the perimeter of QUAD, we need to know sides QU,QD,UA, and AD. Since we

know the area, one side, and one angle in each of the triangles, we can use K = 12 bc sin A to figure out QU

and UA.

5.64 =

1

2

(7.8)(QU) sin 31 6.39 = 1

2

(7.8)UA sin 40

2.8 ≈ QU 2.5 ≈ UA

Now that we know QU and UA, we know two sides and the included angle in each triangle (SAS). This

means that we can use the Law of Cosines to find the other two sides, QD and DA. First we will find QD

and DA.

QD2 = 2.82 + 7.82 − 2(2.8)(7.8) cos 31 DA2 = 2.52 + 7.82 − 2(2.5)(7.8) cos 40

QD2 = 31.23893231 DA2 = 37.21426672

QD ≈ 5.6 DA ≈ 6.1

Finally, we can calculate the perimeter since we have found all four sides of the quadrilateral.

pQUAD = 2.8 + 5.6 + 6.1 + 2.5 = 17

Points to Consider

• Why can’t s (half of the perimeter) in Heron’s Formula be smaller than any of the three sides in the

triangle?

• How could we find the area of a triangle is AAS, SSA, and ASA cases?

• Is it possible to figure out the length of the third side of a triangle if we know the other two sides

and the area?

Review Questions

1. Using the figures and given information below, determine which formula you would need to use in

order to find the area of each triangle (A = 12 bh,K = 12 bc sin A, or Heron’s Formula).

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Table 5.2:

Given Figure Formula

a. CF = 3, FM = 8, and CO = 5

b. HC = 4.1,CE = 7.4, and

HE = 9.6

c. AP = 59.8, PH =

86.3, ∠APH = 103◦

d. RX = 11.1, XE = 18.9, ∠R =

41◦

2. Find the area of all of the triangles in the chart above to the nearest tenth.

3. Using the given information and the figures below, decide which area formula you would need to use

to find each side, angle, or area.

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Table 5.3:

Given Figure Find Formula

a. Area = 1618.98, b =

36.3

h

b. Area = 387.6, b =

25.6, c = 32.9

∠A

c. Area 4ABD =

16.96, AD =

3.2, ∠DBC = 49.6◦

Area of 4ABC

4. Using the figures and information from the table above, find the angle, side, or area requested.

5. The Pyramid Hotel is planning on repainting the exterior of the building. The building has four sides

that are isosceles triangles with bases measuring 590 ft and legs measuring 375 ft.

(a) What is the total area that needs to be painted?

(b) If one gallon of paint covers 25 square feet, how many gallons of paint are needed?

6. A contractor needs to replace a triangular section of roof on the front of a house. The sides of the

triangle are 8.2 feet, 14.6 feet, and 16.3 feet. If one bundle of shingles covers 33 13 square feet and

costs $15.45, how many bundles does he need to purchase? How much will the shingles cost him?

How much of the bundle will go to waste?

7. A farmer needs to replant a triangular section of crops that died unexpectedly. One side of the

triangle measures 186 yards, another measures 205 yards, and the angle formed by these two sides is

148◦.

(a) What is the area of the section of crops that needs to be replanted?

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(b) The farmer goes out a few days later to discover that more crops have died. The side that

used to measure 205 yards now measures 288 yards. How much has the area that needs to be

replanted increased by?

8. Find the perimeter of the quadrilateral at the left If the area of 4DEG = 56.5 and the area of

4EGF = 84.7.

9. In 4ABC, BD is an altitude from B to AC. The area of 4ABC = 232.96, AB = 16.2, and AD = 14.4.

Find DC.

10. Show that in any triangle DEF, d2 + e2 + f 2 = 2(e f cosD+ d f cos E + de cos F).

Review Answers

1. (a) A = 12 bh

(b) Heron’s formula

(c) K = 12 bc sin A

(d) A = 12 bh

2. (a) A = 22

(b) A = 14.3

(c) A = 2514.2

(d) A = 144.7

3. (a) A = 12 bh

(b) K = 12 bc sin A

(c) A = 12 bh

4. (a) h = 89.2

(b) ∠A = 67◦

(c) Area of 4ABC = 82.5

5. (a) Use Heron’s Formula, then multiply your answer by 4, for the 4 sides. s = 12(375 + 375 + 590) =

670, A =

√

670(670 − 375)(670 − 375)(670 − 590) = 5623.5, multiplied by 4 = 22494.1 total square

feet.

(b) 22494.125 ≈ 900 gallons of paint are needed.

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6. Using Heron’s Formula, s and the area are: s = 12(8.2+14.6+16.3) = 19.55 and A =

√

19.55(19.55 − 8.2)(19.55 − 14.6)(19.55 − 16.3) =

59.75 sq. f t. He will need 2 bundles

(

59.75

33.3 = 1.8

)

. The shingles will cost him 2∗$15.45 = $30.90 and

6.92 square feet will go to waste (66.67 − 59.75 = 6.92).

7. (a) Use K = 12 bc sin A,K = 12(186)(205) sin 148◦. So, the area that needs to be replaced is 10102.9

square yards.

(b) K = 12(186)(288) sin 148◦ = 14193.4, the area has increased by 4090.5 yards.

8. You need to use the K = 12 bc sin A formula to find DE and GF.

56.5 =

1

2

(13.6)DE sin 39◦ → DE = 13.2 84.7 = 1

2

(13.6)EF sin 60◦ → EF = 14.4

Second, you need to find sides DG and GF using the Law of Cosines.

DG2 = 13.22 + 13.62 − 2 · 13.2 · 13.6 · cos 39◦ → DG = 8.95

GF2 = 14.42 + 13.62 − 2 · 14.4 · 13.6 · cos 60◦ → GF = 14.0

The perimeter of the quadrilateral is 50.55.

9. First, find BD by using the Pythagorean Theorem. BD =

√

16.22 − 14.42 = 7.42. Then, using the area

and formula (A = 12bh), you can find AC. 232.96 = 12(7.42)AC → AC = 62.78. DC = 62.78 − 14.4 =

48.38.10.

d2 = e2 + f 2 − 2e f cosD

e2 = d2 + f 2 − 2d f cos E All three versions of the Law of Cosines

f 2 = d2 + e2 − 2de cos F

Add the three formulas together, we get:

d2 + e2 + f 2 = e2 + f 2 − 2e f cosD+ d2 + f 2 − 2d f cos E + d2 + e2 − 2de cos F

d2 + e2 + f 2 = 2(d2 + e2 + f 2) − 2(e f cosD+ d f cos E + de cos F)

−(d2 + e2 + f 2) = −2(e f cosD+ d f cos E + de cos F)

d2 + e2 + f 2 = 2(e f cosD+ d f cos E + de cos F)

5.3 The Law of Sines

Learning Objectives

• Understand how both forms of the Law of Sines are obtained.

• Apply the Law of Sines when you know two angles and a non-included side and if you know two

angles and the included side.

• Use the Law of Sines in real-world and applied problems.

We have learned about the Law of Cosines, which is a generalization of the Pythagorean Theorem for

non-right triangles. We know that we can use the Law of Cosines when:

1. We know two sides of a triangle and the included angle (SAS) or

2. We know all three sides of the triangle (SSS)

But, what happens if the triangle we are working with doesn’t fit either of those scenarios? Here we

introduce the Law of Sines.

The Law of Sines is a statement about the relationship between the sides and the angles in any triangle.

While the Law of Sines will yield one correct answer in many situations, there are times when it is

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ambiguous, meaning that it can produce more than one answer. We will explore the ambiguity of the Law

of Sines in the next section.

We can use the Law of Sines when:

1. We know two angles and a non-included side (AAS) or

2. We know two angles and the included side (ASA)

Deriving the Law of Sines

4ABC contains altitude CE, which extends from C and intersects AB. We will refer to the length of altitude

CE as x.

We know that sin A = xb and sin B = xa , by the definition of sine. If we cross-multiply both equations and

substitute, we will have the Law of Sines.

b(sin A) = x and a(sin B) = x

↘ ↙

b(sin A) = a(sin B)

sin A

a

=

sin B

b

or asin A =

b

sin B

Extending these ratios to angle C and side c, we arrive at both forms of the Law of Sines:

Form 1 : sin A

a

=

sin B

b

=

sinC

c

(sines over sides)

Form 2 : asin A =

b

sin B =

c

sinC

(sides over sines)

AAS (Angle-Angle-Side)

One case where we can to use the Law of Sines is when we know two of the angles in a triangle and a

non-included side (AAS).

Example 1: Using 4GMN, ∠G = 42◦, ∠N = 73◦ and g = 12. Find n.

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Since we know two angles and one non-included side (g), we can find the other non-included side (n).

sin 73◦

n

=

sin 42◦

12

n sin 42◦ = 12 sin 73◦

n =

12 sin 73◦

sin 42◦

n ≈ 17.15

Example 2: Continuing on from Example 1, find ∠M and m.

Solution: ∠M is simply 180◦ − 42◦ − 73◦ = 65◦. To find side m, you can now use either the Law of Sines

or Law of Cosines. Considering that the Law of Sines is a bit simpler and new, let’s use it. It does not

matter which side and opposite angle you use in the ratio with ∠M and m.

Option 1: ∠G and g

sin 65◦

m

=

sin 42◦

12

m sin 42◦ = 12 sin 65◦

m =

12 sin 65◦

sin 42◦

m ≈ 16.25

Option 2: ∠N and n

sin 65◦

m

=

sin 73◦

17.15

m sin 73◦ = 17.15 sin 65◦

m =

17.15 sin 65◦

sin 73◦

m ≈ 16.25

Example 3: A business group wants to build a golf course on a plot of land that was once a farm. The

deed to the land is old and information about the land is incomplete. If AB is 5382 feet, BC is 3862 feet,

∠AEB is 101◦, ∠BDC is 74◦, ∠EAB is 41◦ and ∠DCB is 32◦, what are the lengths of the sides of each triangular

piece of land? What is the total area of the land?

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Solution: Before we can figure out the area of the land, we need to figure out the length of each side. In

triangle ABE, we know two angles and a non-included side. This is the AAS case. First, we will find the

third angle in triangle ABE by using the Triangle Sum Theorem. Then, we can use the Law of Sines to

find both AE and EB.

∠ABE = 180 − (41 + 101) = 38◦

sin 101

5382

=

sin 38

AE

sin 101

5382

=

sin 41

EB

AE(sin 101) = 5382(sin 38) EB(sin 101) = 5382(sin 41)

AE =

5382(sin 38)

sin 101 EB =

5382(sin 41)

sin 101

AE = 3375.5 f eet EB ≈ 3597.0 f eet

Next, we need to find the missing side lengths in triangle DCB. In this triangle, we again know two angles

and a non-included side (AAS), which means we can use the Law of Sines. First, let’s find ∠DBC =

180− (74+32) = 74◦. Since both ∠BDC and ∠DBC measure 74◦, triangle DCB is an isosceles triangle. This

means that since BC is 3862 feet, DC is also 3862 feet. All we have left to find now is DB.

sin 74

3862

=

sin 32

DB

DB(sin 74) = 3862(sin 32)

DB =

3862(sin 32)

sin 74

DB ≈ 2129.0 f eet

Finally, we need to calculate the area of each triangle and then add the two areas together to get the total

area. From the last section, we learned two area formulas, K = 12 bc sin A and Heron’s Formula. In this

case, since we have enough information to use either formula, we will use K = 12 bc sin A since it is less

computationally intense.

First, we will find the area of triangle ABE.

Triangle ABE:

K =

1

2

(3375.5)(5382) sin 41

K = 5, 959, 292.8 f t2

Triangle DBC:

K =

1

2

(3862)(3862) sin 32

K = 3, 951, 884.6 f t2

The total area is 5, 959, 292.8 + 3, 951, 884.6 = 9, 911, 177.4 f t2.

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ASA (Angle-Side-Angle)

The second case where we use the Law of Sines is when we know two angles in a triangle and the included

side (ASA). For instance, in 4TRI:

∠T, ∠R, and i are known

∠T, ∠I,and r are known

∠R, ∠I, and t are known

In this case, the Law of Sines allows us to find either of the non-included sides.

Example 4: (Use the picture above) In 4TRI, ∠T = 83◦, ∠R = 24◦, and i = 18.5. Find the measure of t.

Solution: Since we know two angles and the included side, we can find either of the non-included sides

using the Law of Sines. Since we already know two of the angles in the triangle, we can find the third

angle using the fact that the sum of all of the angles in a triangle must equal 180◦.

∠I = 180 − (83 + 24)

∠I = 180 − 107

∠I = 73◦

Now that we know ∠I = 73◦, we can use the Law of Sines to find t.

sin 73

18.5

=

sin 83

t

t(sin 73) = 18.5(sin 83)

t =

18.5(sin 83)

sin 73

t ≈ 19.2

Notice how we wait until the last step to input the values into the calculator. This is so our answer is as

accurate as possible.

Example 5: In order to avoid a large and dangerous snowstorm on a flight from Chicago to Buffalo, pilot

John starts out 27◦ off of the normal flight path. After flying 412 miles in this direction, he turns the plane

toward Buffalo. The angle formed by the first flight course and the second flight course is 88◦. For the

pilot, two issues are pressing:

1. What is the total distance of the modified flight path?

2. How much further did he travel than if he had stayed on course?

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Solution, Part 1: In order to find the total distance of the modified flight path, we need to know side x.

To find side x, we will need to use the Law of Sines. Since we know two angles and the included side, this

is an ASA case. Remember that in the ASA case, we need to first find the third angle in the triangle.

MissingAngle = 180 − (27 + 88) = 65◦ The sum of angles in a triangle is 180

sin 65

412

=

sin 27

x

Law of Sines

x(sin 65) = 412(sin 27) Cross multiply

x =

412(sin 27)

sin 65 Divide by sin 65

x ≈ 206.4 miles

The total distance of the modified flight path is 412 + 206.4 = 618.4 miles.

Solution, Part 2: To find how much farther John had to travel, we need to know the distance of the

original flight path, y. We can use the Law of Sines again to find y.

sin 65

412

=

sin 88

y

Law of Sines

y(sin 65) = 412(sin 88) Cross multiply

y =

412(sin 88)

sin 65 Divide by sin 65

y ≈ 454.3 miles

John had to travel 618.4 − 454.3 = 164.1 miles farther.

Solving Triangles

The Law of Sines can be applied in many ways. Below are some examples of the different ways and

situations to which we may apply the Law of Sines. In many ways, the Law of Sines is much easier to use

than the Law of Cosines since there is much less computation involved.

Example 6: In the figure below, ∠C = 22◦, BC = 12,DC = 14.3, ∠BDA = 65◦, and ∠ABD = 11◦. Find AB.

Solution: In order to find AB, we need to know one side in 4ABD. In 4BCD, we know two sides and an

angle, which means we can use the Law of Cosines to find BD. In this case, we will refer to side BD as c.

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c2 = 122 + 14.32 − 2(12)(14.3) cos 22 Law of Cosines

c2 ≈ 30.28

c ≈ 5.5

Now that we know BD ≈ 5.5, we can use the Law of Sines to find AB. In this case, we will refer to AB as x.

∠A = 180 − (11 + 65) = 104◦ Triangle Sum Theorem

sin 104

5.5

=

sin 65

x

Law of Sines

x =

5.5 sin 65

sin 104 Cross multiply and divide by sin 104

x ≈ 5.14

Example 7: A group of forest rangers are hiking through Denali National Park towards Mt. McKinley,

the tallest mountain in North America. From their campsite, they can see Mt. McKinley, and the angle

of elevation from their campsite to the summit is 21◦. They know that the slope of mountain forms a

127◦ angle with ground and that the vertical height of Mt. McKinley is 20,320 feet. How far away is their

campsite from the base of the mountain? If they can hike 2.9 miles in an hour, how long will it take them

to get the base?

Solution: As you can see from the figure above, we have two triangles to deal with here: a right triangle

(4MON) and non-right triangle (4MOU). In order to find the distance from the campsite to the base

of the mountain, y, we first need to find one side of our non-right triangle, 4MOU. If we look at ∠M in

4MNO, we can see that side ON is our opposite side and side x is our hypotenuse. Remember that the

sine function is opposite/hypotenuse. Therefore we can find side x using the sine function.

sin 21◦ = 20320

x

x sin 21◦ = 20320

x =

20320

sin 21◦

x ≈ 56701.5

Now that we know side x, we know two angles and the non-included side in 4MOU. We can use the Law

of Sines to solve for side y. First, ∠MOU = 180◦ − 127◦ − 21◦ = 32◦ by the Triangle Sum Theorem.

sin 127◦

56701.5

=

sin 32◦

y

y sin 127◦ = 56701.5 sin 32◦

y =

56701.5 sin 32◦

sin 127◦

x ≈ 37623.2 or 7.1 miles

If they can hike 2.9 miles per hour, then they will hike the 7.1 miles in 2.45 hours, or 2 hours and 27

minutes.

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Points to Consider

• Are there any situations where we might not be able to use the Law of Sines or the Law of Cosines?

• Considering what you already know about the sine function, is it possible for two angles to have the

same sine? How might this affect using the Law of Sines to solve for an angle?

• By using both the Law of Sines and the Law of Cosines, it is possible to solve any triangle we are

given?

Review Questions

1. In the table below, you are given a figure and information known about that figure. Decide if each

situation represents the AAS case or the ASA case.

Table 5.4:

Given Figure Case

a. b = 16, A = 11.7◦,C = 23.8◦

b. e = 214.9,D = 39.7◦, E =

41.3◦

c. G = 22◦, I = 18◦,H = 140◦

d. k = 6.3, J = 16.2◦, L = 40.3◦

e. M = 31◦,O = 9◦,m = 15

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Table 5.4: (continued)

Given Figure Case

f. Q = 127◦,R = 21.8◦, r = 3.62

2. Even though the triangles and given information in the table above represent two different cases of

the Law of Sines, what do they all have in common?

3. Using the figures and the given information from the table above, find the following if possible:

(a) side a

(b) side d

(c) side i

(d) side l

(e) side o

(f) side q

4. In 4GHI, ∠I = 21.3◦, ∠H = 62.1◦, and i = 108. Find g and h.

5. Use the Law of Sines to show that ab = sin Asin B is true.

6. Use the Law of Sines, the Law of Cosines, and trigonometry functions to solve for x.

(a)

(b)

7. In order to avoid a storm, a pilot starts out 11◦ off path. After he has flown 218 miles, he turns the

plane toward his destination. The angle formed between his first path and his second path is 105◦.

If the plane traveled at an average speed of 495 miles per hour, how much longer did the modified

flight take?

8. A delivery truck driver has three stops to make before she must return to the warehouse to pick up

more packages. The warehouse, Stop A, and Stop B are all on First Street. Stop A is on the corner

of First Street and Route 52, which intersect at a 41◦ angle. Stop B is on the corner of First Street

and Main Street, which intersect at a 103◦ angle. Stop C is at the intersection of Main Street and

Route 52. The driver knows that Stop A and Stop B are 12.3 miles apart and that the warehouse is

1.1 miles from Stop A. If she must be back to the warehouse by 10:00 a.m., travels at a speed of 45

MPH, and takes 2 minutes to deliver each package, at what time must she leave?

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Review Answers

1. (a) ASA

(b) AAS

(c) neither

(d) ASA

(e) AAS

(f) AAS

2. Student answers will vary but they should notice that in both cases you know or can find an angle

and the side across from it.

3. (a) sin 11.7◦a = sin 144.5

◦

16 , a = 5.6

(b) sin 39.7◦214.9 = sin 41.3

◦

d , d = 222.0

(c) not enough information

(d) sin 40.3◦l = sin 123.5

◦

6.3 , l = 4.9

(e) sin 9◦o = sin 31

◦

15 , o = 4.6

(f) sin 127◦q = sin 21.8

◦

3.62 , q = 7.8

4. ∠G = 180◦ − 62.1◦ − 21.3◦ = 96.6◦sin 96.6◦

g

=

sin 21 3

108

, g = 295.3

sin 62.1◦

h

=

sin 21.3◦

108

, h = 262.85.

sin A

a

=

sin B

b

Law of Sines

a(sin B) = b(sin A) Cross multiply

a

b

=

sin A

sin B Divide by b(sin B)

6. (a) tan 54◦ = h7.15 → h = 9.8, cos 67◦ = 9.8x → x = 25.2

(b) The angle we are finding is the one at the far left side of the triangle.

8.92 = 11.22 + 12.62 − 2 · 11.2 · 12.6 cos A→ A = 43.4◦, sin 43.4

◦

x

=

sin 31

11.2

→ x = 14.9.

7. First we need to find the other two sides in the triangle. sin 64◦218 = sin 11

◦

x =

sin 105◦

y , x = 46.3, y = 234.3,

where y is the length of the original fight plan. The modified flight plan is 218+46.3 = 264.3. Dividing

both by 495 mi/hr, we get 32 min (modified) and 28.4 min (original). Therefore, the modified flight

plan is 3.6 minutes longer.

8. First, we need to find the distance between Stop B (B) and Stop C (C). sin 36◦12.3 = sin 41

◦

B =

sin 103◦

C B =

13.7,C = 20.4. The total length of her route is 1.1 + 12.3 + 13.7 + 20.4 + 1.1 = 48.6 miles. Dividing

this by 45 mi/hr, we get that it will take her 1.08 hours or 64.8 minutes. Subtracting this from 10:00

am, she will need to leave by 8:49.

5.4 The Ambiguous Case

Learning Objectives

• Find possible triangles given two sides and an angle (SSA).

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• Use the Law of Cosines and Sines in various ambiguous cases.

In previous sections, we learned about the Law of Cosines and the Law of Sines. We learned that we can

use the Law of Cosines when:

1. we know all three sides of a triangle (SSS) and

2. we know two sides and the included angle (SAS).

We learned that we can use the Law of Sines when:

1. we know two angles and a non-included side (AAS) and

2. we know two angles and the included side (ASA).

However, we have not explored how to approach a triangle when we know two sides and a non-included

angle (SSA). In this section, we will look at why the SSA case is called the ambiguous case, the possible

triangles formed by the SSA case, and how to apply the Law of Sines and the Law of Cosines when we

encounter the SSA case.

Possible Triangles with SSA

In Geometry, you learned that two sides and a non-included angle do not necessarily define a unique

triangle. Consider the following cases given a, b, and ∠A:

Case 1: No triangle exists (a < b)

In this case a < b and side a is too short to reach the base of the triangle. Since no triangle exists, there

is no solution.

Case 2: One triangle exists (a < b)

In this case, a < b and side a is perpendicular to the base of the triangle. Since this

situation yields exactly one triangle, there is exactly one solution.

Case 3: Two triangles exist (a < b)

In this case, a < b and side a meets the base at exactly two points. Since two triangles

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exist, there are two solutions.

Case 4: One triangle exists (a = b)

In this case a = b and side a meets the base at exactly one point. Since there is exactly

one triangle, there is one solution.

Case 5: One triangle exists (a > b)

In this case, a > b and side a meets the base at exactly one point. Since there is exactly

one triangle, there is one solution.

Case 3 is referred to as the Ambiguous Case because there are two possible triangles and two possible

solutions. One way to check to see how many possible solutions (if any) a triangle will have is to compare

sides a and b. If you are faced with the first situation, where a < b, we can still tell how many solutions

there will be by using a and b sin A.

Table 5.5:

If: Then:

a. a < b No solution, one solution, two so-

lutions

i. a < b sin A No solution

ii. a = b sin A One solution

iii. a > b sin A Two solutions

b. a = b One solution

c. a > b One solution

Example 1: Determine if the sides and angle given determine no, one or two triangles. All sets contain

an angle, its opposite side and the side between them.

a. a = 5, b = 8, A = 62.19◦

b. c = 10, b = 14, B = 15.45◦

c. d = 16, g = 11,D = 44.94◦

d. a = 9, b = 7, B = 51.06◦

Solution: Even though a, b and ∠A are not used in every example, follow the same pattern from the table

by multiplying the non-opposite side (of the angle) by the angle.

a. 5 < 8, 8 sin 62.19◦ = 7.076. So 5 < 7.076, which means there is no solution.

b. 10 < 14, 14 sin 15.45◦ = 3.73. So 10 > 3.73, which means there are two solutions.

c. 16 > 11, there is one solution.

d. 7 < 9, 9 sin 51.06◦ = 7.00. So 7 = 7, which means there is one solution.

In the next two sections we will look at how to use the Law of Cosines and the Law of Sines when faced

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with the various cases above.

Using the Law of Sines

In triangle ABC below, we know two sides and a non-included angle. Remember that the Law of Sines

states: sin Aa = sin Bb . Since we know a, b, and ∠A, we can use the Law of Sines to find ∠B. However, since

this is the SSA case, we have to watch out for the Ambiguous case. Since a < b, we could be faced with

either Case 1, Case 2, or Case 3 above.

Example 2: Find ∠B.

Solution: Use the Law of Sines to determine the angle.sin 41

12

=

sin B

23

23 sin 41 = 12 sin B

23 sin 41

12

= sin B

1.257446472 = sin B

Since no angle exists with a sine greater than 1, there is no solution to this problem.

We also could have compared a and b sin A beforehand to see how many solutions there were to this triangle.

a = 12, b sin A = 15.1: since 12 < 15.1, a < b sin A which tells us there are no solutions.

Example 3: In triangle ABC, a = 15, b = 20, and ∠A = 30◦. Find ∠B.

Solution: Again in this case, a < b and we know two sides and a non-included angle. By comparing a and

b sin A, we find that a = 15, b sin A = 10. Since 15 > 10 we know that there will be two solutions to this

problem.

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sin 30

15

=

sin B

20

20 sin 30 = 15 sin B

20 sin 30

15

= sin B

0.6666667 = sin B

∠B = 41.8◦

There are two angles less than 180◦ with a sine of 0.6666667, however. We found the first one, 41.8◦, by

using the inverse sine function. To find the second one, we will subtract 41.8◦ from 180◦, ∠B = 180◦−41.8◦ =

138.2◦.

To check to make sure 138.2◦ is a solution, we will use the Triangle Sum Theorem to find the third angle.

Remember that all three angles must add up to 180◦.

180◦ − (30◦ + 41.8◦) = 108.2◦ or 180◦ − (30◦ + 138.2◦) = 11.8◦

This problem yields two solutions. Either ∠B = 41.8◦ or 138.2◦.

Example 4: A boat leaves lighthouse A and travels 63km. It is spotted from lighthouse B, which is 82km

away from lighthouse A. The boat forms an angle of 65.1◦ with both lighthouses. How far is the boat from

lighthouse B?

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Solution: In this problem, we again have the SSA angle case. In order to find the distance from the boat

to the lighthouse (a) we will first need to find the measure of angle A. In order to find angle A, we must

first use the Law of Sines to find angle B. Since c > b, this situation will yield exactly one answer for the

measure of angle B.

sin 65.1◦

82

=

sin B

63

63 sin 65.1◦

82

= sin B

0.6969 ≈ sin B

∠B = 44.2◦

Now that we know the measure of angle B, we can find the measure of angle A, ∠A = 180◦ − 65.1◦ − 44.2◦ =

70.7◦. Finally, we can use ∠A to find side a.

sin 65.1◦

82

=

sin 70.7◦

a

82 sin 70.7◦

sin 65.1◦ = a

a = 85.3

The boat is approximately 85.3 km away from lighthouse B.

Using the Law of Cosines

Example 5: In a game of pool, a player must put the eight ball into the bottom left pocket of the table.

Currently, the eight ball is 6.8 feet away from the bottom left pocket. However, due to the position of the

cue ball, she must bank the shot off of the right side bumper. If the eight ball is 2.1 feet away from the

spot on the bumper she needs to hit and forms a 168◦ angle with the pocket and the spot on the bumper,

at what angle does the ball need to leave the bumper?

Solution: In the scenario above, we have the SAS case, which means that we need to use the Law of

Cosines to begin solving this problem. The Law of Cosines will allow us to find the distance from the spot

on the bumper to the pocket (y). Once we know y, we can use the Law of Sines to find the angle (X).

y2 = 6.82 + 2.12 − 2(6.8)(2.1) cos 168◦

y2 = 78.59

y = 8.86 f eet

The distance from the spot on the bumper to the pocket is 8.86 feet. We can now use this distance and

the Law of Sines to find angle X. Since we are finding an angle, we are faced with the SSA case, which

means we could have no solution, one solution, or two solutions. However, since we know all three sides

this problem will yield only one solution.

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sin 168◦

8.86

=

sin X

6.8

6.8 sin 168◦

8.86

= sin X

0.1596 ≈ sin B

∠B = 8.77◦

In the previous example, we looked at how we can use the Law of Sines and the Law of Cosines together

to solve a problem involving the SSA case. In this section, we will look at situations where we can use

not only the Law of Sines and the Law of Cosines, but also the Pythagorean Theorem and trigonometric

ratios. We will also look at another real-world application involving the SSA case.

Example 6: Three scientists are out setting up equipment to gather data on a local mountain. Person 1

is 131.5 yards away from Person 2, who is 67.8 yards away from Person 3. Person 1 is 72.6 yards away from

the mountain. The mountains forms a 103◦ angle with Person 1 and Person 3, while Person 2 forms a 92.7◦

angle with Person 1 and Person 3. Find the angle formed by Person 3 with Person 1 and the mountain.

Solution: In the triangle formed by the three people, we know two sides and the included angle (SAS).

We can use the Law of Cosines to find the remaining side of this triangle, which we will call x. Once we

know x, we will two sides and the non-included angle (SSA) in the triangle formed by Person 1, Person 2,

and the mountain. We will then be able to use the Law of Sines to calculate the angle formed by Person

3 with Person 1 and the mountain, which we will refer to as Y.

To find x:

x2 = 131.52 + 67.82 − 2(131.5)(67.8) cos 92.7

x2 = 22729.06397

x = 150.8 yds

Now that we know x = 150.8, we can use the Law of Sines to find Y. Since this is the SSA case, we need

to check to see if we will have no solution, one solution, or two solutions. Since 150.8 > 72.6, we know that

we will have only one solution to this problem. sin 103

150.8

=

sin Y

72.6

72.6 sin 103

150.8

= sinY

0.4690932805 = sinY

28.0 ≈ ∠Y

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Points to Consider

• Why is there only one possible solution to the SSA case if a > b?

• Explain why a > b sin A yields two possible solutions to a triangle.

• If we have a SSA angle case with two possible solutions, how can we check both solutions to make

sure they are correct?

Review Questions

1. Using the table below, determine how many solutions there would be to each problem based on the

given information and by calculating b sin A and comparing it with a. Sketch an approximate diagram

for each problem in the box labeled “diagram.”

Table 5.6:

Given a >,=, or < b sin A Diagram Number of solutions

a. A = 32.5◦, a =

26, b = 37

b. A = 42.3◦, a =

16, b = 26

c. A = 47.8◦, a =

13.48, b = 18.2

d. A = 51.5◦, a =

3.4, b = 4.2

2. Using the information in the table above, find all possible measures of angle B if any exist.

3. Prove using the Law of Sines: a−cc = sin A−sinCsinC

4. Give the measure of a non-included angle and the lengths of two sides so that two triangles exist.

Explain why two triangles exist for the measures you came up with.

5. If a = 22 and b = 31, find the values of A so that:

(a) There is no solution

(b) There is one solution

(c) There are two solutions

6. In the figure below, AB = 13.7, AD = 9.8, and ∠C = 42.6◦. Find ∠A, ∠B, BC, and AC.

7. In the figure below, ∠C = 21.8◦, BE = 9.9, BD = 10.2, ED = 7.6, and ∠ABC = 109.6◦. Find the

following:

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(a) ∠EBD

(b) ∠BED

(c) ∠DEB

(d) ∠BDC

(e) ∠BEA

(f) ∠DBC

(g) ∠ABE

(h) ∠BAE

(i) BC

(j) AB

(k) AE

(l) DC

(m) AC

8. Radio detection sensors for tracking animals have been placed at three different points in a wildlife

preserve. The distance between Sensor 1 and Sensor 2 is 4500ft. The distance between Sensor 1 and

Sensor 3 is 4000ft. The angle formed by Sensor 3 with Sensors 1 and 2 is 56◦. If the range of Sensor

3 is 6000ft, will it be able to detect all movement from its location to the location of Sensor 2?

9. In problem 8 above, a fourth sensor is placed in the wildlife preserve. Sensor 2 forms a 36 angle with

Sensors 3 and 4, and Sensor 3 forms a 49 angle with Sensors 2 and 4. How far away is Sensor 4 from

Sensors 2 and 3?

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Review Answers

1.

Table 5.7:

Given a >,=, or < b sin A Diagram Number of solutions

a. A = 32.5◦, a =

26, b = 37

26 > 19.9 2

b. A = 42.3◦, a =

16, b = 26

16 < 17.5 0

c. A = 47.8◦, a =

13.48, b = 18.2

13.48 = 13.48 1

d. A = 51.5◦, a =

3.4, b = 4.2

3.4 > 3.3 2

2. (a) sin 32.5◦26 = sin B37 → B = 49.9◦ or 180◦ − 49.9◦ = 130.1◦

(b) no solution

(c) sin 47.8◦13.48 = sin B18.2 → B = 90◦

(d) sin 51.5◦3.4 = sin B4.2 → B = 75.2◦ or 180◦ − 75.2◦ = 104.8◦3.

sin A

a

=

sinC

c

c sin A = a sinC

c sin A − c sinC = a sinC − c sinC

c(sin A − sinC) = sinC(a − c)

sin A − sinC

sinC =

a − c

c

4. Student answers will vary. Student should mention using a > b sin A in their explanation.

5. a < b sin A→ ab < sin A→ 2231 < sin A→ A = 45.2◦

(a) ∠A > 45.2◦

(b) ∠A = 45.2◦

(c) ∠A < 45.2◦

6. This problem can be done entirely with right triangle trig, but there are several different ways to

solve this particular problem.

BC =

√

13.72 − 9.82 = 9.6 tan 42.6◦ = 9.8

DC

→ DC = 10.7

sin 42.6◦ = 9.8

AC

→ AC = 14.5 BC = 9.6 + 10.7 = 20.3

sin B = 9.8

13.7

→ ∠B = 45.7◦ ∠A = 180◦ − 45.7◦ − 42.6◦ = 91.7◦

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7. (a) ∠EBD⇒ 7.62 = 9.92 + 10.22 − 2 · 9.9 · 10.2 cos EBD⇒ 44.4◦

(b) ∠BDE ⇒ sin BDE9.9 = sin 44.4

◦

7.6 ⇒ 65.7◦

(c) ∠DEB⇒ 180◦ − 65.7◦ − 44.4◦ ⇒ 69.9◦

(d) ∠BDC ⇒ 180◦ − 65.7◦ ⇒ 114.3◦

(e) ∠BEA⇒ 180◦ − 69.9◦ ⇒ 110.1◦

(f) ∠DBC ⇒ 180◦ − 114.3◦ − 21.8◦ ⇒ 43.9◦

(g) ∠ABE ⇒ 109.6◦ − 43.9◦ − 44.4◦ ⇒ 21.3◦

(h) ∠BAE ⇒ 180◦ − 21.3◦ − 110.1◦ ⇒ 48.6◦

(i) BC ⇒ sin 114.3◦BC = sin 21.8

◦

10.2 ⇒ 25.0

(j) AB⇒ sin 110.1◦AB = sin 48.6

◦

9.9 ⇒ 12.4

(k) AE ⇒ sin 21.3◦AE = sin 48.6

◦

9.9 ⇒ 4.8

(l) DC ⇒ sin 43.9◦DC = sin 21.8

◦

9.9 ⇒ 19.0

(m) AC = 19 + 4.8 + 7.6 = 31.4

8. We need to find the distance between sensors 2 and 3. If it is less than 6000 ft, then the sensor will

be able to detect all motion between the two. First, 4500 > 4000, so there is going to be one solution.

sin S 2

4000

=

sin 56◦

4500

→ S 2 = 47.47◦ sin 47.47

◦

x

=

sin 56◦

4500

→ x = 4000

The sensor is in range.

9. The length from S 3 to S 4 is x and from S 4 to S 2 is y. 180◦ − 36◦ − 49◦ = 95◦, which is the angle at

S 4.

sin 95◦

4000

=

sin 36◦

x

=

sin 49◦

y

x = 2360.1, y = 3030.4

5.5 General Solutions of Triangles

Learning Objectives

• Use the Pythagorean Theorem, trigonometry functions, the Law of Sines, and the Law of Cosines to

solve various triangles.

• Understand when it is appropriate to use each method.

• Apply the methods above in real-world and applied problems.

In the previous sections we have discussed a number of methods for finding a missing side or angle in a

triangle. Previously, we only knew how to do this in right triangles, but now we know how to find missing

sides and angles in oblique triangles as well. By combining all of the methods we’ve learned up until this

point, it is possible for us to find all missing sides and angles in any triangle we are given.

Summary of Triangle Techniques

Below is a chart summarizing the triangle techniques that we have learned up to this point. This chart

describes the type of triangle (either right or oblique), the given information, the appropriate technique to

use, and what we can find using each technique.

Table 5.8:

Type of Triangle: Given Information: Technique: What we can find:

Right Two sides Pythagorean Theorem Third side

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Table 5.8: (continued)

Type of Triangle: Given Information: Technique: What we can find:

Right One angle and one side Trigonometric ratios Either of the other two

sides

Right Two sides Trigonometric ratios Either of the other two

angles

Oblique 2 angles and a non-

included side (AAS)

Law of Sines The other non-included

side

Oblique 2 angles and the in-

cluded side (ASA)

Law of Sines Either of the non-

included sides

Oblique 2 sides and the angle

opposite one of those

sides (SSA) – Ambigu-

ous case

Law of Sines The angle opposite the

other side (can yield no,

one, or two solutions)

Oblique 2 sides and the included

angle (SAS)

Law of Cosines The third side

Oblique 3 sides Law of Cosines Any of the three

angles

Using the Law of Cosines

It is possible for us to completely solve a triangle using the Law of Cosines. In order to do this, we will

need to apply the Law of Cosines multiple times to find all of the sides and/or angles we are missing.

Example 1: In triangle ABC, a = 12, b = 13, c = 8. Solve the triangle.

Solution: Since we are given all three sides in the triangle, we can use the Law of Cosines. Before we can

solve the triangle, it is important to know what information we are missing. In this case, we do not know

any of the angles, so we are solving for ∠A, ∠B, and ∠C. We will begin by finding ∠A.

122 = 82 + 132 − 2(8)(13) cos A

144 = 233 − 208 cos A

−89 = −208 cos A

0.4278846154 = cos A

64.7 ≈ ∠A

Now, we will find ∠B by using the Law of Cosines. Keep in mind that you can now also use the Law of

Sines to find ∠B. Use whatever method you feel more comfortable with.

132 = 82 + 122 − 2(8)(12) cos B

169 = 208 − 192 cos B

−39 = −192 cos B

0.2031 = cos B

78.3◦ ≈ ∠B

We can now quickly find ∠C by using the Triangle Sum Theorem, 180◦ − 64.7◦ − 78.3◦ = 37◦

Example 2: In triangle DEF, d = 43, e = 37, and ∠F = 124◦. Solve the triangle.

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Solution: In this triangle, we have the SAS case because we know two sides and the included angle. This

means that we can use the Law of Cosines to solve the triangle. In order to solve this triangle, we need to

find side f , ∠D, and ∠E. First, we will need to find side f using the Law of Cosines.

f 2 = 432 + 372 − 2(43)(37) cos 124

f 2 = 4997.351819

f ≈ 70.7

Now that we know f , we know all three sides of the triangle. This means that we can use the Law of

Cosines to find either angle D or angle E. We will find angle D first.

432 = 70.72 + 372 − 2(70.7)(37) cosD

1849 = 6367.49 − 5231.8 cosD

−4518.49 = −5231.8 cosD

0.863658779 = cosD

30.3◦ ≈ ∠D

To find angle E, we need only to use the Triangle Sum Theorem, ∠E = 180 − (124 + 30.3) = 25.7◦.

Example 3: A control tower is receiving signals from two microchips implanted in wild tigers. Microchip

1 is 135 miles from the control tower and microchip 2 is 182 miles from the control tower. If the control

tower forms a 119◦ angle with both microchips, how far apart are the two tigers?

Solution: To find the distance between the two tigers, we need to find the distance between the two

microchips. We will call this distance x. Since we know two sides and the included angle, we can use the

Law of Cosines to find x.

x2 = 1352 + 1822 − 2(135)(182) cos 119

x2 = 75172.54474

x = 274.2 miles

The two tigers are 274.2 miles apart.

Using the Law of Sines

It is also possible for us to completely solve a triangle using the Law of Sines if we begin with the ASA

case, the AAS case, or the SSA case. We must remember that when given the SSA case, it is possible that

we may encounter the Ambiguous Case.

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Example 4: In triangle ABC, A = 43◦, B = 82◦, and c = 10.3. Solve the triangle.

Solution: This is an example of the ASA case, which means that we can use the Law of Sines to solve

the triangle. In order to use the Law of Sines, we must first know angle C, which we can find using the

Triangle Sum Theorem, ∠C = 180◦ − (43◦ + 82◦) = 55◦.

Now that we know ∠C, we can use the Law of Sines to find either side a or side b.

sin 55

10.3

=

sin 43

a

sin 55

10.3

=

sin 82

b

a =

10.3 sin 43

sin 55 b =

10.3 sin 82

sin 55

a = 8.6 b = 12.5

Example 5: A cruise ship is based at Island 1, but makes trips to Island 2 and Island 3 during the day.

If the distance from Island 1 to Island 2 is 28.3 miles, from Island 2 to 3 is 52.4 miles, and Island 3 to 1 is

59.8 miles, what heading (angle) must the captain:

a. Leave Island 1

b. Leave Island 2

c. Leave Island 3

∗Remember that when using a compass, 0◦ is due North and 180◦ is due South which means we must

convert our angle measures from the traditional x− and y−axis measures.

Solution: In order to find all three angles in the triangle, we must use the Law of Cosines because we are

dealing with the SSS case. Once we find one angle using the Law of Cosines, we can use the Law of Sines

to find a second angle. Then, we can use the Triangle Sum Theorem to find the third angle.

We will begin by finding ∠B because it is the largest angle.

59.82 = 52.42 + 28.32 − 2(52.4)(28.3) cos B

3576.04 = 3546.65 − 2965.84 cos B

29.39 = −2965.84 cos B

−0.0099095029 = cos B

B = 90.6◦

Now that we know ∠B, we can find either ∠A or ∠C. We will find ∠C first since it is the second largest

angle.

sin 90.6

59.8

=

sinC

52.4

52.4 sin 90.6

59.8

= sinC

0.876203135 = sinC

∠C = 61.2◦

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Now that we know ∠B and ∠C, we can use the Triangle Sum Theorem to find ∠A = 180◦ − (61.2◦+90.6◦) =

28.2◦.

Now, we must convert our angles into headings. See the figures below.

In the first figure, we see that 61.2◦ is a heading of 28.8◦NE. In the second figure, we see that 90.6◦ is a

heading of 0.6◦NW. In the third figure, we see that 28.2◦ is a heading of 61.8◦NE.

Points to Consider

• Is there ever a situation where you would need to use the Law of Sines before using the Law of

Cosines?

• In what situation might you consider using the Law of Cosines instead of Law of Sines if both were

applicable?

• Why do we only have to use the Law of Cosines one time before we can switch to using the Law of

Sines?

Review Questions

1. Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA),

and whether you would use the Law of Sines or the Law of Cosines to find the requested side or

angle. Make an approximate drawing of each triangle and label the given information. Also, state

how many solutions (if any) each triangle would have. If a triangle has no solution or two solutions,

explain why.

Table 5.9:

Given Drawing Case Law Number of Solu-

tions

a. A = 69◦, B =

12◦, a = 22.3, find

b.

b. a = 1.4, b =

2.3,C = 58◦, find

c.

c. a = 3.3, b =

6.1, c = 4.8, find A.

d. a = 15, b =

25, A = 58◦, find B.

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Table 5.9: (continued)

Given Drawing Case Law Number of Solu-

tions

e. a = 45, b =

60, A = 47◦, find B.

2. Using the information in the chart above, solve for the requested side or angle.

3. Using the information in the chart in question 1 and your answers from question 2, determine what

information you are still missing from each triangle. Then, solve for each piece, solving each triangle.

4. The side of a rhombus is 12 cm and the longer diagonal is 21.5cm. Find the area of the rhombus and

the measures of the angles in the rhombus.

5. Find the area of the pentagon below.

6. In the figure drawn below, angle T is 56.8◦ and RT = 38. Using the figure below, find the length of

the altitude draw to the longest side, the area of the two triangles formed by this altitude, RI and

angle I.

7. Refer back to Example 5, the island hopping problem. Suppose there is a fourth island that tourists

can visit. Island 4 is 22.6 miles away from Island 1 and the heading from Island 1 to Island 4 is 86.2◦.

(a) What is the distance from Island 3 to Island 4?

(b) What is the angle formed by Island 3 with Islands 1 and 4?

(c) What is the angle formed by Island 4 with Island 1 and 3?

8. A golfer is standing on the tee of a golf hole that has a 115◦ bend to the left. The distance from the

tee to the bend is 218 yards. The distance from the bend to the green is 187 yards.

(a) How far would the golfer need to hit the ball if he wanted to make it to the green in one shot?

(b) At what angle would he need to hit the ball?

9. A golfer is standing on the tee, which is 320 yards from the cup on the green. After he hits his first

shot, which is sliced to the right, his ball forms a 162.2◦ angle the tee and the cup, and the cup forms

a 14.2◦ angle with his ball and the tee.

(a) What is the degree of his slice?

(b) How far was his first shot?

(c) How far away from the cup is he?

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Review Answers

1. (a) AAS, Law of Sines, one solution

(b) SAS, Law of Cosines, one solution

(c) SSS, Law of Cosines, one solution

(d) SSA, Law of Sines, no solution (15 < 25 sin 58◦)

(e) SSA, Law of Sines, two solutions (45 > 60 sin 47◦)

2. (a) sin 69◦22.3 = sin 12

◦

b , b = 4.97

(b) c2 = 1.42 + 2.32 − 2(1.4)(2.3) cos 58◦, c = 2.0

(c) 3.32 = 6.12 + 4.82 − 2(6.1)(4.8) cos A, A = 32.6◦

(d) No solution

(e) sin 47◦45 = sin B60 , B = 77.2◦ or 180◦ − 77.2◦ = 102.8◦

3. (a) need angle C and side c. C = 180◦ − 69◦ − 12◦ = 99◦

sin 99◦

c

=

sin 69◦

22.3

, c = 23.6

(b) need angle A and angle B.

sin 58◦

2

=

sin A

1.4

, A = 36.4◦

180◦ − 36.4◦ − 58◦, B = 85.6◦

(c) need angle B and angle C.

sin 32.6◦

3.3

=

sin B

6.1

, B = 84.8◦

180◦ − 32.6◦ − 84.8◦,C = 62.6◦

(d) No solution

(e) Both cases need angle C and side c.

Case 1 : C = 180◦ − 47◦ − 77.2◦ = 55.8◦, sin 55.8

◦

c

=

sin 47◦

45

, c = 50.9

Case 2 : C = 180◦ − 47◦ − 102.8◦ = 30.2◦, sin 30.2

◦

c

=

sin 47◦

45

, c = 30.95

4. To find the area of the rhombus, use the formula K = 12 bc sin A and then multiply that by 2. We

first need to find one of the angles that are opposite the given diagonal (they are both the same

measurement). We will call it angle A. 21.52 = 122 + 122 − 2(12)(12) sin A, A = 127.2◦, which means

the other two angles are both 52.8◦(360◦ − 127.2◦ − 127.2◦ and then divide by 2).

K = 2

(1

2

(12)(12) sin 127.2◦

)

= 114.7

5. Divide the pentagon into three triangles, drawing segments from ∠2 to ∠5, called x below, and ∠2 to

∠4, called y below. With these three triangles, only the middle triangle needs us to find two sides

and the angle between them (called ∠Z below) to use K = 12 bc sin A (the outer two triangles already

have two sides and an angle that fit this criteria).x2 = 1922 + 190.52 − 2(192)(190.5) cos 81◦ → x = 248.4

y2 = 1462 + 173.82 − 2(146)(173.8) cos 73◦ → x = 191.5

1182 = 248.42 + 191.52 − 2(248.4)(191.5) cosZ → ∠Z = 27.4◦

Areas: K = 12(190.5)(192) sin 81◦ = 18062.8

k =

1

2

(248.4)(191.5) sin 27.4◦ = 10945.5

K =

1

2

(173.8)(146) sin 73◦ = 12133.0 Total Area : 41141.3

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6. altitude, x: sin 56.8◦ = x38 → x = 31.8

GT =

√

382 − 31.82 = 20.8 GI = 88 − 20.8 = 67.2

Asmall =

1

2

(20.8)(31.8) = 330.8 Abig =

1

2

(67.2)(31.8) = 1068.5

RI =

√

67.22 + 31.82 = 74.3 ∠I → sin I = 31.8

74.3

→ 25.3◦

7.

8. First, recall that the heading is from vertical, so 86.2◦ heading is actually 3.8◦.

(a) x2 = 22.62 + 59.82 − 2(22.6)(59.8) cos 3.8, 37.3 miles

(b) sin 3.8◦37.3 = sin B22.6 , B = 2.3◦

(c) 180◦ − 3.8◦ − 2.3◦ = 173.9◦

9. (a) a2 = 1872 + 2182 − 2(187)(218) cos 115◦, he would need to hit the ball 342.0 yards.

(b) sin 115◦342 = sin B187 , he would have to hit the ball at a 29.7◦ angle.

10. (a) 180◦ − 14.2◦ − 162.2◦ = 3.6◦

(b) sin 14.2◦b = sin 162.2

◦

320 , 256.8 yards

(c) sin 162.2◦320 = sin 3.6

◦

c , 65.7 yards

5.6 Vectors

Learning Objectives

• Understand directed line segments, equal vectors, and absolute value in relation to vectors.

• Perform vector addition and subtraction.

• Find the resultant vector of two displacements.

In previous examples, we could simply use triangles to represent direction and distance. In real-life, there

are typically other factors involved, such as the speed of the object (that is moving in the given direction

and distance) and wind. We need another tool to represent not only direction but also magnitude (length)

or force. This is why we need vectors. Vectors capture the interactions of real world velocities, forces and

distance changes.

Any application in which direction is specified requires the use of vectors. A vector is any quantity

having direction and magnitude. Vectors are very common in science, particularly physics, engineering,

electronics, and chemistry in which one must consider an object’s motion (either velocity or acceleration)

and the direction of that motion.

In this section, we will look at how and when to use vectors. We will also explore vector addition,

subtraction, and the resultant of two displacements. In addition we will look at real-world problems and

application involving vectors.

Directed Line Segments, Equal Vectors, and Absolute Value

A vector is represented diagrammatically by a directed line segment or arrow. A directed line segment

has both magnitude and direction. Magnitude refers to the length of the directed line segment and is

usually based on a scale. The vector quantity represented, such as influence of the wind or water current

may be completely invisible.

A 25 mph wind is blowing from the northwest. If 1 cm = 5 mph, then the vector would look like this:

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An object affected by this wind would travel in a southeast direction at 25 mph.

A vector is said to be in standard position if its initial point is at the origin. The initial point is where

the vector begins and the terminal point is where it ends. The axes are arbitrary. They just give a place

to draw the vector.

If we know the coordinates of a vector’s initial point and terminal point, we can use these coordinates to

find the magnitude and direction of the vector.

All vectors have magnitude. This measures the total distance moved, total velocity, force or acceleration.

“Distance” here applies to the magnitude of the vector even though the vector is a measure of velocity,

force, or acceleration. In order to find the magnitude of a vector, we use the distance formula. A vector

can have a negative magnitude. A force acting on a block pushing it at 20 lbs north can be also written as

vector acting on the block from the south with a magnitude of -20 lbs. Such negative magnitudes can be

confusing; making a diagram helps. The -20 lbs south can be re-written as +20 lbs north without changing

the vector. Magnitude is also called the absolute value of a vector.

Example 1: If we know the coordinates of the initial point and the terminal point, we can find the

magnitude by using the distance formula. Initial point (0,0) and terminal point (3,5).

Solution: |~v| = √(3 − 0)2 + (5 − 0)2 = √9 + 25 = 5.8 The magnitude of~v is 5.8.

If we don’t know the coordinates of the vector, we must use a ruler and the given scale to find the magnitude.

Also notice the notation of a vector, which is usually a lower case letter (typically u, v, or w) in italics, with

an arrow over it, which indicates direction. If a vector is in standard position, we can use trigonometric

ratios such as sine, cosine and tangent to find the direction of that vector.

Example 2: If a vector is in standard position and its terminal point has coordinates of (12, 9) what is

the direction?

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Solution: The horizontal distance is 12 while the vertical distance is 9. We can use the tangent function

since we know the opposite and adjacent sides of our triangle.

tan θ = 9

12

tan−1 9

12

= 36.9◦

So, the direction of the vector is 36.9◦.

If the vector isn’t in standard position and we don’t know the coordinates of the terminal point, we must

a protractor to find the direction.

Two vectors are equal if they have the same magnitude and direction. Look at the figures below for a

visual understanding of equal vectors.

Example 3: Determine if the two vectors are equal.

~a is in standard position with terminal point (-4, 12)

~b has an initial point of (7, -6) and terminal point (3, 6)

Solution: You need to determine if both the magnitude and the direction are the same.

Magnitude : |~a| =

√

(0 − (−4))2 + (0 − 12)2 = √16 + 144 = √160 = 4√10

|~b| =

√

(7 − 3)2 + (−6 − 6)2 = √16 + 144 = √160 = 4√10

Direction : ~a→ tan θ = 12−4 → θ = 108.43

◦

~b→ tan θ = −6 − 6

7 − 3 =

−12

4

→ θ = 108.43◦

Because the magnitude and the direction are the same, we can conclude that the two vectors are equal.

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Vector Addition

The sum of two or more vectors is called the resultant of the vectors. There are two methods we can use

to find the resultant: the triangle method and the parallelogram method.

The Triangle Method: To use the triangle method, we draw the vectors one after another and place the

initial point of the second vector at the terminal point of the first vector. Then, we draw the resultant

vector from the initial point of the first vector to the terminal point of the second vector. This method is

also referred to as the tip-to-tail method.

To find the sum of the resultant vector we would use a ruler and a protractor to find the magnitude and

direction.

The resultant vector can be much longer than either ~a or ~b, or it can be shorter. Below are some more

examples of the triangle method.

Example 4:

Example 5:

The Parallelogram Method: Another method we could use is the parallelogram method. To use the

parallelogram method, we draw the vectors so that their initial points meet. Then, we draw in lines to

form a parallelogram. The resultant is the diagonal from the initial point to the opposite vertex of the

parallelogram. It is important to note that we cannot use the parallelogram method to find the sum of a

vector and itself.

To find the sum of the resultant vector, we would again use a ruler and a protractor to find the magnitude

and direction.

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If you look closely, you’ll notice that the parallelogram method is really a version of the triangle or tip-to-tail

method. If you look at the top portion of the figure above, you can see that one side of our parallelogram

is really vector b translated.

Vector Subtraction

As you know from Algebra, A − B = A+ (−B). When we think of vector subtraction, we must think about

it in terms of adding a negative vector. A negative vector is the same magnitude of the original vector,

but its direction is opposite.

In order to subtract two vectors, we can use either the triangle method or the parallelogram method from

above. The only difference is that instead of adding vectors A and B, we will be adding A and −B.

Example 6: Using the triangle method for subtraction.

Resultant of Two Displacements

We can use vectors to find direction, velocity, and force of moving objects. In this section we will look at

a few applications where we will use resultants of vectors to find speed, direction, and other quantities. A

displacement is a distance considered as a vector. If one is 10 ft away from a point, then any point at a

radius of 10 ft from that point satisfies the condition. If one is 28 degrees to the east of north, then only

one point satisfies this.

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Example 7: A cruise ship is traveling south at 22 mph. A westward wind is blowing the ship eastward

at 7 mph. What speed is the ship traveling at and in what direction is it moving?

Solution: In order to find the direction and the speed the boat is traveling, we must find the resultant

of the two vectors representing 22 mph south and 7 mph east. Since these two vectors form a right angle,

we can use the Pythagorean Theorem and trigonometric ratios to find the magnitude and direction of the

resultant vector.

First, we will find the speed.

222 + 72 = x2

533 = x2

23.1 = x

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The ship is traveling at a speed of 23.1mph.

To find the direction, we will use tangent, since we know the opposite and adjacent sides of our triangle.

tan θ = 7

22

tan−1 7

22

= 17.7◦

The ship’s direction is 17.7◦ SE.

Example 8: A hot air balloon is rising at a rate of 13 ft/sec, while a wind is blowing at a rate of 22 ft/sec.

Find speed at which the balloon is traveling as well as the angle its angle of elevation.

First, we will find the speed at which our balloon is rising. Since we have a right triangle, we can use the

Pythagorean Theorem to find calculate the magnitude of the resultant.

x2 = 132 + 222

x2 = 653

x = 25.6 f t/ sec

The balloon is traveling at rate of 25.6 feet per second.

To find the angle of elevation of the balloon, we need to find the angle it makes with the horizontal. We

will find the angle A in the triangle and then we will subtract it from 90◦.

tan A = 22

13

A = tan−1 22

13

A = 59.4◦

Angle with the horizontal = 90 − 59.4 = 30.6◦.

The balloon has an angle of elevation of 30.6◦.

Example 9: Continuing on with the previous example, find:

a. How far from the lift off point is the balloon in 2 hours? Assume constant rise and constant wind speed.

(this is total displacement)

b. How far must the support crew travel on the ground to get under the balloon? (horizontal displacement)

c. If the balloon stops rising after 2 hours and floats for another 2 hours, how far did it travel total? How

far away does the crew have to go to be under the balloon when it lands?

Solution:

a. After two hours, the balloon will be 184,320 feet from the lift off point (25.6 ft/sec multiplied by 7200

seconds in two hours).

b. After two hours, the horizontal displacement will be 158,400 feet (22ft/sec multiplied by 7200 seconds

in two hours).

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c. After two hours, the balloon will have risen 93,600 feet, from part a. After an additional two hours of

floating (horizontally only) in the 22ft/sec wind, the balloon will have traveled 316,800 feet horizontally

(22ft/second times 14,400 seconds in four hours). This horizontal distance must be added to the original

horizontal distance the balloon moved in the wind while rising, which was 158,400 ft.

We must recalculate our resultant vector using Pythagorean Theorem.

x =

√

936002 + (316800 + 158400)2 = 484330 f t.

The balloon has traveled 484330 feet from its lift off point. The crew will have to travel 475200 feet or 90

miles (horizontal displacement) to be under the balloon when it lands.

Points to Consider

• Is it possible to find the magnitude and direction of resultants without using a protractor and ruler

and without using right triangles?

• How can we use the Law of Cosines and the Law of Sines to help us find magnitude and direction of

resultants?

Review Questions

1. Vectors ~m and ~n are perpendicular. Make a diagram of each addition, find the magnitude and

direction (with respect to ~m and ~n) of their resultant if:

(a) |~m| = 29.8|~n| = 37.7

(b) |~m| = 2.8|~n| = 5.4

(c) |~m| = 11.9|~n| = 9.4

2. For ~a, ~b,~c, and ~d below, make a diagram of each addition or subtraction. |~a| = 6cm, direction = 45◦

|~b| = 3.2cm, direction = 30◦

|~c| = 1.3cm, direction = 110◦

|~d| = 4.8cm, direction = 80◦

(a) ~a+ ~b

(b) ~a+ ~d

(c) ~c+ ~d

(d) ~a − ~d

(e) ~b − ~a

(f) ~d − ~c

3. Does |~a+ ~b| = |~a|+ |~b|? Explain your answer.

4. A plane is traveling north at a speed of 225 mph while an easterly wind is blowing the plane west at

18 mph. What is the direction and the speed of the plane?

5. Two workers are pulling on ropes attached to a tree stump. One worker is pulling the stump east

with 330 Newtons of forces while the second working is pulling the stump north with 410 Newtons

of force. Find the magnitude and direction of the resultant force on the tree stump.

6. Assume ~a is in standard position. For each terminal point is given, find the magnitude and direction

of each vector.

(a) (12, 18)

(b) (-3, 6)

7. Given the initial and terminal coordinates of ~a, find the magnitude and direction.

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(a) initial ( 2, 4) terminal (8, 6)

(b) initial (5, -2) terminal (3, 1)

8. The magnitudes of vectors ~a and ~b are given along with the angle between them theta. Find the

magnitude of the resultant and the angle it makes with a.

(a) |~a| = 31, |~b| = 31, θ = 132◦

(b) |~a| = 29, |~b| = 44, θ = 26◦

Review Answers

1. For each problem below, use the Pythagorean Theorem to find the magnitude and tan θ = |~n||~m|

(a) magnitude = 48.1, direction = 51.7◦

(b) magnitude = 6.1, direction = 62.6◦

(c) magnitude = 15.2, direction = 38.3◦

2. (a)

(b)

(c)

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(d)

(e)

(f)

3. This is always true because magnitude cannot be negative.

4. Speed (magnitude):

√

182 + 2252 = 225.7 and its direction is tan θ = 18225 = 4.6◦NE.

5. The magnitude is

√

3302 + 4102 = 526.3 Newtons and the direction is tan−1

(

410

330

)

= 51.2◦NE.

6. (a) |~a| = √122 + 182 = 21.6, direction = tan−1

(

18

12

)

= 56.3◦.

(b) |~a| = √(−3)2 + 62 = 6.7, direction = tan−1 ( 6−3 ) = 116.6◦.

7. (a) |~a| = √(2 − 8)2 + (4 − 6)2 = 6.3, direction = tan−1 (4−62−8 ) = 18.4◦.

(b) |~a| = √(5 − 3)2 + (−2 − 1)2 = 3.6, direction = tan−1 (−2−15−3 ) = 123.7◦.

8. In both a and b, we have the SAS case, so you can do the Law of Cosines, followed by the Law of

Sines.

(a) (~a+ ~b)2 = 312 + 312 − 2(31)(31) cos 132, ~a+ ~b = 56.6, sin 13256.6 = sin x31 , x = 24◦

(b) (~a+ ~b)2 = 292 + 442 − 2(29)(44) cos 26, ~a+ ~b = 22, sin x44 = sin 2622 , x = 61.3◦

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5.7 Component Vectors

Learning Objectives

• Perform scalar multiplication with vectors.

• Find the resultant as a sum of two components.

• Find the resultant as magnitude and direction.

• Use component vectors to solve real-world and applied problems.

A car has traveled 216 miles in a direction of 46◦ north of east. How far east of its initial point has it

traveled? How far north has the car traveled?

The car traveled on a vector distance called a displacement. It moved in a line to a particular distance

from the starting point. Having two components in their expression, vectors are confusing to some. A

diagram helps sort out confusion. Looking at vectors by separating them into components allows us to deal

with many real-world problems. The components often relate to very different elements of the problem,

such as wind speed in one direction and speed supplied by a motor in another.

In order to find how far the car has traveled east and how far it has traveled north, we will need to find

the horizontal and vertical components of the vector. To find ~x, we use cosine and to find ~y we use sine.

cos 46 = |~x|

216

=

x

216

sin 46 = |~y|

216

=

y

216

cos 46 = x

216

sin 46 = y

216

216 cos 46 = x 216 sin 46 = y

x = 150.0 y = 155.4

In this section, we will learn about component vectors and how to find them. We will also explore other

ways of finding the magnitude and direction of a resultant of two or more vectors. We will be using many

of the tools we learned in the previous sections dealing with right and oblique triangles.

Vector Multiplied by a Scalar

In working with vectors there are two kinds of quantities employed. The first is the vector, a quantity

that has both magnitude and direction. The second quantity is a scalar. Scalars are just numbers. The

magnitude of a vector is a scalar quantity. A vector can be multiplied by a real number. This real number

is called a scalar. The product of a vector ~a and a scalar k is a vector, written ~ka. It has the same direction

as ~a with a magnitude of k|~a| if k > 0. If k < 0, the vector has the opposite direction of ~a and a magnitude

of k|~a|.

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Example 1: The speed of the wind before a hurricane arrived was 20 mph from the SSE (135◦ on the

compass). It quadrupled when the hurricane arrived. What is the current vector for wind velocity?

Solution: The wind is coming now at 80 mph from the same direction.

Example 2: A sailboat was traveling at 15 knots due north. After realizing he had overshot his destination,

the captain turned the boat around and began traveling twice as fast due south. What is the current velocity

vector of the ship?

Solution: The ship is traveling at 30 knots in the opposite direction.

If the vector is expressed in coordinates with the starting end of the vector at the origin, this is called

standard form. To perform a scalar multiplication, we multiply our scalar by both the coordinates of

our vector. The word scalar comes from “scale.” Multiplying by a scalar just makes the vectors longer or

shorter, but doesn’t change their direction.

Example 3: Consider the vector from the origin to (4, 6). What would the representation of a vector

that had three times the magnitude be?

Solution: Here k = 3 and ~v is the directed segment from (0,0) to (4, 6).

Multiply each of the components in the vector by 3.

~kv = (0, 0) to (12, 18)

The new coordinates of the directed segment are (0, 0), (12, 18).

Example 4: Consider the vector from the origin to (3, 5). What would the representation of a vector

that had -2 times the magnitude be?

Solution: Here, k = −2 and ~v is the directed segment from (0, 0) to (3, 5).

~kv = (−2(3),−2(5)) = (−6,−10)

Since k < 0, our result would be a directed segment that is twice and long but in the opposite direction of

our original vector.

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Translation of Vectors and Slope

What would happen if we performed scalar multiplication on a vector that didn’t start at the origin?

Example 5: Consider the vector from (4, 7) to (12, 11). What would the representation of a vector that

had 2.5 times the magnitude be?

Solution: Here, k = 2.5 and ~v = the directed segment from (4, 7) to (12, 11).

Mathematically, two vectors are equal if their direction and magnitude are the same. The positions of

the vectors do not matter. This means that if we have a vector that is not in standard position, we can

translate it to the origin. The initial point of ~v is (4, 7). In order to translate this to the origin, we would

need to add (-4, -7) to both the initial and terminal points of the vector.

Initial point: (4, 7) + (−4,−7) = (0, 0)

Terminal point: (12, 11) + (−4,−7) = (8, 4)

Now, to calculate ~kv :

~kv = (2.5(8), 2.5(4))

~kv = (20, 10)

The new coordinates of the directed segment are (0, 0) and (20, 10). To translate this back to our original

terminal point:

Initial point: (0, 0) + (4, 7) = (4, 7)

Terminal point: (20, 10) + (4, 7) = (24, 17)

The new coordinates of the directed segment are (4, 7) and (24, 17).

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Vectors with the same magnitude and direction are equal. This means that the same ordered pair could

represent many different vectors. For instance, the ordered pair (4, 8) can represent a vector in standard

position where the initial point is at the origin and the terminal point is at (4, 8). This vector could be

thought of as the resultant of a horizontal vector with a magnitude or 4 units and a vertical vector with a

magnitude of 8 units. Therefore, any vector with a horizontal component of 4 and vertical component of

8 could also be represented by the ordered pair (4, 8).

If you think back to Algebra, you know that the slope of a line is the change in y over the change in x, or

the vertical change over the horizontal change. Looking at our vectors above, since they all have the same

horizontal and vertical components, they all have the same slope, even though they do not all start at the

origin.

Unit Vectors and Components

A unit vector is a vector that has a magnitude of one unit and can have any direction. Traditionally î

(read “i hat”) is the unit vector in the x direction and ĵ (read “ j hat”) is the unit vector in the y direction.

|̂i| = 1 and | ĵ| = 1. Unit vectors on perpendicular axes can be used to express all vectors in that plane.

Vectors are used to express position and motion in three dimensions with k̂ (“k hat”) as the unit vector

in the z direction. We are not studying 3D space in this course. The unit vector notation may seem

burdensome but one must distinguish between a vector and the components of that vector in the direction

of the x− or y−axis. The unit vectors carry the meaning for the direction of the vector in each of the

coordinate directions. The number in front of the unit vector shows its magnitude or length. Unit vectors

are convenient if one wishes to express a 2D or 3D vector as a sum of two or three orthogonal components,

such as x− and y−axes, or the z−axis.

Component vectors of a given vector are two or more vectors whose sum is the given vector. The sum

is viewed as equivalent to the original vector. Since component vectors can have any direction, it is useful

to have them perpendicular to one another. Commonly one chooses the x and y axis as the basis for the

unit vectors. Component vectors do not have to be orthogonal.

A vector from the origin (0, 0) to the point (8, 0) is written as 8̂i. A vector from the origin to the point

(0, 6) is written as 6 ĵ.

The reason for having the component vectors perpendicular to one another is that this condition allows

us to use the Pythagorean Theorem and trigonometric ratios to find the magnitude and direction of the

components. One can solve vector problems without use of unit vectors if specific information about

orientation or direction in space such as N, E, S or W are not part of the problem.

Resultant as the Sum of Two Components

We can look at any vector as the resultant of two perpendicular components. If we generalize the figure

above, |~r|̂i is the horizontal component of a vector ~q and |~s| ĵ is the vertical component of ~q. Therefore ~r is

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a magnitude, |~r|, times the unit vector in the x direction and ~s is its magnitude, |~s|, times the unit vector

in the y direction. The sum of ~r plus ~s is: ~r + ~s = ~q. This addition can also be written as |~r|̂i+ |~s| ĵ = ~q.

If we are given the vector ~q, we can find the components of ~q,~r, and ~s using trigonometric ratios if we know

the magnitude and direction of ~q.

Example 6: If |~q| = 19.6 and its direction is 73◦, find the horizontal and vertical components.

Solution: If we know an angle and a side of a right triangle, we can find the other remaining sides using

trigonometric ratios. In this case, ~q is the hypotenuse of our triangle, ~r is the side adjacent to our 73◦

angle, ~s is the side opposite our 73◦ angle, and ~r is directed along the x−axis.

To find ~r, we will use cosine and to find ~s we will use sine. Notice this is a scalar equation so all quantities

are just numbers. It is written as the quotient of the magnitudes, not the vectors.

cos 73 = |~r||~q| =

r

q

sin 73 = |~s||~q| =

s

q

cos 73 = r

19.6

sin 73 = s

19.6

r = 19.6 cos 73 s = 19.6 sin 73

r = 5.7 s = 18.7

The horizontal component is 5.7 and the vertical component is 18.7. One can rewrite this in vector notation

as 5.7̂i + 18.7 ĵ = ~q. The components can also be written ~q =

〈

5.7, 18.7

〉

, with the horizontal component

first, followed by the vertical component. Be careful not to confuse this with the notation for plotted

points.

Resultant as Magnitude and Direction

If we don’t have two perpendicular vectors, we can still find the magnitude and direction of the resultant

without a graphic estimate with a construction using a compass and ruler. This can be accomplished using

both the Law of Sines and the Law of Cosines.

Example 7: ~A makes a 54◦ angle with ~B. The magnitude of ~A is 13.2. The magnitude of ~B is 16.7. Find

the magnitude and direction the resultant makes with the smaller vector.

There is no preferred orientation such as a compass direction or any necessary use of x and y coordinates.

The problem can be solved without the use of unit vectors.

Solution: In order to solve this problem, we will need to use the parallelogram method. Since vectors only

have magnitude and direction, one can move them on the plane to any position one wishes, as long as the

magnitude and direction remain the same. First, we will complete the parallelogram: Label the vectors.

Move ~b so its tail is on the tip of ~a. Move ~a so its tail is on the tip of ~b. This makes a parallelogram because

the angles did not change during the translation. Put in labels for the vertices of the parallelogram.

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Since opposite angles in a parallelogram are congruent, we can find angle A.

∠CBD+ ∠CAD+ ∠ACB+ ∠BDA = 360

2∠CBD+ 2∠ACB = 360

∠ACB = 54◦

2∠CBD = 360 − 2(54)

∠CBD =

360 − 2(54)

2

= 126

Now, we know two sides and the included angle in an oblique triangle. This means we can use the Law of

Cosines to find the magnitude of our resultant.

x2 = 13.22 + 16.72 − 2(13.2)(16.7) cos 126

x2 = 712.272762

x = 26.7

To find the direction, we can use the Law of Sines since we now know an angle and a side across from it.

We choose the Law of Sines because it is a proportion and less computationally intense than the Law of

Cosines. sin θ

16.7

=

sin 126

26.7

sin θ = 16.7 sin 126

26.7

sin θ = 0.5060143748

θ = sin−1 0.5060 = 30.4◦

The magnitude of the resultant is 26.7 and the direction it makes with the smaller vector is 30.4◦ counter-

clockwise.

We can use a similar method to add three or more vectors.

Example 8: Vector A makes a 45◦ angle with the horizontal and has a magnitude of 3. Vector B makes

a 25◦ angle with the horizontal and has a magnitude of 5. Vector C makes a 65◦ angle with the horizontal

and has a magnitude of 2. Find the magnitude and direction (with the horizontal) of the resultant of all

three vectors.

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Solution: To begin this problem, we will find the resultant using Vector A and Vector B. We will do this

using the parallelogram method like we did above.

Since Vector A makes a 45◦ angle with the horizontal and Vector B makes a 25◦ angle with the horizontal,

we know that the angle between the two (∠ADB) is 20◦.

To find ∠DBE:

2∠ADB+ 2∠DBE = 360

∠ADB = 20◦

2∠DBE = 360 − 2(20)

∠DBE =

360 − 2(20)

2

= 160

Now, we will use the Law of Cosines to find the magnitude of DE.

DE2 = 32 + 52 − 2(3)(5) cos 160

DE2 = 62

DE = 7.9

Next, we will use the Law of Sines to find the measure of angle EDB.

sin 160

7.9

=

sin ∠EDB

3

sin ∠EDB = 3 sin 160

7.9

sin ∠EDB = .1299

∠EDB = sin−1 0.1299 = 7.46◦

We know that Vector B forms a 25◦ angle with the horizontal so we add that value to the measure of ∠EDB

to find the angle DE makes with the horizontal. Therefore, DE makes a 32.46◦ angle with the horizontal.

Next, we will take DE, and we will find the resultant vector of DE and Vector C from above. We will

repeat the same process we used above.

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Vector C makes a 65◦ angle with the horizontal and DE makes a 33◦ angle with the horizontal. This means

that the angle between the two (∠CDE) is 32◦. We will use this information to find the measure of ∠DEF.

2∠CDE + 2∠DEF = 360

∠CDE = 32◦

2∠DEF = 360 − 2(32)

∠DEF =

360 − 2(32)

2

= 148

Now we will use the Law of Cosines to find the magnitude of DF.

DF2 = 7.92 + 22 − 2(7.9)(2) cos 148

DF2 = 93.2

DF = 9.7

Next, we will use the Law of Sines to find ∠FDE.

sin 148

9.7

=

sin ∠FDE

2

sin ∠FDE = 2 sin 148

9.7

sin ∠FDE = .1093

∠FDE = sin−1 0.1093 = 6◦

Finally, we will take the measure of ∠FDE and add it to the 33◦ angle that DE forms with the horizontal.

Therefore, DF forms a 39◦ angle with the horizontal.

Example 9: Two forces of 310 lbs and 460 lbs are acting on an object. The angle between the two forces

is 61.3◦. What is the magnitude of the resultant? What angle does the resultant make with the smaller

force?

Solution: We do not need unit vectors here as there is no preferred direction like a compass direction or a

specific axis. First, to find the magnitude we will need to figure out the other angle in our parallelogram.

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2∠ACB+ 2∠CAD = 360

∠ACB = 61.3◦

2∠CAD = 360 − 2(61.3)

∠CAD =

360 − 2(61.3)

2

= 118.7

Now that we know the other angle, we can find the magnitude using the Law of Cosines.

x2 = 4602 + 3102 − 2(460)(310) cos 118.7◦

x2 = 444659.7415

x = 667

To find the angle the resultant makes with the smaller force, we will use the Law of Sines.

sin θ

460

=

sin 118.7

666.8

sin θ = 460 sin 118.7

666.8

sin θ = .6049283888

θ = sin−1 0.6049 = 37.2◦

Example 10: Two trucks are pulling a large chunk of stone. Truck 1 is pulling with a force of 635 lbs

at a 53◦ angle from the horizontal while Truck 2 is pulling with a force of 592 lbs at a 41◦ angle from the

horizontal. What is the magnitude and direction of the resultant force?

Solution: Since Truck 1 has a direction of 53◦ and Truck 2 has a direction of 41◦, we can see that the

angle between the two forces is 12◦. We need this angle measurement in order to figure out the other angles

in our parallelogram.

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2∠ACB+ 2∠CAD = 360

∠ACB = 12◦

2∠CAD = 360 − 2(12)

∠CAD =

360 − 2(12)

2

= 168

Now, use the Law of Cosines to find the magnitude of the resultant.

x2 = 6352 + 5922 − 2(635)(592) cos 168◦

x2 = 1489099

x = 1220.3 lbs

Now to find the direction we will use the Law of Sines.sin θ

635

=

sin 168

1220.3

sin θ = 635 sin 168

1220.3

sin θ = 0.1082

θ = sin−1 0.1082 = 6◦

Since we want the direction we need to add the 6◦ to the 41◦ from the smaller force. The magnitude is

1220 lbs and 47◦ counterclockwise from the horizontal.

Example 11: Two tractors are being used to pull down the framework of an old building. One tractor

is pulling on the frame with a force of 1675 pounds and is headed directly north. The second tractor is

pulling on the frame with a force of 1880 pounds and is headed 33◦ north of east. What is the magnitude

of the resultant force on the building? What is the direction of the result force?

Soultion: We are asked to find the resultant force and direction, which means we are dealing with vectors.

In order to complete our diagram, we will need to connect our two vectors and draw in our resultant. We

will refer to the magnitude of our resultant as x and the direction of our resultant as θ.

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When finding the resultant of two vectors, we can choose from either the triangle method or the parallel-

ogram method. We will solve this problem using the parallelogram method. Looking at the diagram, we

can see that the two vectors form an angle of 57, (90 - 33). This means that the angle opposite the angle

formed by our two vectors is also 57. To find the other two angles in our parallelogram, we know that the

sum of all the angles must add up to 360 and that opposite angles must be congruent, 360−(57+57)2 = 123.

Now, we can use two sides of our parallelogram and our resultant to form a triangle in which we know two

sides and the included angle (SAS).

This means that we can use the Law of Cosines to find the magnitude (x) of the resultant.

x2 = 16752 + 18802 − 2(1675)(1880) cos 123

x2 = 9770161.643

x = 3125.7

To find the direction (θ), we can use the Law of Sines since we now know an angle and the side opposite it.

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sin 123

3125.7

=

sin θ

1675

1675 sin 123

3125.7

= sin θ

0.449427 = sin θ

26.71 = θ

Now that we know θ, in order to find the angle of the resultant, we must add the 33◦ from the x−axis to

θ, 33◦ + 26.71◦ = 59.71◦.

Points to Consider

• How you can verify if your answers to problems involving vectors that are not perpendicular are

correct?

• In what ways are solving problems with oblique triangles and solving problems involving vectors

similar?

• In what ways are they different?

• When is it appropriate to use vectors instead of oblique triangles to solve problems?

• When is it helpful to use unit vectors? When can one solve a problem without explicitly using them?

Review Questions

1. Find the resulting ordered pair that represents ~a in each equation if you are given and ~b = (0, 0) to (5, 4)

and ~c = (0, 0) to (−3, 7).

(a) ~a = 2~b

(b) ~a = −12~c

(c) ~a = 0.6~b

(d) ~a = −3~b

2. Find the magnitude of the horizontal and vertical components of the following vectors given that the

coordinates of their initial and terminal points.

(a) initial = (−3, 8) terminal = (2,−1)

(b) initial = (7, 13) terminal = (11, 19)

(c) initial = (4.2,−6.8) terminal = (−1.3,−9.4)

3. Find the magnitude of the horizontal and vertical components if the resultant vector’s magnitude

and direction are given.

(a) magnitude = 75 direction = 35◦

(b) magnitude = 3.4 direction = 162◦

(c) magnitude = 15.9 direction = 12◦

4. Two forces of 8.50 Newtons and 32.1 Newtons act on an object at right angles. Find the magnitude

of the resultant and the angle that it makes with the smaller force.

5. Forces of 140 Newtons and 186 Newtons act on an object. The angle between the forces is 43◦. Find

the magnitude of the resultant and the angle it makes with the larger force.

6. An incline ramp is 12 feet long and forms an angle of 28.2◦ with the ground. Find the horizontal and

vertical components of the ramp.

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7. An airplane is traveling at a speed of 155 km/h. It needs to head in a direction of 83◦ while there is

a 42.0 km/h wind from 325◦. What should the airplane’s heading be?

8. A speedboat is capable of traveling at 10.0 mph, but is in a river that has a current of 2.00 mph. In

order to cross the river at right angle, in what direction should the boat be heading?

9. If −→AB is any vector, what is −→AB+ −→BA?

Review Answers

1. (a) 2~b = 2

〈

5, 4

〉

=

〈

10, 8

〉

= 10i+ 8 j

(b) −12~c = −12

〈

−3, 7

〉

=

〈

1.5,−3.5

〉

= 1.5i − 3.5 j

(c) 0.6~b = 0.6

〈

5, 4

〉

=

〈

3, 2.4

〉

= 3i+ 2.4 j

(d) −3~b = −3

〈

5, 4

〉

=

〈

−15,−12

〉

= −15i − 12 j

2. All of these need to be translated to (0,0). Also, recall that magnitudes are always positive. (a)

(−3, 8) + (3,−8) = (0, 0) (2,−1) + (3,−8) = (5,−9)

horizontal = 5, vertical = 9

(b) (7, 13) + (−7,−13) = (0, 0) (11, 19) + (−7,−13) = (4, 6)

horizontal = 4, vertical = 6

(c) (4.2,−6.8) + (−4.2, 6.8) = (0, 0) (−1.3,−9.4) + (−4.2, 6.8) = (−5.5,−2.6)

horizontal = 5.5, vertical = 2.6

3. (a) cos 35◦ = x75 , sin 35◦ =

y

75 , x = 61.4, y = 43

(b) cos 162◦ = x3.4 , sin 162◦ =

y

3.4 , x = 1.1, y = 3.2

(c) cos 12◦ = x15.9 , sin 12◦ =

y

15.9 , x = 15.6, y = 3.3

4. magnitude = 33.2, direction = 75.2◦ from the horizontal

5. magnitude = 304, 18.3◦ between resulant and larger force

6. y = 12 sin 28.2◦ = 5.7, x = 12 cos 28.2◦ = 10.6

7. Recall that headings and angles in triangles are complementary. So, an 83◦ heading translates to 7◦

from the horizontal. Adding that to 35◦ (325◦ from 360◦) we get 42◦ for two of the angles in the

parallelogram. So, the other angles in the parallelogram measure 138◦ each, 360−2(42)2 . Using 138◦

in the Law of Cosines, we can find the diagonal or resultant, x2 = 422 + 1552 − 2(42)(155) cos 138,

so x = 188.3. We then need to find the angle between the resultant and the speed using the Law of

Sines. sin a42 = sin 138188.3 , so a = 8.6◦. To find the correct heading, this number needs to be added to 90◦,

getting 98.6◦.

8. The heading is just tan θ = 210 , or 11.3◦ against the current.

9. BA is the same vector as AB, but because it starts with B it is in the opposite direction. Therefore,

when you add the two together, you will get (0,0).

5.8 Chapter Review

Chapter Summary

This chapter has taught us how to solve any kind of triangle, using the Law of Cosines, Law of Sines and

vectors. We also discovered two additional formulas for finding the area, Heron’s Formula and 12 bc sin A.

Then, the ambiguous case for the Law of Sines was introduced. This is when you are given two sides and

the non-included angle and have to solve for another angle in the triangle. There can be no, one or two

solutions and you need to compare the two given sides to determine which option it is. Finally, vectors

were discussed. We learned how to add, subtract and multiply them by a scalar. Vectors are very useful

for representing speed, wind velocity and force.

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Vocabulary

Ambiguous case A situation that occurs when applying the Law of Sines in an oblique triangle when two

sides and a non-included angle are known. The ambiguous case can yield no solution, one solution,

or two solutions.

component vectors Two or more vectors whose vector sum, the resultant, is the given vector. Compo-

nents can be on axes or more generally in space.

directed line segment A line segment having both magnitude and direction, often used to represent a

vector.

displacement When an object moves a certain distance in a certain direction.

equal vectors Vectors with the same magnitude and direction.

force When an object is pushed or pulled in a certain direction.

Heron’s Formula A formula used to calculate the area of a triangle when all three sides are known.

included angle The angle in between two known sides of a triangle.

included side The side in between two known sides of a triangle.

initial point The starting point of a vector

Law of Cosines A general statement relating the lengths of the sides of a general triangle to the cosine

of one of its angles.

Law of Sines A statement about the relationship between the sides and the angles in any triangle.

magnitude Length of a vector.

negative vector A vector with the same magnitude as the original vector but with the opposite direction.

non-included angle An angle that is not in between two known sides of a triangle.

non-included side A side that is not in between two known sides of a triangle.

oblique triangle A non-right triangle.

resultant The sum of two or more vectors

scalar A real number by which a vector can be multiplied. The magnitude of a vector is always a scalar.

standard position A vector with its initial point at the origin of a coordinate plane.

terminal point The ending point of a vector.

unit vector A vector that has a magnitude of one unit. These generally point on coordinate axes.

vector Any quantity having magnitude and direction, often represented by an arrow.

velocity When an object travels at a certain speed in a certain direction.

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Review Questions

1. Use the Law of Cosines to determine whether or not the following triangle is drawn accurately. If

not, determine how much ∠B is off by.

2. An artist is making a large sculpture for the lobby of a new building. She has drawn out what she

wants the sculpture to look like at the left. If she wants BC = 51.4 f eet, BD = 32.6 f eet, AD = 37.3 f eet

and ∠DBC = 27◦, verify that the length of AB would be 34.3 feet. If not, figure out the correct

measure.

3. A family’s farm plot is a trapezoid with dimensions: the longer leg is 3,000 ft and the shortest side

is 2,100 ft. The other leg is 2,400 ft. The shorter diagonal is 2,200 ft. What is the area of the land

in square feet?

4. A biomechanics class is designing a functioning artificial arm for an adult. They are using a hydraulic

cylinder (fluid filled) to be the bicep’s muscle. The elbow is at point E. The forearm dimension EH

is 24 cm. The upper arm dimension EA is 21 cm. The cylinder attaches from the top of the upper

arm at point A and to a point on the lower arm 4 cm from the mechanical elbow at point B. When

fluid is pumped out of the cylinder the distance AB is shortened. The forearm goes up raising the

hand at point H.

Some fluid is pumped out of the cylinder to make the distance AB 5 cm shorter. What is the new

angle of the arm, ∠AEH?

5. For each figure below, use the Law of Sines, the Law of Cosines, or the the trig functions to solve for

x.

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(a)

(b)

(c)

(d)

6. A surveyor has the job of determining the distance across the Palo Duro Canyon in Amarillo, Texas,

the second largest canyon in the United States. Standing on one side of the canyon, he measures the

angle formed by the edge of the canyon and the line of sight to a large boulder on the other side of

the canyon. He then walks 12 ft and measures the angle formed by the edge of the canyon and the

new line of sight to the boulder.

(a) If the first angle formed is 29◦ and the second angle formed is 3◦, find the distance across the

canyon.

(b) The surveyor spots another boulder while he is at his second spot, and finds that it forms a 37◦

angle with his line of sight. He then walks 15 feet further and finds that the boulder forms a

25◦ angle with the line of sight. What is the distance between the two boulders?

7. Two cell phone companies have towers along Route 47. Company 1’s tower is 38 miles from one point

on Route 47 and 47 miles from another point. This tower’s signal forms a 72.8◦ angle. Company 2’s

tower is 58 miles from one point of Route 47 and 59 miles from another. Company 2’s signal forms

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a 12◦ angle with the road at the point that is 58 miles from the tower. For how many miles would

a person driving along Route 47 have service with Company A? Company B? If the signals start to

overlap 32 miles from Tower 1 (along the same line as the 47 mile), how long does the two coverages

overlap over the highway?

8. Find the magnitude and direction of each resultant vector (addition). ~m and ~n are perpendicular.

(a) |~m| = 48.3, |~n| = 47.6

(b) |~m| = 18.6, |~n| = 17.5

9. Given the initial and terminal coordinates of ~a, find the magnitude and direction.

(a) initial (-4, 19) terminal (12, 1)

(b) initial (11, -21) terminal (21, -11)

10. A golfer tees off at the 16th hole. The pin is 425 yards from tee-off and his ball is 16◦ off of the

straight line to the hole. If his ball is 137 yards from the hole, how far did he hit the ball?

11. Street A runs north and south and intersects with Street B, which runs east and west. Street C

intersects both A and B, and it intersects Street A at a 36◦ angle. There are stoplights at each of

these intersections. If the distance between the two stoplights on Street C is 0.5miles, what is the

distance between the two stoplights on Street A?

12. During a baseball game, a ball is hit into right field. The angle from the ball to home to 2nd base is

18◦. The angle from the ball to 2nd to home is 127◦. The distance from home to 2nd base is 127.3 ft.

How far was the ball hit? How far is the 2nd baseman from the ball?

13. The military is testing out a new infrared sensor that can detect movement up to thirty miles away.

Will the sensor be able to detect the second target? If not, how far out of the range of the sensor is

Target 2?

14. An environmentalist is sampling the water in a local lake and finds a strain of bacteria that lives on

the surface of the lake. In a one square foot area, he found 5.2× 1013 bacteria. There are three docks

in a certain section of the lake. If Dock 3 is 396 ft from Dock 1, how many bacteria are living on the

surface of the water between the three docks?

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Reviwew Answers

1. Use Law of Cosines to solve for angle B. 17.62 = 152 + 20.92 − 2(15)(20.9) cos B, B = 55.8◦, which

means the picture was drawn accurately.2.

CD2 = 32.62 + 51.42 − 2(32.6)(51.4) cos 27,CD = 26.8

∠C, ∠C = 33.5◦

AB2 = 64.12 + 51.42 − 2(64.1)(51.4) cos 33.5, AB = 35.4

This means that the artist’s drawing is off by 1.1 foot.

3. The area of a trapezoid is A = 12h(b1 + b2). In the problem, we were given both bases, a side and a

diagonal. So, we need to find the height. In order to do this, we first need to find the angle between

2200 and 2400(x), use that to find it’s complement, then we can take the cosine to find the height.

21002 = 24002 + 22002 − 2(2400)(2200) cos x

x = 54.1◦ → 90◦ − 54.1◦ = 35.9◦

cos 35.9◦ = h

2200

→ h = 1782.1

A =

1

2

(1782.1)(2400 + 3000) = 4, 811, 670 sq. f t.

4. AB2 = 42 + 212 − 2(4)(21) cos 120, AB = 23.3 minus 5, new AB is 18.3 new ∠E is 18.32 = 42 + 212 −

2(4)(21) cos E, ∠E = 42.6◦

5. (a) x2 = 13.32 + 17.62 − 2(13.3)(17.6) cos 132, x = 28.3

(b) cos 52 = x19.3 , x = 11.9

(c) sin 4218.6 = sin x15.9 , x = 34.9◦

(d) sin 4692.7 = sin 12x , x = 26.8

6. (a) We will call the distance across the canyon h. sin 3212 = sin 87x , x = 22.6. sin 61 = h22.6 , 19.8.

(b) We will call the distance between the two boulders y. y = a + b, where a is the distance from

boulder 1 to the first stop (that the surveyor took) and b is the distance from that spot to

boulder 2.

tan 87 = 19.8

a

, a = 1.0,

sin 62

15

=

sin 65

z

, z = 15.4→ cos 53 = b

15.4

, b = 9.3.

So, a+ b = 1.0 + 9.3 = 10.3.

7. Company A coverage: x2 = 382 + 472 − 2(38)(47) cos 72.8, x = 51Company B coverage: sin x58 =sin 12

59 , x = 11.8

◦ → sin 1259 = sin 156.2b , b = 114.52

Overlap: sin a38 = sin 72.851 , a = 45.4◦ → sin 1215 = sin 122.6o , o = 60.8

8. (a) magnitude =

√

48.32 + 47.62 = 67.8, direction = tan θ = 47.648.3 → 44.6◦

(b) magnitude =

√

18.62 + 17.52 = 25.5, direction = tan θ = 47.648.3 → 43.3◦

9. (a) magnitude =

√

(19 − 1)2 + (−4 − 12)2 = 24.1, direction = tan θ = −1618 → 138.4◦

(b) magnitude =

√

(−21 + 11)2 + (11 − 21)2 = 14.1, direction = tan θ = −10−10 → 45◦

10. This is the SSA or ambiguous case. Because 137 > 425 sin 16◦, we will have two solutions or two

different lengths that the golfer could have hit the ball. sin 16137 = sin x425 , x = 58.8◦ or 121.2◦ (this is the

angle at the ball)

Case 1: Use 58.8◦, the angle at the pin is 105.2◦

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sin 105.2

d =

sin 16

137 , distance is 479.6 yards.

Case 2: Use 121.2◦, the angle at the pin is 42.8◦

sin 42.8

d =

sin 16

137 , distance is 337.7 yards.

It is safe to say that the golfer did not hit the ball 479.6 yards, considering that Tiger Woods longest

recorded drive is 425 yards. So, we can logically rule out Case 1 and our answer is that the golfer’s

drive was 337.7 yards.

11. cos 36◦ = x0.5 , x = 0.4 miles

12. The third angle in the triangle is 35◦, from the Triangle Sum Theorem.sin 35

127.3

=

sin 127

x

, x = 177.2

13. The distance between the tower and target 2 is: x2 = 372 +182 − 2(37)(18) cos 67.2◦, 34.3 miles. This

means that the second target is out of range by 4.3 miles.

14. We need to find area, use Heron’s Formula. s = 12(587 + 247 + 396) = 615

A =

√

615(615 − 587)(615 − 247)(615 − 396) = 37, 253.1, times 5.2 × 1013 = 1.9 × 1018 bacteria.

Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator

activities designed to supplement the objectives for some of the lessons in this chapter. See

http://www.ck12.org/flexr/chapter/9703.

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Chapter 6

The Polar System - 2nd edition

6.1 Polar Coordinates

Introduction

This chapter introduces and explores the polar coordinate system, which is based on a radius and theta.

Students will learn how to plot points and basic graphs in this system as well as convert x and y coordinates

into polar coordinates and vise versa. We will explore the different graphs that can be generated in the

polar system and also use polar coordinates to better understand different aspects of complex numbers.

Learning Objectives

• Distinguish between and understand the difference between a rectangular coordinate system and a

polar coordinate system.

• Plot points with polar coordinates on a polar plane.

Plotting Polar Coordinates

The graph paper that you have used for plotting points and sketching graphs has been rectangular grid

paper. All points were plotted in a rectangular form (x, y) by referring to a perpendicular x− and y−axis.

In this section you will discover an alternative to graphing on rectangular grid paper – graphing on circular

grid paper.

Look at the two options below:

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You are all familiar with the rectangular grid paper shown above. However, the circular paper lends itself

to new discoveries. The paper consists of a series of concentric circles-circles that share a common center.

The common center O, is known as the pole or origin and the polar axis is the horizontal line r that is

drawn from the pole in a positive direction. The point P that is plotted is described as a directed distance

r from the pole and by the angle that OP makes with the polar axis. The coordinates of P are (r, θ).

These coordinates are the result of assuming that the angle is rotated counterclockwise. If the angle were

rotated clockwise then the coordinates of P would be (r,−θ). These values for P are called polar coordinates

and are of the form P(r, θ) where r is the absolute value of the distance from the pole to P and θ is the

angle formed by the polar axis and the terminal arm OP.

Example 1: Plot the point A(5,−255◦) and the point B(3, 60◦)

Solution, A: To plot A, move from the pole to the circle that has r = 5 and then rotate 255◦ clockwise

from the polar axis and plot the point on the circle. Label it A.

Solution, B: To plot B, move from the pole to the circle that has r = 3 and then rotate 60◦ counter

clockwise from the polar axis and plot the point on the circle. Label it B.

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These points that you have plotted have r values that are greater than zero. How would you plot a polar

point in which the value of r is less than zero? How could you plot these points if you did not have polar

paper? If you were asked to plot the point (−1, 135◦) or

(

−1, 3pi4

)

you would rotate the terminal arm OP

counterclockwise 135◦ or 3pi4 . (Remember that the angle can be expressed in either degrees or radians). To

accommodate r = −1, extend the terminal arm OP in the opposite direction the number of units equal to

|r|. Label this point M or whatever letter you choose. The point can be plotted, without polar paper, as a

rotation about the pole as shown below.

The point is reflected across the pole to point.

There are multiple representations for the coordinates of a polar point P(r, θ). If the point P has polar

coordinates (r, θ), then P can also be represented by polar coordinates (r, θ+360k◦) or (−r, θ+[2k+1]180◦)

if θ is measured in degrees or by (r, θ + 2pik) or (−r, θ + [2k + 1]pi) if θ is measured in radians. Remember

that k is any integer and represents the number of rotations around the pole. Unless there is a restriction

placed upon θ, there will be an infinite number of polar coordinates for P(r, θ).

Example 2: Determine four pairs of polar coordinates that represent the following point P(r, θ) such that

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−360◦ ≤ θ ≤ 360◦.

Solution: Pair 1 → (4, 120◦). Pair 2 → (4,−240◦) comes from using k = −1 and (r, θ + 360◦k), (4, 120◦ +

360(−1)). Pair 3 → (−4, 300◦) comes from using k = 0 and (−r, θ+ [2k+1]180◦), (−4, 120◦+ [2(0)+1]180◦).

Pair 4 → (−4,−60◦) comes from using k = −1 and (−r, θ + [2k + 1]180◦), (−4, 120◦ + [2(−1) + 1]180◦).

These four pairs of polar coordinates all represent the same point P. You can apply the same procedure

to determine polar coordinates of points that have θ measured in radians. This will be an exercise for you

to do at the end of the lesson.

The Distance between Two Polar Coordinates

Just like the Distance Formula for x and y coordinates, there is a way to find the distance between

two polar coordinates. One way that we know how to find distance, or length, is the Law of Cosines,

a2 = b2 + c2 − 2bc cos A or a = √b2 + c2 − 2bc cos A. If we have two points (r1, θ1) and (r2, θ2), we can

easily substitute r1 for b and r2 for c. As for A, it needs to be the angle between the two radii, or (θ2 − θ1).

Finally, a is now distance and you have d =

√

r21 + r

2

2 − 2r1r2 cos(θ2 − θ1).

Example 3: Find the distance between (3, 60◦) and (5, 145◦).

Solution: After graphing these two points, we have a triangle. Using the new Polar Distance Formula,

we have d =

√

32 + 52 − 2(3)(5) cos 85◦ ≈ 5.6.

Example 4: Find the distance between (9,−45◦) and (−4, 70◦).

Solution: This one is a little trickier than the last example because we have negatives. The first

point would be plotted in the fourth quadrant and is equivalent to (9, 315◦). The second point would

be (4, 70◦) reflected across the pole, or (4, 250◦). Use these two values of θ for the formula. Also,

the radii should always be positive when put into the formula. That being said, the distance is d =√

92 + 42 − 2(9)(4) cos(315 − 250)◦ ≈ 8.16.

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Points to Consider

• How is the polar coordinate system similar/different from the rectangular coordinate system?

• How do you plot a point on a polar coordinate grid?

• How do you determine the coordinates of a point on a polar grid?

• How do you calculate the distance between two points that have polar coordinates?

Review Questions

1. Graph each point:

(a) M(2.5,−210◦)

(b) S

(

−3.5, 5pi6

)

(c) A

(

1, 3pi4

)

(d) Y

(

−5.25,−pi3

)

2. For the given point A

(

−4, pi4

)

, list three different pairs of polar coordinates that represent this point

such that −2pi ≤ θ ≤ 2pi.

3. For the given point B(2, 120◦), list three different pairs of polar coordinates that represent this point

such that −2pi < θ < 2pi.

4. Given P1 and P2, calculate the distance between the points.

(a) P1(1, 30◦) and P2(6, 135◦)

(b) P1(2,−65◦) and P2(9, 85◦)

(c) P1(−3, 142◦) and P2(10,−88◦)

(d) P1(5,−160◦) and P2(16,−335◦)

Review Answers

1. (a)

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(b)

(c)

(d)

2. (

−4, pi

4

)

all positive→

(

4,

5pi

4

)

both negative→

(

−4, −7pi

4

)

negative angle→

(

4,

−3pi

4

)

3.

(2, 120◦) negative angle only→ (2,−240◦)

negative radius only→ (−2, 300◦)

both negative→ (−2,−60◦)

4. Use P1P2 =

√

r21 + r

2

2 − 2r1r2 cos(θ2 − θ1).(a)

P1P2 =

√

12 + 62 − 2(1)(6) cos(135◦ − 30◦)

P1P2 ≈ 6.33 units(b)

P1P2 =

√

22 + 92 − 2(2)(9) cos 150◦

= 10.78(c)

P1P2 =

√

32 + 102 − 2(3)(10) cos(322 − 272)◦

= 8.39(d)

P1P2 =

√

52 + 162 − 2(5)(16) cos(200 − 25)◦

= 20.99

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6.2 Graphing Basic Polar Equations

Learning Objectives

• Graph polar equations.

• Graph and recognize limaçons and cardioids.

• Determine the shape of a limaçon from the polar equation.

Just as in graphing on a rectangular grid, you can also graph polar equations on a polar grid. These

equations may be simple or complex. To begin, you should try something simple like r = k or θ = k where

k is a constant. The solution for r = 1.5 is simply all ordered pairs such that r = 1.5 and θ is any real

number. The same is true for the solution of θ = 30◦. The ordered pairs will be any real number for r and

θ will equal 30◦. Here are the graphs for each of these polar equations.

Example 1: On a polar plane, graph the equation r = 1.5

Solution: The solution is all ordered pairs of (r, θ) such that r is always 1.5. This means that it doesn’t

matter what θ is, so the graph is a circle with radius 1.5 and centered at the origin.

Example 2: On a polar plane, graph the equation θ = 30◦

Solution: For this example, the r value, or radius, is arbitrary. θ must equal 30◦, so the result is a straight

line, with an angle of elevation of 30◦.

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To begin graphing more complicated polar equations, we will make a table of values for y = sin θ or in this

case r = sin θ. When the table has been completed, the graph will be drawn on a polar plane by using the

coordinates (r, θ).

Example 3: Create a table of values for r = sin θ such that 0◦ ≤ θ ≤ 360◦ and plot the ordered pairs.

(Note: Students can be directed to use intervals of 30◦ or allow them to create their own tables.)

Table 6.1:

θ 0◦ 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦

sin θ 0 0.5 0.9 1 0.9 0.5 0 -0.5 -0.9 -1 -0.9 -0.5 0

Remember that the values of sin θ are the r−values.

This is a sinusoid curve of one revolution.

We will now repeat the process for r = cos θ.

Example 4: Create a table of values for r = cos θ such that 0◦ ≤ θ ≤ 360◦ and plot the ordered pairs.

(Note: Students can be directed to use intervals of 30◦ or allow them to create their own tables.)

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Table 6.2:

θ 0◦ 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦

cos θ 1 0.9 0.5 0 -0.5 -0.9 -1 -0.9 -0.5 0 0.5 0.9 1

Remember that the values of cos θ are the r−values.

This is also a sinusoid curve of one revolution.

Notice that both graphs are circles that pass through the pole and have a diameter of one unit. These

graphs can be altered by adding a number to the function or by multiplying the function by a constant or

by doing both. We will explore the results of these alterations by first creating a table of values and then

by graphing the resulting coordinates (r, θ).

Example 5: Create a table of values for r = 2 + 3 sin θ such that 0 ≤ θ ≤ 2pi and plot the ordered pairs.

Remember that the values of 2 + 3 sin θ are the r−values.

Table 6.3:

θ 0 pi6

pi

3

pi

2

2pi

3

5pi

6 pi

7pi

6

4pi

3

3pi

2

5pi

3

11pi

6 2pi

2 +

3 sin θ

2.0 3.5 4.6 5.0 4.6 3.5 2.0 0.5 -0.6 -1.0 -0.6 0.5 2.0

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This sinusoid curve is called a limaçon. It has r = a± b sin θ or r = a± b cos θ as its polar equation. Not all

limaçons have the inner loop as a part of the shape. Some may curve to a point, have a simple indentation

(known as a dimple) or curve outward. The shape of the limaçon depends upon the ratio of ab where a is

a constant and b is the coefficient of the trigonometric function. In example 5, the ratio of ab = 23 which is

< 1. All limaçons that meet this criterion will have an inner loop.

Using the same format as was used in the examples above, the following limaçons were graphed. If you

like, you may create the table of values for each of these functions.

i) r = 4 + 3 cos θ such that 0 ≤ θ ≤ 2pi

a

b =

4

3 which is > 1 but < 2

This is an example of a dimpled limaçon.

ii) r = 4 + 2 sin θ such that 0 ≤ θ ≤ 2pi

a

b =

4

2 which is ≥ 2

This is an example of a convex limaçon.

Example 7: Create a table of values for r = 2 + 2 cos θ such that 0 ≤ θ ≤ 2pi and plot the ordered pairs.

Remember that the values of 2 + 2 cos θ are the r−values.

Table 6.4:

θ 0 pi6

pi

3

pi

2

2pi

3

5pi

6 pi

7pi

6

4pi

3

3pi

2

5pi

3

11pi

6 2pi

2 +

2 cos

4.0 3.7 3.0 2.0 1.0 0.27 0 .27 1.0 2.0 3.0 3.7 4.0

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This type of curve is called a cardioid. It is a special type of limaçon that has r = a+a cos θ or r = a+a sin θ

as its polar equation. The ratio of ab = 22 which is equal to 1.

Examples 3 and 4 were shown with θ measured in degrees while examples 5 and 7 were shown with θ

measured in radians. The results in the tables and the resulting graphs will be the same in both units.

Now that you are familiar with the limaçon and the cardioid, also called classical curves, it is time to

examine the polar pattern of the cardioid microphone. The polar pattern is modeled by the polar equation

r = 2.5+2.5 cos θ. The values of a and b are equal which means that the ratio ab = 1. Therefore the limaçon

will be a cardioid.

Create a table of values for r = 2.5 + 2.5 cos θ such that 0◦ ≤ θ ≤ 360◦ and graph the results.

Table 6.5:

θ 0◦ 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦

2.5+

2.5 cos θ

5.0 4.7 3.8 2.5 1.3 0.3 0 0.3 1.3 2.5 3.8 4.7 5.0

What does this pattern tell you about the cardioid microphone?

This pattern reveals that the microphone will pick up loud sounds behind it but softer sounds in front.

Transformations of Polar Graphs

Equations of limaçons have two general forms:

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r = a ± b sin θ and r = a ± b cos θ

The values of “a” and “b” will determine the shape of the graph and whether or not it passes through

the origin. When the values of “a” and “b” are equal, the graph will be a rounded heart-shape called a

cardioid. The general polar equation of a cardioid can be written as r = a(1 ± sin θ) and r = a(1 ± cos θ)

Example 8: Graph the following polar equations on the same polar grid and compare the graphs.

r = 5 + 5 sin θ r = 5 − 5 sin θ

r = 5(1 + sin θ) r = 5(1 − sin θ)

Solution:

The cardioid is symmetrical about the positive y−axis and the point of indentation is at the pole. The

result of changing + to - is a reflection in the x−axis. The cardioid is symmetrical about the negative

y−axis and the point of indentation is at the pole.

Changing the value of “a” to a negative did not change the graph of the cardioid.

Example 9: What effect will changing the values of a and b have on the cardioid if a > b? We can discover

the answer to this question by plotting the graph of r = 5 + 3 sin θ.

Solution:

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The cardioid is symmetrical about the positive y−axis and the point of indentation is pulled away from

the pole.

Example 8: What effect will changing the values of a and b or changing the function have on the cardioid

if a < b? We can discover the answer to this question by plotting the graph of r = 2 + 3 sin θ.

Solution:

The cardioid is now a looped limaçon symmetrical about the positive y−axis. The loop crosses the pole.

r = 2 + 3 cos θ

The cardioid is now a looped limaçon symmetrical about the positive x−axis. The loop crosses the pole.

Changing the function to cosine rotated the limaçon 90◦ clockwise.

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As you have seen from all of the graphs, transformations can be performed by making changes in the

constants and/or the functions of the polar equations.

Applications

In this subsection we will explore examples of real-world problems that use polar coordinates and polar

equations as their solutions.

Example 8: A local charity is sponsoring an outdoor concert to raise money for the children’s hospital.

To accommodate as many patrons as possible, they are importing bleachers so that all the fans will be

seated during the performance. The seats will be placed in an area such that −pi3 ≤ θ ≤ pi3 and 0 ≤ r ≤ 4,

where r is measured in hundreds of feet. The stage will be placed at the origin (pole) and the performer

will face the audience in the direction of the polar axis (r).

a. Create a polar graph of this area.

b. If all the seats are occupied and each seat takes up 5 square feet of space, how many people will be

seated in the bleachers?

Solution: Now that the region has been graphed, the next step is to calculate the area of this sector. To

do this, use the formula A = 12r2θ.

A =

1

2

r2θ

A =

1

2

(400)2

(2pi

3

)

A ≈ 167552 f t2.

167552 f t2. ÷ 5 f t2. ≈ 33510

The number of people in the bleachers is 33510.

Example 9: When Valentine’s Day arrives, hearts can be seen everywhere. As an alternative to purchasing

a greeting card, use a computer to create a heart shape. Write an equation that could be used to create

this heart and be careful to ensure that it is displayed in the correct position.

Solution: The classical curve that resembles a heart is a cordioid. You may have to experiment with the

equation to create a heart shape that is displayed in the correct direction. One example of an equation

that produces a proper heart shape is r = −2 − 2 sin θ.

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You can create other hearts by replacing the number 2 in the equation. Another equation is r = −3−3 sin θ.

Example 10: For centuries, people have been making quilts. These are frequently created by sewing a

uniform fabric pattern onto designated locations on the quilt. Using the equation that models a rose curve,

create three patterns that could be used for a quilt. Write the equation for each rose and sketch its graph.

Explain why the patterns have different numbers of petals. Can you create a sample quilt?

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Solution: The rose curve is a graph of the polar equation of the form r = a cos nθ or r = a sin nθ. If n is

odd, then the number of petals will be equal to n. If n is even, then the number of petals will be equal to

2n.

A Sample Quilt:

Graphin Polar Equations on the Calculator

You can use technology, the TI graphing calculator, to create these graphs. However, there are steps that

must be followed in order to graph polar equations correctly on the graphing calculator. We will go through

the step by step process to plot the polar equation r = 3 cos θ.

Example 11: Graph r = 3 cos θ using the TI-83 graphing calculator.

Solution: Press the MODE button. Scroll down to Func and over to highlight Pol. Also, while

on this screen, make sure that Radian is highlighted. Now you must edit the axes for the graph. Press

WINDOW 0 ENTER 2nd [pi] ENTER .05 ENTER (−) 4 ENTER 4 ENTER 1 ENTER (−) 3 ENTER 3 ENTER 1 ENTER .

When you have completed these steps, the screen should look like this:

The secondWINDOW shows part of the first screen since you had to scroll down to access the remaining

items.

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Enter the equation. Press Y = 3 cos X,T, θ, n) Press GRAPH .

Sometimes the polar equation you graph will look more like an ellipse than a circle. If this happens, press

ZOOM 5 to set a square viewing window. This will make the graph appear like a circle.

Review Questions

1. Name the classical curve in each of the following diagrams and explain why you feel you’re your

answer is correct. Also, find the equation of each curve.

(a)

(b)

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(c)

2. Graph each curve below. Comparing your answers from part one, determine if you can find a pattern

for how to find the equation of a classical curve from its graph.

(a) r = −3 − 3 cos θ

(b) r = 2 + 4 sin θ

(c) r = 4

(d) θ = pi2

(e) r = 5 + 3 cos θ

(f) r = −6 − 5 sin θ

3. Another classical curve we saw is called a rose and it is modeled by the function r = a cos nθ or

r = a sin nθ where n is any positive integer. Graph r = 4 cos 2θ and r = 4 cos 3θ. Is there a difference

in the curves? Explain.

4. Graph the roses below. Determine if you can find a pattern for how to find the equation of a rose

from its graph.

(a) r = 3 sin 4θ

(b) r = 2 sin 5θ

(c) r = 3 cos 3θ

(d) r = −4 sin 2θ

(e) r = 5 cos 4θ

(f) r = −2 cos 6θ

Review Answers

1. (a) a limaçon with an innerloop. y = 2 − 3 sin θ

(b) a cardioid y = 2 + 2 sin θ

(c) a dimpled limaçon y = 5 + 4.5 sin θ

2. (a) r = −3 − 3 cos θ

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(b) r = 2 + 4 sin θ

(c) r = 4

(d) θ = pi2

(e) r = 5 + 3 cos θ

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(f) r = −6 − 5 sin θ

vertical line on the “y” axis

To determine the equation of these curves, first notice that cosine curves are along the horizontal

axis and sine curves are along the vertical axis. Second, where the curve passes through the axis on

the non-dimpled side is a in r = a + b sin θ. b is a little harder to see, but it is the average of the

two intercepts where the curve crosses the axis on the dimpled axis. If there is an inner loop, use the

innermost value of the loop.

3.

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Table 6.6:

θ 0◦ 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦

4 cos 2θ 4 2 -2 -4 -2 2 4 2 -2 -4 -2 2 4

Table 6.7:

θ 0◦ 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦

4 cos 3θ 4 0 -4 0 4 0 -4 0 4 0 -4 0 4

In the graph of r = 4 cos 2θ, the rose has four petals on it but the graph of r = 4 cos 3θ has only three

petals. It appears, that if n is an even positive integer, the rose will have an even number of petals and if

n is an odd positive integer, the rose will have an odd number of petals.

4. (a) r = 3 sin 4θ

(b) r = 2 sin 5θ

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(c) r = 3 cos 3θ

(d) r = −4 sin 2θ

(e) r = 5 cos 4θ

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(f) r = −2 cos 6θ

For roses, the general equation is r = a sin nθ or r = a cos nθ. a indicates how long each petal is, and

depending on if n is even or odd indicates the number of pedals. If n is ood, there are n pedals and

if n is even there are 2n pedals.

6.3 Converting Between Systems

Learning Objectives

• Convert rectangular coordinates to polar coordinates.

• Convert equations given in rectangular form to equations in polar form and vice versa.

Polar to Rectangular

Just as x and y are usually used to designate the rectangular coordinates of a point, r and θ are usually used

to designate the polar coordinates of the point. r is the distance of the point to the origin. θ is the angle

that the line from the origin to the point makes with the positive x−axis. The diagram below shows both

polar and Cartesian coordinates applied to a point P. By applying trigonometry, we can obtain equations

that will show the relationship between polar coordinates (r, θ) and the rectangular coordinates (x, y)

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The point P has the polar coordinates (r, θ) and the rectangular coordinates (x, y).

Therefore

x = r cos θ r2 = x2 + y2

y = r sin θ tan θ = y

x

These equations, also known as coordinate conversion equations, will enable you to convert from polar to

rectangular form.

Example 1: Given the following polar coordinates, find the corresponding rectangular coordinates of the

points: W(4,−200◦),H

(

4, pi3

)

Solution:

a) For W(4,−200◦), r = 4 and θ = −200◦

x = r cos θ y = r sin θ

x = 4 cos(−200◦) y = 4 sin(−200◦)

x = 4(−.9396) y = 4(.3420)

x ≈ −3.76 y ≈ 1.37

The rectangular coordinates of W are approximately (−3.76, 1.37).

b) For H

(

4, pi3

)

, r = 4 and θ = pi3

x = r cos θ y = r sin θ

x = 4 cos pi

3

y = 4 sin pi

3

x = 4

(1

2

)

y = 4

√32

x = 2 y = 2

√

3

The rectangular coordinates of H are (2, 2

√

3) or approximately (2, 3.46).

In addition to writing polar coordinates in rectangular form, the coordinate conversion equations can also

be used to write polar equations in rectangular form.

Example 2: Write the polar equation r = 4 cos θ in rectangular form.

Solution:

r = 4 cos θ

r2 = 4r cos θ Multiply both sides by r.

x2 + y2 = 4x r2 = x2 + y2 and x = r cos θ

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The equation is now in rectangular form. The r2 and θ have been replaced. However, the equation, as

it appears, does not model any shape with which we are familiar. Therefore, we must continue with the

conversion.

x2 − 4x+ y2 = 0

x2 − 4x+ 4 + y2 = 4 Complete the square f or x2 − 4x.

(x − 2)2 + y2 = 4 Factor x2 − 4x+ 4.

The rectangular form of the polar equation represents a circle with its centre at (2, 0) and a radius of 2

units.

This is the graph represented by the polar equation r = 4 cos θ for 0 ≤ θ ≤ 2pi or the rectangular form

(x − 2)2 + y2 = 4.

Example 3: Write the polar equation r = 3 csc θ in rectangular form and graph the result.

Solution:

r = 3 csc θ

r

csc θ = 3 divide by csc θ

r · 1csc θ = 3

r sin θ = 3 sin θ = 1csc θ

y = 3 y = r sin θ

Rectangular to Polar

When converting rectangular coordinates to polar coordinates, we must remember that there are many

possible polar coordinates. We will agree that when converting from rectangular coordinates to polar

coordinates, one set of polar coordinates will be sufficient for each set of rectangular coordinates. Most

graphing calculators are programmed to complete the conversions and they too provide one set of coor-

dinates for each conversion. The conversion of rectangular coordinates to polar coordinates is done using

the Pythagorean Theorem and the Arctangent function. The Arctangent function only calculates angles in

the first and fourth quadrants so pi radians must be added to the value of θ for all points with rectangular

coordinates in the second and third quadrants.

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In addition to these formulas, r =

√

x2 + y2 is also used in converting rectangular coordinates to polar

form.

Example 4: Convert the following rectangular coordinates to polar form.

P(3,−5) and Q(−9,−12)

Solution: For P(3,−5) x = 3 and y = −5. The point is located in the fourth quadrant and x > 0.

r =

√

x2 + y2 θ = Arc tan y

x

r =

√

(3)2 + (−5)2 θ = tan−1

(

−5

3

)

r =

√

34 θ ≈ −1.03

r ≈ 5.83

The polar coordinates of P(3,−5) are P(5.83,−1.03).

For Q(−9,−12) x = −9 and y = −5. The point is located in the third quadrant and x < 0.

r =

√

x2 + y2 θ = Arc tan y

x

+ pi

r =

√

(−9)2 + (−12)2 θ = tan−1

(−12

−9

)

+ pi

r =

√

225 θ ≈ 4.07

r = 15

The polar coordinates of Q(−9,−12) are Q(15, 4.07)

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Converting Equations

To write a rectangular equation in polar form, the conversion equations of x = r cos θ and y = r sin θ are

used.

Example 5: Write the rectangular equation x2 + y2 = 2x in polar form.

Solution: Remember r =

√

x2 + y2, r2 = x2 + y2 and x = r cos θ.

x2 + y2 = 2x

r2 = 2(r cos θ) Pythagorean Theorem and x = r cos θ

r2 = 2r cos θ

r = 2 cos θ Divide each side by r

Example 6: Write the rectangular equation (x − 2)2 + y2 = 4 in polar form.

Solution: Remember x = r cos θ and y = r sin θ.

(x − 2)2 + y2 = 4

(r cos θ − 2)2 + (r sin θ)2 = 4 x = r cos θ and y = r sin θ

r2 cos2 θ − 4r cos θ + 4 + r2 sin2 θ = 4 expand the terms

r2 cos2 θ − 4r cos θ + r2 sin2 θ = 0 subtract 4 f rom each side

r2 cos2 θ + r2 sin2 θ = 4r cos θ isolate the squared terms

r2(cos2 θ + sin2 θ) = 4r cos θ f actor r2 − a common f actor

r2 = 4r cos θ Pythagorean Identity

r = 4 cos θ Divide each side by r

If the graph of the polar equation is the same as the graph of the rectangular equation, then the conversion

has been determined correctly.

(x − 2)2 + y2 = 4

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The rectangular equation (x − 2)2 + y2 = 4 represents a circle with center (2, 0) and a radius of 2 units.

The polar equation r = 4 cos θ is a circle with center (2, 0) and a radius of 2 units.

Converting Using the Graphing Calculator

You have learned how to convert back and forth between polar coordinates and rectangular coordinates

by using the various formulae presented in this lesson. The TI graphing calculator allows you to use the

angle function to convert coordinates quickly from one form to the other. The calculator will provide you

with only one pair of polar coordinates for each pair of rectangular coordinates.

Example 7: Express the rectangular coordinates of A(−3, 7) as polar coordinates.

Polar coordinates are expressed in the form (r, θ). An angle can be measured in either degrees or radians,

and the calculator will express the result in the form selected in the MODE menu of the calculator.

Press MODE and cursor down to Radian Degree. Highlight radian. Press 2nd mode to return to

home screen. To access the angle menu of the calculator press 2nd APPS and this screen will appear:

Cursor down to 5 and press ENTER . The following screen will appear . Press -3, 7) ENTER

and the value of r will appear . Access the angle menu again by pressing 2nd APPS .

When the angle menu screen appears, cursor down to 6 and pres ENTER or press 6 on the calculator.

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The screen will appear. Press -3, 7) ENTER and the value of θ will appear.

This procedure can be repeated to determine the rectangular coordinates in degrees. Before starting, press

MODE and cursor down to Radian Degree and highlight degree.

Example 8: Express the polar coordinates of (300, 70◦) in rectangular form.

The angle θ is given in degrees so the mode menu of the calculator should also be set in degree. Therefore,

press MODE and cursor down to Radian Degree and highlight degree. Press 2nd mode to return to

home screen. To access the angle menu of the calculator press 2nd APPS and this screen will appear:

Cursor down to 7 and press ENTER or press 7 on the calculator. The following screen will screen will

appear: Press 300, 70) and the value of x will appear Access the angle

menu again by pressing 2nd APPS . When the angle menu screen appears, cursor down to 8 and pres

ENTER or press 8 on the calculator. The screen will appear. Press 300,70) ENTER and

the value of y will appear .

Points to Consider

• When we convert coordinates from polar form to rectangular form, the process is very straightforward.

However, when converting a coordinate from rectangular form to polar form there are some choices

to make. For example the point 0,1 could translate to (1, 2pi) or to (1,−4pi), and so on.

• Are there any advantages to using polar coordinates instead of rectangular coordinates? List any

situations in which this is the case. What types of curves are easier to draw with polar coordinates?

• List situations in which rectangular coordinates are preferable.

Review Questions

1. For the following polar coordinates that are shown on the graph, determine the rectangular coordi-

nates for each point.

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2. Write the following polar equations in rectangular form.

(a) r = 6 cos θ

(b) r sin θ = −3

(c) r = 2 sin θ

(d) r sin2 θ = 3 cos θ

3. Write the following rectangular points in polar form.

(a) A(−2, 3)

(b) B(5,−4)

(c) C(1, 9)

(d) D(−12,−5)

4. Write the rectangular equations in polar form.

(a) (x − 4)2 + (y − 3)2 = 25

(b) 3x − 2y = 1

(c) x2 + y2 − 4x+ 2y = 0

(d) x3 = 4y2

Review Answers

1. For A, r = −4 and θ = 5pi4x = r cos θ y = r sin θ

x = −4 cos 5pi

4

y = −4 sin 5pi

4

x = −4

− √22

y = −4 − √22

x = 2

√

2 y = 2

√

2

For B, r = −3 and θ = 135◦

x = r cos θ y = r sin θ

x = −3 cos 135◦ y = −3 sin 135◦

x = −3 −

√

2

2

y = −3

√

2

2

x =

3

√

2

2

y =

−3√2

2

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For C, r = 5 and θ =

(

2pi

3

)

x = r cos θ y = r sin θ

x = 5 cos 2pi

3

y = 5 sin 2pi

3

x = 5

(

−1

2

)

y = 5

√32

x = −2.5 y = 5

√

3

22. (a)

r = 6 cos θ

r2 = 6r cos θ

x2 + y2 = 6x

x2 − 6x+ y2 = 0

x2 − 6x+ 9 + y2 = 9

(x − 3)2 + y2 = 9(b)

r sin θ = −3

y = −3

(c) r = 2 sin θ

r2 = 2r sin θ

x2 + y2 = 2y

y2 − 2y = −x2

y2 − 2y+ 1 = −x2 + 1 use both the positive and negative parts when graphing

(y − 1)2 = −x2 + 1

y − 1 = ±

√

x2 + 1

y = 1 ±

√

x2 + 1(d)

r sin2 θ = 3 cos θ

r2 sin2 θ = 3r cos θ

y2 = 3x

3. (a) For A(−2, 5)x = −2 and y = 3. The point is located in the second quadrant and x < 0.

r =

√

(−2)2 + (5)2 = √29 ≈ 5.39, θ = Arc tan 5−2 + pi = 1.95.

The polar coordinates for the rectangular coordinates A(−2, 5) are A(5.39, 1.95)

(b) For B(5,−4)x = 5 and y = −4. The point is located in the fourth quadrant and x > 0.

r =

√

(5)2 + (−4)2 = √41 ≈ 6.4, θ = tan−1

(−4

5

)

≈ −0.67

The polar coordinates for the rectangular coordinates B(5,−4) are A(6.40,−0.67)

(c) C(1, 9) is located in the first quadrant.

r =

√

12 + 92 =

√

82 ≈ 9.06, θ = tan−1 9

1

≈ 83.66◦.

(d) D(−12,−5) is located in the third quadrant and x < 0.

r =

√

(−12)2 + (−5)2 = √169 = 13, θ = tan−1 5

12

+ pi ≈ 202.6◦.

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4. (a)

(x − 4)2 + (y − 3)2 = 25

x2 − 8x+ 16 + y2 − 6y+ 9 = 25

x2 − 8x+ y2 − 6y+ 25 = 25

x2 − 8x+ y2 − 6y = 0

x2 + y2 − 8x − 6y = 0 From graphing r − 8 cos θ − 6 sin θ = 0, we see that the

r2 − 8(r cos θ) − 6(r sin θ) = 0 additional solutions are 0 and 8.

r2 − 8r cos θ − 6r sin θ = 0

r(r − 8 cos θ − 6 sin θ) = 0

r = 0 or r − 8 cos θ − 6 sin θ = 0

(b)

3x − 2y = 1

3r cos θ − 2r sin θ = 1

r(3 cos θ − 2 sin θ) = 1

r =

1

3 cos θ − 2 sin θ(c)

x2 + y2 − 4x+ 2y = 0

r2 cos2 θ + r2 sin2 θ − 4r cos θ + 2r sin θ = 0

r2(sin2 θ + cos2 θ) − 4r cos θ + 2r sin θ = 0

r(r − 4 cos θ + 2 sin θ) = 0

r − 4 cos θ + 2 sin θ = 0

r = 4 cos θ − 2 sin θ

(d)

x3 = 4y2

(r cos θ)3 = 4(r sin θ)2

r3 cos3 θ = 4r2 sin2 θ

4r2 sin2 θ

r3 cos3 θ = 1

4 tan2 θ sec θ

r

= 1

4 tan2 θ sec θ = r

6.4 More with Polar Curves

Learning Objectives

• Graph polar curves to see the points of intersection of the curves.

• Graph equivalent polar curves.

• Recognize equivalent polar curves from their equations.

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Intersections of Polar Curves

When you worked with a system of linear equations with two unknowns, finding the point of intersection

of the equations meant finding the coordinates of the point that satisfied both equations. If the equations

are rectangular equations for curves, determining the point(s) of intersection of the curves involves solving

the equations algebraically since each point will have one ordered pair of coordinates associated with it.

Example 1: Solve the following system of equations algebraically:

x2 + 4y2 − 36 = 0

x2 + y = 3

Solution: Before solving the system, graph the equations to determine the number of points of intersection.

The graph of x2 + 4y2 − 36 = 0 is an ellipse and the graph represented by x2 + y = 3 is a parabola. There

are three points of intersection. To determine the exact values of these points, algebra must be used.

x2 + 4y2 − 36 = 0→ x2 + 4y2 = 36 x2 + 4y2 + 0y = 36 x2 + 4y2 + 0y = 36

x2 + y = 3→ x2 + 0y2 + y = 3 − 1(x2 + 0y2 + y = 3) − x2 − 0y2 − y = −3

4y2 − y = 33

Using the quadratic formula, a = 4 b = −1 c = −33

y =

−(−1) ± √(−1)2 − 4(4)(−33)

2(4)

y =

1 + 23

8

= 3 y =

1 − 23

8

= −2.75

These values must be substituted into one of the original equations.

x2 + y = 3 x2 + y = 3

x2 + 3 = 3 x2 + (−2.75) = 3

x2 = 0 x2 = 5.75

x = 0 x = ±√5.75 ≈ 2.4

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The three points of intersection as determined algebraically in Cartesian representation are A(0, 3),B(2.4,−2.75)

and C(2.4, 2.75).

If we are working with polar equations to determine the polar coordinates of a point of intersection, we must

remember that there are many polar coordinates that represent the same point. Remember that switching

to polar form changes a great deal more than the notation. Unlike the Cartesian system which has one

name for each point, the polar system has an infinite number of names for each point. One option would be

to convert the polar coordinates to rectangular form and then to convert the coordinates for the intersection

points back to polar form. Perhaps the best option would be to explore some examples. As these examples

are presented, be sure to use your graphing calculator to create your own visual representations of the

equations presented.

Example 2: Determine the polar coordinates for the intersection point(s) of the following polar equations:

r = 1 and r = 2 cos θ .

Solution: Begin with the graph. Using the process described in the technology section in this chapter;

create the graph of these polar equations on your graphing calculator. Once the graphs are on the screen,

use the trace function and the arrow keys to move the cursor around each graph. As the cursor is moved,

you will notice that the equation of the curve is shown in the upper left corner and the values of θ, x, y are

shown (in decimal form) at the bottom of the screen. The values change as the cursor is moved.

r = 1 2 cos θ = 1

r = 2 cos θ cos θ = 1

2

cos−1(cos θ) = cos−1 1

2

θ = pi3 in the first quadrant and θ = 5pi3 in the fourth quadrant.

The points of intersection are

(

1, pi3

)

and

(

1, 5pi3

)

. However, these two solutions only cover the possible values

0 ≤ θ ≤ 2pi. If you consider that cos θ = 12 is true for an infinite number of theta these solutions must be

extended to include

(

1, pi3

)

and

(

1, 5pi3

)

+ 2pik, kεZ. Now the solutions include all possible rotations.

This example was solved as any system of rectangular equations would be solved. Does this approach work

all the time?

Example 3: Find the intersection of the graphs of r = sin θ and r = 1 − sin θ

Solution: Begin with the graph. You can create these graphs using your graphing calculator.

373 www.ck12.org

r = sin θ sin θ = 1 − sin θ

r = 1 − sin θ 2 sin θ = 1

sin θ = 1

2

r = sin θ θ = pi

6

in the first quadrant and θ = 5pi

6

in the second quadrant.

r = sin pi

6

The intersection points are

(1

2

,

pi

6

)

and

(1

2

,

5pi

6

)

r =

1

2

Another intersection point seems to be the origin (0, 0).

If you consider that sin θ = 12 is true for an infinite number of theta as was cos θ = 12 in the previous

example, the same consideration must be applied to include all possible solutions. To prove if the origin is

indeed an intersection point, we must determine whether or not both curves pass through (0, 0).

r = sin θ r = 1 − sin θ

0 = sin θ 0 = 1 − sin θ

r = 0 1 = sin θ

pi

2

= 0

From this investigation, the point (0, 0) was on the curve r = sin θ and the point

(

0, pi2

)

was on the curve

r = 1 − sin θ. Because the second coordinates are different, it seems that they are two different points.

However, the coordinates represent the same point (0,0). The intersection points are

(

1

2 ,

pi

6

)

,

(

1

2 ,

5pi

6

)

and

(0,0).

Sometimes it is helpful to convert the equations to rectangular form, solve the system and then convert

the polar coordinates back to polar form.

Example 4: Find the intersection of the graphs of r = 2 cos θ and r = 1 + cos θ

Solution: Begin with the graph:

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r = 2 cos θ expressed in rectangular form

r = 2 cos θ

r2 = 2r cos θ Multiply by r

x2 + y2 = 2x S ubstitution

r = 1 + cosθ expressed in rectangular form

r = 1 + cos θ

r2 = r + r cos θ Multiply by r

x2 + y2 =

√

x2 + y2 + x S ubstitution

The equations are now in rectangular form. Solve the system of equations.

x2 + y2 = 2x

x2 + y2 =

√

x2 + y2 + x

2x =

√

2x+ x

x =

√

2x

x2 = 2x

x2 − 2x = 0

x(x − 2) = 0

x = 0 x − 2 = 0

x = 2

Substituting these values into the first equation:

x2 + y2 = 2x x2 + y2 = 2x

(0)2 + y2 = 2(0) (2)2 + y2 = 2(2)

y2 = 0 4 + y2 = 4

y = 0 y2 = 0

y = 0

The points of intersection are (0, 0) and (2, 0).

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The rectangular coordinates are (0, 0) and (2, 0). Converting these coordinates to polar coordinates give

the same coordinates in polar form. The points can be converted by using the angle menu of the TI

calculator.

Example 5: Josie is drawing a mural with polar equations. One mural is represented by the equation

r = 3 cos θ and the other by r = 2 − cos θ. She wants to see where they will intersect before she transfers

her image onto the wall where she is painting.

Solution: To determine where they will intersect, we will begin with a graph.

r = 3 cos θ

r = 2 − cos θ

3 cos θ = 2 − cos θ

3 cos θ + cos θ = 2

4 cos θ = 2

cos θ = 2

4

=

1

2

θ =

pi

3

and θ = 5pi

3

r = 3 cos θ r = 3 cos θ

r = 3 cos pi

3

r = 3 cos 5pi

3

r = 3 · 1

2

=

3

2

r = 3 · 1

2

=

3

2

Josie’s murals would intersect and two points

(

3

2 ,

pi

3

)

and

(

3

2 ,

5pi

3

)

.

Equivalent Polar Curves

The expression “same only different” comes into play in this lesson. We will graph two distinct polar

equations that will produce two equivalent graphs. Use your graphing calculator and create these curves

as the equations are presented.

Previously, graphs were generated of a limaçon, a dimpled limaçon, a looped limaçon and a cardioid. All of

these were of the form r = a±b sin θ or r = a±b cos θ. The easiest way to see what polar equations produce

equivalent curves is to use either a graphing calculator or a software program to generate the graphs of

various polar equations.

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Example 6: Plot the following polar equations and compare the graphs.

a)

r = 1 + 2 sin θ

r = −1 + 2 sin θ

b)

r = 5 cos 90

r = 2 cos(−90)

Solution: By looking at the graphs, the result is the same. So, even though a is different in both, they

have the same graph. We can assume that the sign of a does not matter.

b) These functions also result in the same graph. Here, θ differed by a negative. So we can assume that

the sign of θ does not change the appearance of the graph.

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Example 7:

Graph the equations x2 + y2 = 16. Describe the graphs.

r = 4

Solution:

Both equations, one in rectangular form and one in polar form, are circles with a radius of 4 and center at

the origin.

Example 8: Graph the equations (x − 2)2 + (y+ 2)2 = 8 Describe the graphs.

r = 4 cos θ − 4 sin θ

There is not a visual representation shown here, but on your calculator you should see that the graphs are

circles centered at (2, -2) with a radius 2

√

2 ≈ 2.8.

Example 9: Stephanie is making a quilt. In each block, she is sewing a rose with 4 petals and adding

a sheer, metallic overlay on top of the rose. She plans to repeat this pattern in every fourth block of her

quilt. To keep the pattern repeating in a perfect manner, Stephanie must decide the exact position of the

overlay on the rose. If she knows this, she can be certain that every fourth block will repeat exactly. The

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limaçon, which is the shape of the overlay, was designed by using the equation r = 3 + 2 cos θ, while the

shape of the rose was designed by using the equation r = 5 sin 2θ. Create a graphic representation of this

design so you can explain the intersection points to Stephanie.

Solution: There appear to be 8 intersection points between the limaçon, and the rose. However, the true

points of intersection are the two points in the first quadrant and the two points in the third quadrant. At

the other four intersection points, the r−values on the rose are negative.

From graphing software, we get:

D : (3.88,−0.36pi radians)

E : (4.65,−0.19pi radians)

F : (4.64, 0.19pi radians)

G : (3.85, 0.36pi radians)

Points to Consider

• When looking for intersections, which representation is easier to work with? Look over the examples

and find some in which doing the algebra in polar coordinates is more direct than finding intersections

in Cartesian form.

• Will polar curves always intersect?

• If not, when will intersection not occur?

• If two polar curves have different equations, can they be the same curve?

Review Questions

1. Find the intersection of the graphs of r = sin 3θ and r = 3 sin θ.

2. Find the intersection of the graphs of r = 2 + 2 sin θ and r = 2 − 2 cos θ

3. Find the intersection of the graphs of r = sin 2θ and r = cos 2θ.

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4. Write the rectangular equation x2 + y2 = 6x in polar form and graph both equations. Should they

be equivalent?

5. Determine if r = −2 + sin θ and r = 2 − sin θ are equivalent without graphing.

6. Determine if r = −3 + 4 cos(−pi) and r = 3 + 4 cos pi are equivalent without graphing.

7. Graph the equations r = 7 − 3 cos pi3 . Are they equivalent? r = 7 − 3 cos

(

−pi3

)

8. Formulate a theorem about equivalent polar curves. What can be different to yield the same graph?

What must be the same? Explain your answer and show graphs to support your conclusions.

9. Determine two polar curves that will never intersect.

Review Answers

1.

There appears to be one point of intersection.

r = sin 3 θ r = 3 sin θ

0 = sin 3 θ 0 = 3 sin θ

0 = θ 0 = sin θ

0 = θ

The point of intersection is (0, 0)

2.

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r = 2 + 2 sin θ r = 2 + 2 sin θ

r = 2 + 2 sin

(3pi

4

)

r = 2 + 2 sin 7pi

4

r ≈ 3.4 r ≈ 0.59

r = 2 + 2 sin θ r = 2 − 2 cos θ

r = 2 + 2 sin(0) r = 2 − 2 cos 0

r = 2 r = 0

The coordinates represent the same point (0, 0).

r = 2 + 2 sin θ

r = 2 − 2 cos θ

2 + 2 sin θ = 2 − 2 cos θ

2 sin θ = −2 cos θ

2 sin θ

2 cos θ = −

2 cos θ

2 cos θ

sin θ

cos θ = −1

tan θ = −1

θ =

3pi

4

and θ = 7pi

4

The points of intersection are

(

3.4, 3pi4

)

,

(

0.59, 7pi4

)

and (0, 0)

3. Set the two functions equal to each other. sin 2θ = cos 2θ This is true when 2θ = pi4 , 5pi4 , 9pi4 , and 13pi4 so

θ = pi8 ,

5pi

8 ,

9pi

8 , and 13pi8 . Plug these values in for θ to find r.

r = sin pi

4

=

√

2

2

≈ 0.7 r = sin 5pi

4

= −

√

2

2

≈ −0.7

r = sin 9pi

4

=

√

2

2

≈ 0.7 r = sin 13pi

4

= −

√

2

2

≈ −0.7

So, the points of intersection are: (0, 0), (±pi8 , 0.707), (±5pi8 ,−0.707), (±9pi8 , 0.707), (±13pi8 ,−0.707).

4.

381 www.ck12.org

x2 + y2 = 6x

r2 = 6(r cos θ) r2 = x2 + y2 and x = y cos θ

r = 6 cos θ divide by r

Both equations produced a circle with center (3, 0) and a radius of 3.

5. r = −2+sin θ and r = 2−sin θ are not equivalent because the sine has the opposite sign. r = −2+sin θ

will be primarily above the horizontal axis and r = 2 − sin θ will be mostly below. However, the two

do have the same pole axis intercepts.

6. r = −3 + 4 cos(−pi) and r = 3+ 4 cos pi are equivalent because the sign of a does not matter, nor does

the sign of θ.

7. Yes, the equations produced the same graph so they are equivalent.

8. Students answers will vary, but they need to include that b must be the same sign. They should also

mention that the sign of a does not matter, nor does the sign of θ.

9. There are several answers here. The most obvious are any two pairs of circles, for example r = 3 and

r = 9.

6.5 The Trigonometric Form of Complex Num-

bers

Learning Objectives

• Understand the relationship between the rectangular form of complex numbers and their correspond-

ing polar form.

• Convert complex numbers from standard form to polar form and vice versa.

Despite their names, complex numbers and imaginary numbers have very real and significant applications

in both mathematics and in the real world. Complex numbers are useful in pure mathematics, providing

a more consistent and flexible number system that helps solve algebra and calculus problems. We will see

some of these applications in the examples throughout this lesson.

The Trigonometric or Polar Form of a Complex Number

The following diagram will introduce you to the relationship between complex numbers and polar coordi-

nates.

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In the figure above, the point that represents the number x+ yi was plotted and a vector was drawn from

the origin to this point. As a result, an angle in standard position, θ, has been formed. In addition to this,

the point that represents x + yi is r units from the origin. Therefore, any point in the complex plane can

be found if the angle θ and the r− value are known. The following equations relate x, y, r and θ.

x = r cos θ y = r sin θ r2 = x2 + y2 tan θ = y

x

If we apply the first two equations to the point x+ yi the result would be:

x+ yi = r cos θ + ri sin θ → r(cos θ + i sin θ)

The right side of this equation r(cos θ + i sin θ) is called the polar or trigonometric form of a complex

number. A shortened version of this polar form is written as r cis θ. The length r is called the absolute

value or the modulus, and the angle θ is called the argument of the complex number. Therefore, the

following equations define the polar form of a complex number:

r2 = x2 + y2 tan θ = y

x

x+ yi = r(cos θ + i sin θ)

It is now time to implement these equations perform the operation of converting complex numbers in

standard form to complex numbers in polar form. You will use the above equations to do this.

Example 1: Represent the complex number 5 + 7i graphically and express it in its polar form.

Solution: As discussed in the Prerequisite Chapter, here is the graph of 5 + 7i.

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Converting to polar from rectangular, x = 5 and y = 7.

r =

√

52 + 72 = 8.6 tan θ = 7

5

tan−1(tan θ) = tan−1 7

5

θ = 54.5◦

So, the polar form is 8.6(cos 54.5◦ + i sin 54.5◦).

Another widely used notation for the polar form of a complex number is r∠θ = r(cos θ+ i sin θ). Now there

are three ways to write the polar form of a complex number.

x+ yi = r(cos θ + i sin θ) x+ yi = rcisθ x+ yi = r∠θ

Example 2: Express the following polar form of each complex number using the shorthand representations.

a) 4.92(cos 214.6◦ + i sin 214.6◦)

b) 15.6(cos 37◦ + i sin 37◦)

Solution:

a) 4.92∠214.6◦

4.92 cis 214.6◦

b) 15.6∠37◦

15.6 cis 37◦

Example 3: Represent the complex number −3.12 − 4.64i graphically and give two notations of its polar

form.

Solution: From the rectangular form of −3.12 − 4.64i x = −3.12 and y = −4.64

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r =

√

x2 + y2

r =

√

(−3.12)2 + (−4.64)2

r = 5.59

tan θ = y

x

tan θ = −4.64−3.12

θ = 56.1◦

This is the reference angle so now we must determine the measure of the angle in the third quadrant.

56.1◦ + 180◦ = 236.1◦

One polar notation of the point −3.12 − 4.64i is 5.59 (cos 236.1◦ + i sin 236.1◦). Another polar notation of

the point is 5.59∠236.1◦

So far we have expressed all values of theta in degrees. Polar form of a complex number can also have theta

expressed in radian measure. This would be beneficial when plotting the polar form of complex numbers

in the polar plane.

The answer to the above example −3.12 − 4.64i with theta expressed in radian measure would be:

tan θ = −4.64−3.12 tan θ = .9788(reference angle)

0.9788 + 3.14 = 4.12 rad.

5.59(cos 4.12 + i sin 4.12)

Now that we have explored the polar form of complex numbers and the steps for performing these conver-

sions, we will look at an example in circuit analysis that requires a complex number given in polar form

to be expressed in standard form.

Example 4: The impedance Z, in ohms, in an alternating circuit is given by Z = 4650∠ − 35.2◦. Express

the value for Z in standard form. (In electricity, negative angles are often used.)

Solution: The value for Z is given in polar form. From this notation, we know that r = 4650 and

θ = −35.2◦ Using these values, we can write:

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Z = 4650(cos(−35.2◦) + i sin(−35.2◦))

x = 4650 cos(−35.2◦)→ 3800

y = 4650 sin(−35.2◦)→ −2680

Therefore the standard form is Z = 3800 − 2680i ohms.

Points to Consider

• Is it possible to perform basic operations on complex numbers in polar form?

• If operations can be performed, do the processes change for polar form or remain the same as for

standard form?

Review Questions

1. Express the following polar forms of complex numbers in the two other possible ways.

(a) 5 cispi6

(b) 3∠135◦

(c) 2

(

cos 2pi3 + i sin 2pi3

)

2. Express the complex number 6 − 8i graphically and write it in its polar form.

3. Express the following complex numbers in their polar form.

(a) 4 + 3i

(b) −2 + 9i

(c) 7 − i

(d) −5 − 2i

4. Graph the complex number 3(cos pi4 + i sin pi4) and express it in standard form.

5. Find the standard form of each of the complex numbers below.

(a) 2 cispi2

(b) 4∠5pi6

(c) 8

(

cos

(

−pi3

)

+ i sin

(

−pi3

))

Review Answers

1. (a) 5 cispi6 = 5∠pi6 = 5

(

cos pi6 + i sin pi6

)

(b) 3∠135◦ = 3cis135◦ = 3(cos 135◦ + i sin 135◦)

(c) 2

(

cos 2pi3 + i sin 2pi3

)

= 2cis2pi3 = 2∠

2pi

3

2. 6 − 8i

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6 − 8i

x = 6 and y = −8 tan θ = y

x

r =

√

x2 + y2 tan θ = −8

6

r =

√

(6)2 + (−8)2 θ = −53.1◦

r = 10

Since θ is in the fourth quadrant then θ = −53.1◦ + 360◦ = 306.9◦ Expressed in polar form 6 − 8i is

10(cos 306.9◦ + i sin 306.9◦) or 10∠306.9◦

3. (a) 4 + 3i→ x = 4, y = 3

r =

√

42 + 32 = 5, tan θ = 3

4

→ θ = 36.87◦ → 5(cos 36.87◦ + i sin 36.87◦)

(b) −2 + 9i→ x = −2, y = 9

r =

√

(−2)2 + 92 = √85 ≈ 9.22, tan θ = −9

2

→ θ = 102.53◦ → 9.22(cos 102.53◦ + i sin 102.53◦)

(c) 7 − i→ x = 7, y = −1

r =

√

72 + 12 =

√

50 ≈ 7.07, tan θ = −1

7

→ θ = 351.87◦ → 7.07(cos 351.87◦ + i sin 351.87◦)

(d) −5 − 2i→ x = −5, y = −2

r =

√

(−5)2 + (−2)2 = √29 ≈ 5.39, tan θ = 2

5

→ θ = 201.8◦ → 5.39(cos 201.8◦ + i sin 201.8◦)

4.

387 www.ck12.org

3

(

cos pi

4

+ i sin pi

4

)

r = 3

x = cos pi

4

=

√

2

2

y = sin pi

4

=

√

2

2

The standard form of the polar complex number 3

(

cos pi4 + i sin pi4

)

is 3

√

2

2 +

3

√

2

2 i.

5. (a) 2cispi2 → x = cos pi2 = 0, y = sin pi2 = 1→ 2(0) + 2(1i) = 2i

(b) 4∠5pi6 → x = cos 5pi6 = −

√

3

2 , y = sin 5pi6 = 12 → 4

(

−

√

3

2

)

+ 4

(

i12

)

= −2√3 + 2i

(c) 8

(

cos

(

−pi3

)

+ i sin

(

−pi3

))

→ x = cos

(

−pi3

)

= 12 , y = sin

(

−pi3

)

= −

√

3

2 → 8

(

1

2

)

+ 8

(

−

√

3

2 i

)

= 4 − 4i√3

6.6 The Product & Quotient Theorems

Learning Objectives

• Determine the quotient theorem of complex numbers in polar form.

• Determine the product theorem of complex numbers in polar form.

• Solve everyday problems that require you to use the product and/or quotient theorem of complex

numbers in polar form to obtain the correct solution.

The Product Theorem

Multiplication of complex numbers in polar form is similar to the multiplication of complex numbers in

standard form. However, to determine a general rule for multiplication, the trigonometric functions will

be simplified by applying the sum/difference identities for cosine and sine. To obtain a general rule for the

multiplication of complex numbers in polar from, let the first number be r1(cos θ1+ i sin θ1) and the second

number r2(cos θ2 + i sin θ2). Now that the numbers have been designated, proceed with the multiplication

of these binomials.

r1(cos θ1 + i sin θ1) · r2(cos θ2 + i sin θ2)

r1r2(cos θ1 cos θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 + i2 sin θ1 sin θ2)

r1r2[(cos θ1 cos θ2 − sin θ1 sin θ2 + i(sin θ1 cos θ2 + cos θ1 sin θ2)]

r1r2[cos(θ1 + θ2) + i sin(θ1 + θ2)]

To arrive at the general rule, recall that i2 = −1 and the sum identity cosα cos β − sinα sin β = cos(α+ β)

and sinα cos β+ cosα sin β = sin(α+ β) were applied. Therefore:

r1(cos θ1 + i sin θ1) · r2(cos θ2 + i sin θ2) = r1r2[cos(θ1 + θ2) + i sin(θ1 + θ2)]

Quotient Theorem

Division of complex numbers in polar form is similar to the division of complex numbers in standard form.

However, to determine a general rule for division, the denominator must be rationalized by multiplying the

fraction by the complex conjugate of the denominator. In addition, the trigonometric functions must be

www.ck12.org 388

simplified by applying the sum/difference identities for cosine and sine as well as one of the Pythagorean

identities. To obtain a general rule for the division of complex numbers in polar from, let the first number

be r1(cos θ1 + i sin θ1) and the second number r2(cos θ2 + i sin θ2) The conjugate of cos θ2 + i sin θ2 is

cos θ2 − i sin θ2.

r1(cos θ1 + i sin θ1)

r2(cos θ2 + i sin θ2)

r1(cos θ1 + i sin θ1)

r2(cos θ2 + i sin θ2)

· (cos θ2 − i sin θ2)

(cos θ2 − i sin θ2)

r1

r2

· cos θ1 cos θ2 − i cos θ1 sin θ2 + i sin θ1 cos θ2 − i

2 sin θ1 sin θ2

cos2 θ2 − i2 sin2 θ2

r1

r2

· cos θ1 cos θ2 + sin θ1 sin θ2 + i(sin θ1 cos θ2 − cos θ1 sin θ2)cos2 θ2 + sin2 θ2

r1

r2

[cos(θ1 − θ2) + i sin(θ1 − θ2)]

To arrive at the general rule, i2 = −1 the difference identity cosα cos β + sinα sin β = cos(α − β) and

sinα cos β − cosα sin β = sin(α − β) and the Pythagorean identity were applied. In general:

r1(cos θ1 + i sin θ1)

r2(cos θ2 + i sin θ2)

=

r1

r2

[cos(θ1 − θ2) + i sin(θ1 − θ2)]

Using the Quotient and Product Theorem

Now that general rules have been obtained for the multiplication and division of complex numbers in polar

form, they can now be implemented.

Example 1: Find the product of the complex numbers 3.61(cos 56.3◦ + i sin 56.3◦) and 1.41(cos 315◦ +

i sin 315◦)

Solution: Use the Product Theorem, r1(cos θ1+i sin θ1)·r2(cos θ2+i sin θ2) = r1r2[cos(θ1+θ2)+i sin(θ1+θ2)].

3.61(cos 56.3◦ + i sin 56.3◦) · 1.41(cos 315◦ + i sin 315◦)

= (3.61)(1.41)[cos(56.3◦ + 315◦) + i sin(56.3◦ + 315◦)

= 5.09(cos 371.3◦ + i sin 371.3◦)

= 5.09(cos 11.3◦ + i sin 11.3◦)

∗Note: Angles are expressed 0◦ ≤ θ ≤ 360◦ unless otherwise stated.

Example 2: Find the product of 5

(

cos 3pi4 + i sin 3pi4

)

· √3

(

cos pi2 + i sin pi2

)

Solution: First, calculate r1r2 = 5 ·

√

3 = 5

√

3 and θ = θ1 + θ2 = 3pi4 + pi2 = 5pi4

5

√

3

(

cos 5pi

4

+ i sin 5pi

4

)

Example 3: Find the quotient of (

√

3 − i) ÷ (2 − i2√3)

Solution: Express each number in polar form.

389 www.ck12.org

√

3 − i 2 − i2√3

r1 =

√

x2 + y2 r2 =

√

x2 + y2

r1 =

√

(

√

3)2 + (−1)2 r2 =

√

(2)2 + (−2√3)2

r1 =

√

4 = 2 r2 =

√

16 = 4

r1

r2

= .5

θ1 = tan−1

( −1√

3

)

θ2 = tan−1

−2√32

θ = θ1 − θ2

θ1 = 5.75959 rad. θ2 = 5.23599 rad. θ = 5.75959 − 5.23599

θ = 0.5236

Now, plug in what we found to the Quotient Theorem.

r1

r2

[cos(θ1 − θ2) + i sin(θ1 − θ2)] = .5(cos 0.5236 + i sin 0.5236)

Example 4: Find the quotient of the two complex numbers 28∠35◦ and 14∠24◦

Solution:

For 28 ∠35◦ For 14 ∠24◦ r1

r2

=

28

14

= 2

r1 = 28 r2 = 14 θ = θ1 − θ2

θ1 = 35◦ θ2 = 24◦ θ = 35◦ − 24◦ = 11◦

r1∠θ1

r2∠θ2

=

r1

r2

∠(θ1 − θ2)

= 2∠11◦

Points to Consider

• We have performed the basic operations of arithmetic on complex numbers, but we have not dealt

with any exponents other than 2 or any roots other than √. Are these the only powers that exist for

complex numbers?

• How do you think operations like those mentioned above are carried out on complex numbers?

Review Questions

1. Multiply together the following complex numbers. If they are not in polar form, change them before

multiplying.

(a) 2∠56◦, 7∠113◦

(b) 3(cos pi+ i sin pi), 10

(

cos 5pi3 + i sin 5pi3

)

(c) 2 + 3i,−5 + 11i

(d) 6 − i,−20i

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2. Part c from #1 was not in polar form. Mulitply the two complex numbers together without changing

them into polar form. Which method do you think is easier?

3. Use the Product Theorem to find 4

(

cos pi4 + i sin pi4

)2.

4. The electric power (in watts) supplied to an element in a circuit is the product of the voltage e

and the current i (in amps). Find the expression for the power supplied if e = 6.80∠56.3◦ volts and

i = 7.05∠ − 15.8◦ amps. Note: Use the formula P = ei.

5. Divide the following complex numbers. If they are not in polar form, change them before dividing.

In

(a) 2∠56◦7∠113◦

(b) 10(cos

5pi

3 +i sin 5pi3 )

5(cos pi+i sin pi)

(c) 2+3i−5+11i

(d) 6−i1−20i

6. Part c from #5 was not in polar form. Mulitply the two complex numbers together without changing

them into polar form. Which method do you think is easier?

7. Use the Product Theorem to find 4

(

cos pi4 + i sin pi4

)3. Hint: use #3 to help you.

8. Using the Quotient Theorem determine 14cis pi6 .

Review Answers

1. (a) 2∠56◦, 7∠113◦ = (2)(7)∠(56◦ + 113◦) = 14∠169◦

(b) 3(cos pi+ i sin pi), 10

(

cos 5pi3 + i sin 5pi3

)

= (3)(10)cis

(

pi+ 5pi3

)

= 30cis8pi3 = 30cis

2pi

3

(c) 2 + 3i,−5 + 11i→ change to polar

x = 2, y = 3 x = −5, y = 11

r =

√

22 + 32 =

√

13 ≈ 3.61 r =

√

(−5)2 + 112 = √146 ≈ 12.08

tan θ = 3

2

→ θ = 56.31◦ tan θ = −11

5

→ θ = 114.44◦

(3.61)(12.08)∠(56.31◦ + 114.44◦) = 43.61∠170.75◦

(d) 6 − i,−20i→ change to polar

x = 6, y = −1 x = 0, y = −20

r =

√

62 + (−1)2 = √37 ≈ 6.08 r =

√

02 + (−20)2 = √40 = 20

tan θ = −1

6

→ θ = 350.54◦ tan θ = −20

0

= und → θ = 270◦

(6.08)(20)∠(350.54◦ + 270◦) = 121.6∠620.54◦ = 121.6∠260.54◦

2. Without changing complex numbers to polar form, you mulitply by FOIL-ing.

(2 + 3i)(−5 + 11i) = −10 + 22i − 15i+ 33i2 = −10 − 33 + 7i = −43 + 7i

The answer is student opinion, but they seem about equal in the degree of difficulty.3.

4

(

cos pi

4

+ i sin pi

4

)2

= 4

(

cos pi

4

+ i sin pi

4

)

· 4

(

cos pi

4

+ i sin pi

4

)

= 16

(

cos

(

pi

4

+

pi

4

)

+ i sin

(

pi

4

+

pi

4

))

= 16

(

cos pi

2

i sin pi

2

)

391 www.ck12.org

4.

P = (6.80)(7.05)∠(56.3◦ − 15.8◦), P = 47.9∠40.5◦watts

5. (a) 2∠56◦7∠113◦ = 27∠(56◦ − 113◦) = 27∠ − 57◦

(b) 10(cos

5pi

3 +i sin 5pi3 )

5(cos pi+i sin pi) = 2

(

cos

(

5pi

3 − pi

)

+ i sin

(

5pi

3 − pi

))

= 2

(

cos 2pi3 + i sin 2pi3

)

(c) 2+3i−5+11i → change each to polar.

x = 2, y = 3 x = −5, y = 11

r =

√

22 + 32 =

√

13 ≈ 3.61 r =

√

(−5)2 + 112 = √146 ≈ 12.08

tan θ = 3

2

→ θ = 56.31◦ tan θ = −11

5

→ θ = 114.44◦

3.61

12.08∠(56.31

◦ − 114.44◦) = 0.30∠ − 58.13◦

(d) 6−i1−20i → change both to polar

x = 6, y = −1 x = 1, y = −20

r =

√

62 + (−1)2 = √37 ≈ 6.08 r =

√

12 + (−20)2 = √401 = 20.02

tan θ = −1

6

→ θ = 350.54◦ tan θ = −20

1

→ θ = 272.68◦

6.08

20.02∠(350.54

◦ − 272.86◦) = 0.304∠77.68◦6.

2 + 3i

−5 + 11i =

2 + 3i

−5 + 11i ·

−5 − 11i

−5 − 11i =

−10 − 22i − 15i+ 33

25 + 121

=

23 − 37i

146

Again, this is opinion, but in general, using the polar form is “easier.”7.

4

(

cos pi

4

+ i sin pi

4

)3

= 4

(

cos pi

4

+ i sin pi

4

)2

· 4

(

cos pi

4

+ i sin pi

4

)

.

From #3, 4

(

cos pi4 i sin pi4

)2

= 16

(

cos pi2 i sin pi2

)

. So,

4

(

cos pi

4

i sin pi

4

)3

= 16

(

cos pi

2

i sin pi

2

)

· 4

(

cos pi

4

i sin pi

4

)

= 64

(

cos 3pi

4

i sin 3pi

4

)

8. Even though 1 is not a complex number, we can still change it to polar form. 1→ x = 1, y = 0

r =

√

12 + 02 = 1 So, 1

4cispi6

=

1cis0

4cispi6

=

1

4

cis

(

0 − pi

6

)

=

1

4

cis

(

−pi

6

)

.

tan θ = 0

1

= 0→ θ = 0◦

6.7 De Moivre’s and the nth Root Theorems

Learning Objectives

• Use De Moivre’s Theorem to find the powers of complex numbers in polar form.

• Find the nth roots of complex numbers in polar form.

www.ck12.org 392

DeMoivre’s Theorem

The basic operations of addition, subtraction, multiplication and division of complex numbers have all

been explored in this chapter. The addition and subtraction of complex numbers lent themselves best to

numbers expressed in standard form. However multiplication and division were easily performed when the

complex numbers were in polar form. Another operation that is performed using the polar form of complex

numbers is the process of raising a complex number to a power.

The polar form of a complex number is r(cos θ+ i sin θ). If we allow z to equal the polar form of a complex

number, it is very easy to see the development of a pattern when raising a complex number in polar form to

a power. To discover this pattern, it is necessary to perform some basic multiplication of complex numbers

in polar form. Recall #3 and #7 from the Review Questions in the previous section.

If z = r(cos θ + i sin θ) and z2 = z · z then:

z2 = r(cos θ + i sin θ) · r(cos θ + i sin θ)

z2 = r2[cos(θ + θ) + i sin(θ + θ)]

z2 = r2(cos 2θ + i sin 2θ)

Likewise, if z = r(cos θ + i sin θ) and z3 = z2 · z then:

z3 = r2(cos 2θ + i sin 2θ) · r(cos θ + i sin θ)

z3 = r3[cos(2θ + θ) + i sin(2θ + θ)]

z3 = r3(cos 3θ + i sin 3θ)

Again, if z = r(cos θ + i sin θ) and z4 = z3 · z then

z4 = r4(cos 4θ + i sin 4θ)

These examples suggest a general rule valid for all powers of z, or n. We offer this rule and assume

its validity for all n without formal proof, leaving that for later studies. The general rule for raising a

complex number in polar form to a power is called DeMoivre’s Theorem, and has important applications

in engineering, particularly circuit analysis. The rule is as follows:

zn = [r(cos θ + i sin θ)]n = rn(cos nθ + i sin nθ)

Where z = r(cos θ + i sin θ) and let n be a positive integer.

Notice what this rule looks like geometrically. A complex number taken to the nth power has two motions:

First, its distance from the origin is taken to the nth power; second, its angle is multiplied by n. Conversely,

the roots of a number have angles that are evenly spaced about the origin.

Example 1: Find. [2(cos 120◦ + i sin 120◦)]5

Solution: θ = 120◦ = 2pi3 rad, using De Moivre’s Theorem:

zn = [r(cos θ + i sin θ)]n = rn(cos nθ + i sin nθ)

[2(cos 120◦ + i sin 120◦)]5 = 25

[

cos 52pi

3

+ i sin 52pi

3

]

= 32

(

cos 10pi

3

+ i sin 10pi

3

)

= 32

−12 + −

√

3

2

= −16 − 16i√3

393 www.ck12.org

Example 2: Find

(

−12 +

√

3

2

)10

Solution: Change into polar form.

r =

√

x2 + y2 θ = tan−1

√32 · −21

= −pi3

r =

√(−1

2

)2

+

√32

2

r =

√

1

4

+

3

4

r =

√

1 = 1

The polar form of

(

−12 +

√

3

2

)

is 1

(

cos−pi3 + i sin−pi3

)

Now use De Moivre’s Theorem:

zn = [r(cos θ + i sin θ)]n = rn(cos nθ + i sin nθ)−12 +

√

3

2

10 = 110 [cos 10 (−pi3

)

+ i sin 10

(

−pi

3

)]

−12 +

√

3

2

10 = 1 (cos−10pi3 + i sin−10pi3

)

−12 +

√

3

2

10 = −12 + i

√

3

2

→ Standard Form

nth Roots

We have explored all of the basic operations of arithmetic as they apply to complex numbers in standard

form and in polar form. The last discovery is that of taking roots of complex numbers in polar form. Using

DeMoire’s Theorem we can develop another general rule –one for finding the nth root of a complex number

written in polar form.

As before, let z = r(cos θ + i sin θ) and let the nth root of z be v = s(cosα+ i sinα). So, in general, n√z = v

and vn = z.

n√z = v

n

√

r(cos θ + i sin θ) = s(cosα+ i sinα)

[r(cos θ + i sin θ)] 1n = s(cosα+ i sinα)

r

1

n

(

cos 1

n

θ + i sin 1

n

θ

)

= s(cosα+ i sinα)

r

1

n

(

cos θ

n

+ i sin θ

n

)

= s(cosα+ i sinα)

From this derivation, we can conclude that r 1n = s or sn = r and α = θn . Therefore, for any integer

k(0, 1, 2, . . . n − 1), v is an nth root of z if s = n√r and α = θ+2pikn . Therefore, the general rule for finding

the nth roots of a complex number if z = r(cos θ + i sin θ) is: n√r

(

cos θ+2pikn + i sin θ+2pikn

)

. Let’s begin with a

simple example and we will leave θ in degrees.

www.ck12.org 394

Example 3: Find the two square roots of 2i.

Solution: Express 2i in polar form.

r =

√

x2 + y2 cos θ = 0

r =

√

(0)2 + (2)2 θ = 90◦

r =

√

4 = 2

(2i)

1

2 = 2

1

2

(

cos 90

◦

2

+ i sin 90

◦

2

)

=

√

2(cos 45◦ + i sin 45◦) = 1 + i

To find the other root, add 360◦ to θ.

(2i)

1

2 = 2

1

2

(

cos 450

◦

2

+ i sin 450

◦

2

)

=

√

2(cos 225◦ + i sin 225◦) = −1 − i

Example 4: Find the three cube roots of −2 − 2i√3

Solution: Express −2 − 2i√3 in polar form:

r =

√

x2 + y2

r =

√

(−2)2 + (−2√3)2

r =

√

16 = 4 θ = tan−1

−2√3−2

= 4pi3

n√r

(

cos θ + 2pik

n

+ i sin θ + 2pik

n

)

3

√

−2 − 2i√3 = 3√4

cos 4pi3 + 2pik3 + i sin

4pi

3 + 2pik

3

k = 0, 1, 2

z1 =

3√

4

[

cos

(4pi

9

+

0

3

)

+ i sin

(4pi

9

+

0

3

)]

k = 0

=

3√

4

[

cos 4pi

9

+ i sin 4pi

9

]

z2 =

3√

4

[

cos

(4pi

9

+

2pi

3

)

+ i sin

(4pi

9

+

2pi

3

)]

k = 1

=

3√

4

[

cos 10pi

9

+ i sin 10pi

9

]

z3 =

3√

4

[

cos

(4pi

9

+

4pi

3

)

+ i sin

(4pi

9

+

4pi

3

)]

k = 2

=

3√

4

[

cos 16pi

9

+ i sin 16pi

9

]

In standard form: z1 = 0.276 + 1.563i, z2 = −1.492 − 0.543i, z3 = 1.216 − 1.02i.

Solve Equations

The roots of a complex number are cyclic in nature. This means that when the roots are plotted on the

complex plane, the nth roots are equally spaced on the circumference of a circle.

395 www.ck12.org

Since you began Algebra, solving equations has been an extensive topic. Now we will extend the rules to

include complex numbers. The easiest way to explore the process is to actually solve an equation. The

solution can be obtained by using De Moivre’s Theorem.

Example 5:

Consider the equation x5 − 32 = 0. The solution is the same as the solution of x5 = 32. In other words, we

must determine the fifth roots of 32.

Solution:

x5 − 32 = 0 and x5 = 32.

r =

√

x2 + y2

r =

√

(32)2 + (0)2

r = 32

θ = tan−1

( 0

32

)

= 0

Write an expression for determining the fifth roots of 32 = 32 + 0i

32

1

5 = [32(cos(0 + 2pik) + i sin(0 + 2pik)] 15

= 2

(

cos 2pik

5

+ i sin 2pik

5

)

k = 0, 1, 2, 3, 4

x1 = 2

(

cos 0

5

+ i sin 0

5

)

→ 2(cos 0 + i sin 0) = 2 f or k = 0

x2 = 2

(

cos 2pi

5

+ i sin 2pi

5

)

≈ 0.62 + 1.9i f or k = 1

x3 = 2

(

cos 4pi

5

+ i sin 4pi

5

)

≈ −1.62 + 1.18i f or k = 2

x4 = 2

(

cos 6pi

5

+ i sin 6pi

5

)

≈ −1.62 − 1.18i f or k = 3

x5 = 2

(

cos 8pi

5

+ i sin 8pi

5

)

≈ 0.62 − 1.9i f or k = 4

The Geometry of Complex Roots

In the previous example, we have one real and four complex roots. Plot these in the complex plane.

The nth roots of a complex number, when graphed on the complex plane, are

equally spaced around a circle. So, instead of having all the roots, all that is necessary to graph the roots

is one of them and the radius of the circle. For this particular example, the roots are 2pi5 or 72◦ apart (look

www.ck12.org 396

in the root equation in the example, θ increased by 2pi5 ). This goes along with what we know about regular

pentagons. The roots are 2pin degrees apart.

Example 6: Calculate the three cube roots of 1 and represent them graphically.

Solution: In standard form, 1 = 1 + 0i r = 1 and θ = 0. The polar form is 1 + 0i = 1[cos(0 + 2pik) +

i sin(0 + 2pik)]. The expression for determining the cube roots of 1 + 0i is:

(1 + 0i)

1

3 = 1

1

3

(

cos 0 + 2pik

3

+ i sin 0 + 2pik

3

)

When k = 0, k = 1 and k = 2 the three cube roots of 1 are 1,−12 + i

√

3

2 ,−12 − i

√

3

2 . When these three roots

are represented graphically, the three points, on the circle with a radius of 1 (the cubed root of 1 is 1),

form a triangle.

Points to Consider

• If the roots can be determined, will some form of De Moivre’s Theorem be used?

• If the root of a complex number in polar form can be determined, can the solution to an exponential

equation be found in the same way?

• Does the number of roots have anything to do with the shape of the graph?

Review Questions

1. Show that z3 = 1, if z = −12 + i

√

3

2

2. Evaluate:

(a)

[ √

2

2

(

cos pi4 + i sin pi4

)]8

(b)

[

3

(√

3 − i√3

)]4

(c) (

√

5 − i)7

(d)

[

3

(

cos pi6 + i sin pi6

)]12

3. Rewrite the following in rectangular form: [2(cos 315◦ + i sin 315◦)]3

4. Find 3

√

27i.

5. Find the principal root of (1+ i) 15 . Remember the principal root is the positive root i.e.

√

9 = ±3 so

the principal root is +3.

6. Find the fourth roots of 81i.

397 www.ck12.org

7. Solve the equation x4 + 1 = 0. What shape do the roots make?

8. Solve the equation x3 − 64 = 0. What shape do the roots make?

Review Answers

1. Express z in polar form:

r =

√

x2 + y2

r =

√(

−1

2

)2

+

√32

2

r =

√

1

4

+

3

4

= 1

θ = tan−1

− √31

= 120◦

The polar form is z = 1(cos 120◦ + i sin 120◦)

zn = [r(cos θ + i sin θ)n = r(cos nθ + i sin nθ)

z3 = 13[cos 3(120◦) + i sin(120◦)]

z3 = 1(cos 360◦ + i sin 360◦)

z3 = 1(1 + 0i)

z3 = 1

2. (a)

[ √

2

2

(

cos pi4 + i sin pi4

)]8

=

( √

2

2

)8 (

cos 8pi4 + i sin 8pi4

)

= 116 cos 2pi+ i16 sin 2pi = 116b

[

3(

√

3 − i√3)

]4

= (3

√

3 − 3i√3)4

r =

√

(3

√

3)2 + (3

√

3)2 = 3

√

6, tan θ = 3

√

3

3

√

3

= 1→ 45◦

=

(

3

√

6

(

cos pi

4

+ i sin pi

4

))4

= (3

√

6)4

(

cos 4pi

4

+ i sin 4pi

4

)

= 81(36)[−1 + i(0)] = −2936

(c)

(

√

5 − i)7 → r =

√

(

√

5)2 + (−1)2 = √6, tan θ = − 1√

5

→ θ = 335.9◦

[

√

6(cos 335.9◦ + i sin 335.9◦)]7 = (

√

6)7(cos(7 · 335.9◦) + i sin(7 · 335.9◦))

= 216

√

6(cos 2351.3◦ + i sin 2351.3◦)

= 216

√

6(−0.981 + 0.196i)

= −519.04 + 103.7i

(d)

[

3

(

cos pi6 + i sin pi6

)]12

= 312(cos 2pi+ i sin 2pi) = 531, 441

www.ck12.org 398

3.

r = 2 and θ = 315◦ or 7pi

4

.

zn = [r(cos θ + i sin θ)]n = rn(cos nθ + i sin nθ)

z3 = 23

[

(cos 3

(7pi

4

)

+ i sin 3

(7pi

4

)]

z3 = 8

(

cos 21pi

4

+ i sin 21pi

4

)

z3 = 8

− √22 − i

√

2

2

z3 = −4√2 − 4i√2

21pi

4 is in the third quadrant so both are negative.4.

a = 0 and b = 27

3√

27i = (0 + 27i)

1

3 x = 0 and y = 27

Polar From r =

√

x2 + y2 θ =

pi

2

r =

√

(0)2 + (27)2

r = 27

3√

27i =

[

27

(

cos pi

2

+ i sin pi

2

)] 1

3

3√

27i =

3√

27

[

cos

(1

3

) (

pi

2

)

+ i sin

(1

3

) (

pi

2

)]

3√

27i = 3

(

cos pi

6

+ i sin pi

6

)

3√

27i = 3

√32 + 12 i

, 3i,−3i

5.

r =

√

x2 + y2 θ = tan−1

(1

1

)

=

√

2

2

Polar Form =

(√

2,

pi

4

)

r =

√

(1)2 + (1)2

r =

√

2

(1 + i)

1

5 =

[√

2

(

cos pi

4

+ i sin pi

4

)] 1

5

(1 + i)

1

5 =

√

2

1

5

[

cos

(1

5

) (

pi

4

)

+ i sin

(1

5

) (

pi

4

)]

(1 + i)

1

5 =

5√

2

(

cos pi

20

+ i sin pi

20

)

In standard form (1 + i) 15 = (1.06 + 1.06i) and this is the principal root of (1 + i) 15 .

6. 81i in polar form is:

399 www.ck12.org

r =

√

02 + 812 = 81, tan θ = 81

0

= und → θ = pi

2

81

(

cos pi

2

+ i sin pi

2

)

[

81

(

cos

(

pi

2

+ 2pik

)

+ i sin

(

pi

2

+ 2pik

))] 1

4

3

(

cos

( pi

2 + 2pik

4

)

+ i sin

( pi

2 + 2pik

4

))

3

(

cos

(

pi

8

+

pik

2

)

+ i sin

(

pi

8

+

pik

2

))

z1 = 3

(

cos

(

pi

8

+

0pi

2

)

+ i sin

(

pi

8

+

0pi

2

))

= 3 cos pi

8

+ 3i sin pi

8

= 2.77 + 1.15i

z2 = 3

(

cos

(

pi

8

+

pi

2

)

+ i sin

(

pi

8

+

pi

2

))

= 3 cos 5pi

8

+ 3i sin 5pi

8

= −1.15 + 2.77i

z3 = 3

(

cos

(

pi

8

+

2pi

2

)

+ i sin

(

pi

8

+

2pi

2

))

= 3 cos 9pi

8

+ 3i sin 9pi

8

= −2.77 − 1.15i

z4 = 3

(

cos

(

pi

8

+

3pi

2

)

+ i sin

(

pi

8

+

3pi

2

))

= 3 cos 13pi

8

+ 3i sin 13pi

8

= 1.15 − 2.77i

7.

x4 + 1 = 0 r =

√

x2 + y2

x4 = −1 r =

√

(−1)2 + (0)2

x4 = −1 + 0i r = 1

θ = tan−1

( 0

−1

)

+ pi = pi

Write an expression for determining the fourth roots of x4 = −1 + 0i

(−1 + 0i) 14 = [1(cos(pi+ 2pik) + i sin(pi+ 2pik))] 14

(−1 + 0i) 14 = 1 14

(

cos pi+ 2pik

4

+ i sin pi+ 2pik

4

)

x1 = 1

(

cos pi

4

+ i sin pi

4

)

=

√

2

2

+ i

√

2

2

f or k = 0

x2 = 1

(

cos 3pi

4

+ i sin 3pi

4

)

= −

√

2

2

+ i

√

2

2

f or k = 1

x3 = 1

(

cos 5pi

4

+ i sin 5pi

4

)

= −

√

2

2

− i

√

2

2

f or k = 2

x4 = 1

(

cos 7pi

4

+ i sin 7pi

4

)

=

√

2

2

− i

√

2

2

f or k = 3

8.

x3 − 64 = 0→ x3 = 64 + 0i

3√

64

(

cos

(

pi+ 2pik

3

)

+ i sin

(

pi+ 2pik

3

))

z1 = 4

(

cos

(

pi+ 2pi0

3

)

+ i sin

(

pi+ 2pi0

3

))

= 4 cos pi

3

+ 4i sin pi

3

= 2 + 2i

√

3

z2 = 4

(

cos

(

pi+ 2pi

3

)

+ i sin

(

pi+ 2pi

3

))

= 4 cos 3pi

3

+ 4i sin 3pi

3

= −4

z3 = 4

(

cos

(

pi+ 4pi

3

)

+ i sin

(

pi+ 4pi

3

))

= 4 cos 5pi

3

+ 4i sin 5pi

3

= 2 − 2i√3

www.ck12.org 400

6.8 Chapter Review

Chapter Summary

In this chapter we made the connection between complex numbers and trigonometry. First, we started

with the polar system, by graphing and converting equations into polar coordinates. This allowed us to

compare the complex plane with the polar plane and we realized that there are many similarities. Because

of this, we are able to convert complex numbers into polar, or trigonometric, form. Converting complex

numbers to polar form makes it easier to multiply and divide complex numbers by using the Product and

Quotient theorems. These theorems lead to DeMoirve’s Theorem, which is a shortcut for raising complex

numbers to different powers. Finally, we were able manipulate DeMoirve’s Theorem to find all the complex

solutions to different equations.

Vocabulary

Argument In the complex number r(cos θ + i sin θ), the argument is the angle θ.

Modulus In the complex number r(cos θ + i sin θ), the modulus is r. It is the distance from the origin to

the point (x, y) in the complex plane.

Polar coordinate system A method of recording the position of an object by using the distance from a

fixed point and an angle consisting of a fixed ray from that point. Also called a polar plane.

Pole In a polar coordinate system, it is the fixed point or origin.

Polar axis In a polar coordinate system, it is the horizontal ray that begins at the pole and extends in a

positive direction.

Polar coordinates The coordinates of a point plotted on a polar plane (r, θ).

Polar Equation An equation which uses polar coordinates.

Polar Form Also called trigonometric form is the complex number x+yi written as r(cos θ+ i sin θ) where

r =

√

x2 + y2 and tan θ = yx .

Review Questions

1. Plot A

(

−3, 3pi4

)

and find three other equivalent coordinates.

2. Find the distance between (2, 94◦) and (7,−73◦).

3. Graph the following polar curves.

(a) r = 3 sin 5θ

(b) r = 6 − 3 cos θ

(c) r = 2 + 5 cos 9θ

4. Determine the equations of the curves below.

401 www.ck12.org

(a)

(b)

5. Convert each equation or point into polar form.

(a) A(−6, 11)

(b) B(15,−8)

(c) C(9, 40)

(d) x2 + (y − 6)2 = 36

6. Convert each equation or point into rectangular form.

(a) D

(

4,−pi3

)

(b) E(−2, 135◦)

(c) r = 7

(d) r = 8 sin θ

7. Determine where r = 9 + 2 sin θ and r = 1 − 4 cos θ intersect.

8. Change −3 + 8i into polar form.

9. Change 15∠240◦ into rectangular form.

10. Multiply or divide the following complex numbers using polar form.

(a)

(

7cis7pi4

)

·

(

3cispi3

)

(b) 8∠80◦2∠−155◦

11. Expand [4

(

cos pi4 + i sin pi4

)

]6

12. Find the 6th roots of -64 and graph them in the complex plane.

13. Find all the solutions of x4 + 32 = 0.

www.ck12.org 402

Review Answers

1. A

(

−3, 3pi4

)

three equivalent coordinates →

(

3,−pi4

)

,

(

3, 7pi4

)

,

(

−3,−5pi4

)

.

2. (2, 94◦) and (7,−73◦)

d =

√

22 + 72 − 2(2)(7) cos(94◦ − (−73◦))

=

√

4 + 49 − 28 cos 167◦

=

√

80.28 ≈ 8.96

3. a) r = 3 sin 5θ

b) r = 6 − 3 cos θ

c) r = 2 + 5 cos 9θ

403 www.ck12.org

4. (a) r = 2 − 6 cos θ

(b) r = 7 + 3 sin θ

5. (a) A(−6, 11)→ r = √36 + 121 ≈ 12.59, tan θ = −116 , θ = 118.6◦ → 12.59(cos 118.6◦ + i sin 118.6◦)

(b) B(15,−8)→ r = √225 + 64 = 17, tan θ = − 815 , θ = −28.1◦ 17(cos(−28.1)◦ + i sin(−28.1◦))

(c) C(9, 40)→ r = √91 + 1600 = 41, tan θ = 409 , θ = 77.3◦41(cos 77.3◦ + i sin 77.3◦)(d)

x2 + (y − 6)2 = 36

r2 cos2 θ + (r sin θ − 6)2 = 36

r2 cos2 θ + r2 sin2 θ − 12r sin θ + 36 = 36

r2 − 12r sin θ = 0 or

r2 = 12r sin θ

r = 12 sin θ

6. (a) D

(

4,−pi3

)

→ x = 4 cos

(

−pi3

)

= 2, y = 4 sin

(

−pi3

)

= −2√3→ (2,−2√3)

(b) E(−2, 135◦)→ x = −2 cos 135◦ = −√2, x = −2 sin 135◦ = √2→ (−√2, √2)

(c) r = 7→ r2 = 49→ x2 + y2 = 49(d)

r = 8 sin θ

r2 = 8r sin θ

x2 + y2 = 8y

y2 − 8y = −x2

y2 − 8y+ 16 = 16 − x2

(y − 4)2 = 16 − x2

y − 4 = ±

√

16 − x2

y = 4 ±

√

16 − x2

7. r = 6 + 5 sin θ and r = 3 − 4 cos θ

www.ck12.org 404

∗ angle measures in the graph are in radians

8. −3 + 8i, x = −3, y = 8 → r = √(−3)2 + 82 ≈ 8.54, tan θ = −83 → θ = 110.56◦ 8.54(cos 110.56◦ +

i sin 110.56◦

9. 15∠240◦, r = 15, θ = 240◦ → x = 15 cos 240◦ = −7.5, y = 15 sin 240◦ = 15

√

3

2 = 7.5

√

3

10. (a)

(

7cis7pi4

)

·

(

3cispi3

)

= 21cis

(

7pi

4 +

pi

3

)

= 21cis25pi12

(b) 8∠80◦2∠−155◦ = 4∠(80◦ − (−155◦)) = 4∠235◦

11.

[

4

(

cos pi4 + i sin pi4

)]6

= 46

(

cos 6pi4 + i sin 6pi4

)

= 4096

(

cos 3pi2 + i sin 3pi2

)

12. -64 in polar form is 64(cos pi − i sin pi)

[64(cos(pi+ 2pik) + i sin(pi+ 2pik))] 16

2

(

cos

(

pi+ 2pik

6

)

+ i sin

(

pi+ 2pik

6

))

2

(

cos

(

pi

6

+

pik

3

)

+ i sin

(

pi

6

+

pik

3

))

z1 = 2

(

cos

(

pi

6

+

0pi

3

)

+ i sin

(

pi

6

+

0pi

3

))

= 2 cos pi

6

+ 2i sin pi

6

=

2

√

3

2

+

2i

2

=

√

3 + i

z2 = 2

(

cos

(

pi

6

+

pi

3

)

+ i sin

(

pi

6

+

pi

3

))

= 2 cos pi

2

+ 2i sin pi

2

= 2i

z3 = 2

(

cos

(

pi

6

+

2pi

3

)

+ i sin

(

pi

6

+

2pi

3

))

= 2 cos 5pi

6

+ 2i sin 5pi

6

= −2

√

3

2

+

2i

2

= −√3 + i

z4 = 2

(

cos

(

pi

6

+ pi

)

+ i sin

(

pi

6

+ pi

))

= 2 cos 7pi

6

+ 2i sin 7pi

6

= −2

√

3

2

− 2i

2

= −√3 − i

z5 = 2

(

cos

(

pi

6

+

4pi

3

)

+ i sin

(

pi

6

+

4pi

3

))

= 2 cos 3pi

2

+ 2i sin 3pi

2

= −2i

z6 = 2

(

cos

(

pi

6

+

5pi

3

)

+ i sin

(

pi

6

+

5pi

3

))

= 2 cos 11pi

6

+ 2i sin 11pi

6

=

2

√

3

2

− 2i

2

=

√

3 − i

Graph of the solutions:

405 www.ck12.org

13.

x4 + 32 = 0→ x4 = −32 + 0i = −32(cos pi+ i sin pi)

[32(cos(pi+ 2pik) + i sin(pi+ 2pik))] 14

2

4√

2

(

cos

(

pi+ 2pik

4

)

+ i sin

(

pi+ 2pik

4

))

2

4√

2

(

cos

(

pi

4

+

pik

2

)

+ i sin

(

pi

4

+

pik

2

))

z1 = 2

4√

2

(

cos

(

pi

4

)

+ i sin

(

pi

4

))

= 2

4√

2 cos pi

4

+ 2i

4√

2 sin pi

4

=

2 4

√

2

√

2

2

+

2i 4

√

2

√

2

2

=

4√

2

3

+ i

4√

2

3

z2 = 2

4√

2

(

cos

(

pi

4

+

pi

2

)

+ i sin

(

pi

4

+

pi

2

))

= 2

4√

2 cos 3pi

4

+ 2i

4√

2 sin 3pi

4

= −2

4√2√2

2

+

2i 4

√

2

√

2

2

= − 4√23 + i 4√23

z3 = 2

4√

2

(

cos

(

pi

4

+ pi

)

+ i sin

(

pi

4

+ pi

))

= 2

4√

2 cos 5pi

4

+ 2i

4√

2 sin 5pi

4

= −2

4√2√2

2

− 2i

4√2√2

2

= − 4√23 − i 4√23

z4 = 2

4√

2

(

cos

(

pi

4

+

3pi

2

)

+ i sin

(

pi

4

+

3pi

2

))

= 2

4√

2 cos 7pi

4

+ 2i

4√

2 sin 7pi

4

=

2 4

√

2

√

2

2

− 2i

4√2√2

2

=

4√

2

3 − i 4√23

Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator

activities designed to supplement the objectives for some of the lessons in this chapter. See

http://www.ck12.org/flexr/chapter/9704.

www.ck12.org 406

Chapter 7

Solving Systems of Equations

and Inequalities

7.1 Linear Systems by Graphing

Learning Objectives

• Determine whether an ordered pair is a solution to a system of equations.

• Solve a system of equations graphically.

• Solve a system of equations graphically with a graphing calculator.

• Solve word problems using systems of equations.

Introduction

In this lesson, we’ll discover methods to determine if an ordered pair is a solution to a system of two

equations. Then we’ll learn to solve the two equations graphically and confirm that the solution is the

point where the two lines intersect. Finally, we’ll look at real-world problems that can be solved using the

methods described in this chapter.

Determine Whether an Ordered Pair is a Solution to a System

of Equations

A linear system of equations is a set of equations that must be solved together to find the one solution

that fits them both.

Consider this system of equations:

y = x+ 2

y = −2x+ 1

Since the two lines are in a system, we deal with them together by graphing them on the same coordinate

axes. We can use any method to graph them; let’s do it by making a table of values for each line.

Line 1: y = x+ 2

407 www.ck12.org

Table 7.1:

x y

0 2

1 3

Line 2: y = −2x+ 1

Table 7.2:

x y

0 1

1 -1

We already know that any point that lies on a line is a solution to the equation for that line. That means

that any point that lies on both lines in a system is a solution to both equations.

So in this system:

• Point A is not a solution to the system because it does not lie on either of the lines.

• Point B is not a solution to the system because it lies only on the blue line but not on the red line.

• Point C is a solution to the system because it lies on both lines at the same time.

In fact, point C is the only solution to the system, because it is the only point that lies on both lines.

For a system of equations, the geometrical solution is the intersection of the two lines in the system. The

algebraic solution is the ordered pair that solves both equations—in other words, the coordinates of that

intersection point.

You can confirm the solution by plugging it into the system of equations, and checking that the solution

works in each equation.

Example 1

Determine which of the points (1, 3), (0, 2), or (2, 7) is a solution to the following system of equations:

y = 4x − 1

y = 2x+ 3

www.ck12.org 408

Solution

To check if a coordinate point is a solution to the system of equations, we plug each of the x and y values

into the equations to see if they work.

Point (1, 3):

y = 4x − 1

3 ? = ? 4(1) − 1

3 = 3 solution checks

y = 2x+ 3

3 ? = ? 2(1) + 3

3 , 5 solution does not check

Point (1, 3) is on the line y = 4x − 1, but it is not on the line y = 2x + 3, so it is not a solution to the

system.

Point (0, 2):

y = 4x − 1

2 ? = ? 4(0) − 1

2 , −1 solution does not check

Point (0, 2) is not on the line y = 4x− 1, so it is not a solution to the system. Note that it is not necessary

to check the second equation because the point needs to be on both lines for it to be a solution to the

system.

Point (2, 7):

y = 4x − 1

7 ? = ? 4(2) − 1

7 = 7 solution checks

y = 2x+ 3

7 ? = ? 2(2) + 3

7 = 7 solution checks

Point (2, 7) is a solution to the system since it lies on both lines.

The solution to the system is the point (2, 7).

Determine the Solution to a Linear System by Graphing

The solution to a linear system of equations is the point, (if there is one) that lies on both lines. In other

words, the solution is the point where the two lines intersect.

We can solve a system of equations by graphing the lines on the same coordinate plane and reading the

intersection point from the graph.

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This method most often offers only approximate solutions, so it’s not sufficient when you need an exact

answer. However, graphing the system of equations can be a good way to get a sense of what’s really going

on in the problem you’re trying to solve, especially when it’s a real-world problem.

Example 2

Solve the following system of equations by graphing:

y = 3x − 5

y = −2x+ 5

Solution

Graph both lines on the same coordinate axis using any method you like.

In this case, let’s make a table of values for each line.

Line 1: y = 3x − 5

Table 7.3:

x y

1 -2

2 1

Line 2: y = −2x+ 5

Table 7.4:

x y

1 3

2 1

The solution to the system is given by the intersection point of the two lines. The graph shows that the

lines intersect at point (2, 1). So the solution is x = 2, y = 1 or (2, 1).

Example 3

Solve the following system of equations by graphing:

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2x+ 3y = 6

4x − y = −2

Solution

Since the equations are in standard form, this time we’ll graph them by finding the x− and y−intercepts

of each of the lines.

Line 1: 2x+ 3y = 6

x−intercept: set y = 0⇒ 2x = 6⇒ x = 3 so the intercept is (3, 0)

y−intercept: set x = 0⇒ 3y = 6⇒ y = 2 so the intercept is (0, 2)

Line 2: −4x+ y = 2

x−intercept: set y = 0⇒ −4x = 2⇒ x = −12 so the intercept is

(

−12 , 0

)

y−intercept: set x = 0⇒ y = 2 so the intercept is (0, 2)

The graph shows that the lines intersect at (0, 2). Therefore, the solution to the system of equations

is x = 0, y = 2.

Solving a System of Equations Using a Graphing Calculator

As an alternative to graphing by hand, you can use a graphing calculator to find or check solutions to a

system of equations.

Example 4

Solve the following system of equations using a graphing calculator.

x − 3y = 4

2x+ 5y = 8

To input the equations into the calculator, you need to rewrite them in slope-intercept form (that is,

y = mx+ b form).

x − 3y = 4 y = 13 x − 43⇒

2x+ 5y = 8 y = −25 x+ 85

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Press the [y=] button on the graphing calculator and enter the two functions as:

Y1 =

x

3

− 4

3

Y2 =

−2x

5

+

8

5

Now press [GRAPH]. Here’s what the graph should look like on a TI-83 family graphing calculator with

the window set to −5 ≤ x ≤ 10 and −5 ≤ y ≤ 5.

There are a few different ways to find the intersection point.

Option 1: Use [TRACE] and move the cursor with the arrows until it is on top of the intersection point.

The values of the coordinate point will be shown on the bottom of the screen. The second screen above

shows the values to be X = 4.0957447 and Y = 0.03191489.

Use the [ZOOM] function to zoom into the intersection point and find a more accurate result. The third

screen above shows the system of equations after zooming in several times. A more accurate solution

appears to be X = 4 and Y = 0.

Option 2 Look at the table of values by pressing [2nd] [GRAPH]. The first screen below shows a table

of values for this system of equations. Scroll down until the Y−values for the two functions are the same.

In this case this occurs at X = 4 and Y = 0.

(Use the [TBLSET] function to change the starting value for your table of values so that it is close to the

intersection point and you don’t have to scroll too long. You can also improve the accuracy of the solution

by setting the value of ∆ Table smaller.)

Option 3 Using the [2nd] [TRACE] function gives the second screen shown above.

Scroll down and select “intersect.”

The calculator will display the graph with the question [FIRSTCURVE]? Move the cursor along the first

curve until it is close to the intersection and press [ENTER].

The calculator now shows [SECONDCURVE]?

Move the cursor to the second line (if necessary) and press [ENTER].

The calculator displays [GUESS]?

Press [ENTER] and the calculator displays the solution at the bottom of the screen (see the third screen

above).

The point of intersection is X = 4 and Y = 0. Note that with this method, the calculator works out the

intersection point for you, which is generally more accurate than your own visual estimate.

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Solve Real-World Problems Using Graphs of Linear Systems

Consider the following problem:

Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run

at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How

long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with

Peter?

Let’s start by drawing a sketch. Here’s what the race looks like when Nadia starts running; we’ll call this

time t = 0.

Now let’s define two variables in this problem:

t = the time from when Nadia starts running

d = the distance of the runners from the starting point.

Since there are two runners, we need to write equations for each of them. That will be the system of

equations for this problem.

For each equation, we use the formula: distance = speed × time

Nadia’s equation: d = 6t

Peter’s equation: d = 5t + 20

(Remember that Peter was already 20 feet from the starting point when Nadia started running.)

Let’s graph these two equations on the same coordinate axes.

Time should be on the horizontal axis since it is the independent variable. Distance should be on the

vertical axis since it is the dependent variable.

We can use any method for graphing the lines, but in this case we’ll use the slope–intercept method since

it makes more sense physically.

To graph the line that describes Nadia’s run, start by graphing the y−intercept: (0, 0). (If you don’t see

that this is the y−intercept, try plugging in the test-value of x = 0.)

The slope tells us that Nadia runs 6 feet every one second, so another point on the line is (1, 6). Connecting

these points gives us Nadia’s line:

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To graph the line that describes Peter’s run, again start with the y−intercept. In this case this is the point

(0, 20).

The slope tells us that Peter runs 5 feet every one second, so another point on the line is (1, 25). Connecting

these points gives us Peter’s line:

In order to find when and where Nadia and Peter meet, we’ll graph both lines on the same graph and

extend the lines until they cross. The crossing point is the solution to this problem.

The graph shows that Nadia and Peter meet 20 seconds after Nadia starts running, and 120 feet

from the starting point.

These examples are great at demonstrating that the solution to a system of linear equations means the

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point at which the lines intersect. This is, in fact, the greatest strength of the graphing method because it

offers a very visual representation of system of equations and its solution. You can also see, though, that

finding the solution from a graph requires very careful graphing of the lines, and is really only practical

when you’re sure that the solution gives integer values for x and y. Usually, this method can only offer

approximate solutions to systems of equations, so we need to use other methods to get an exact solution.

Review Questions

Determine which ordered pair satisfies the system of linear equations.

1. y = 3x − 2

y = −x

(a) (1, 4)

(b) (2, 9)

(c)

(

1

2 ,

−1

2

)

2. y = 2x − 3

y = x+ 5

(a) (8, 13)

(b) (-7, 6)

(c) (0, 4)

3. 2x+ y = 8

5x+ 2y = 10

(a) (-9, 1)

(b) (-6, 20)

(c) (14, 2)

4. 3x+ 2y = 6

y = 12 x − 3

(a)

(

3, −32

)

(b) (-4, 3)

(c)

(

1

2 , 4

)

5. 2x − y = 10

3x+ y = −5

(a) (4, -2)

(b) (1, -8)

(c) (-2, 5)

Solve the following systems using the graphing method.

6. y = x+ 3

y = −x+ 3

7. y = 3x − 6

y = −x+ 6

8. 2x = 4

y = −3

9. y = −x+ 5

− x+ y = 1

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10. x+ 2y = 8

5x+ 2y = 0

11. 3x+ 2y = 12

4x − y = 5

12. 5x+ 2y = −4

x − y = 2

13. 2x+ 4 = 3y

x − 2y+ 4 = 0

14. y = 12 x − 3

2x − 5y = 5

15. y = 4

x = 8 − 3y

16. Try to solve the following system using the graphing method:

y = 35 x+ 5

y = −2x − 12 .

(a) What does it look like the x−coordinate of the solution should be?

(b) Does that coordinate really give the same y−value when you plug it into both equations?

(c) Why is it difficult to find the real solution to this system?

17. Try to solve the following system using the graphing method:

y = 4x+ 8

y = 5x+ 1.

Use a grid with x−values and y−values ranging from -10 to 10.

(a) Do these lines appear to intersect?

(b) Based on their equations, are they parallel?

(c) What would we have to do to find their intersection point?

18. Try to solve the following system using the graphing method:

y = 12 x+ 4

y = 49 x+

9

2 .

Use the same grid as before.

(a) Can you tell exactly where the lines cross?

(b) What would we have to do to make it clearer?

Solve the following problems by using the graphing method.

19. Mary’s car has broken down and it will cost her $1200 to get it fixed—or, for $4500, she can buy a

new, more efficient car instead. Her present car uses about $2000 worth of gas per year, while gas for

the new car would cost about $1500 per year. After how many years would the total cost of fixing

the car equal the total cost of replacing it?

20. Juan is considering two cell phone plans. The first company charges $120 for the phone and $30 per

month for the calling plan that Juan wants. The second company charges $40 for the same phone

but charges $45 per month for the calling plan that Juan wants. After how many months would the

total cost of the two plans be the same?

21. A tortoise and hare decide to race 30 feet. The hare, being much faster, decides to give the tortoise

a 20 foot head start. The tortoise runs at 0.5 feet/sec and the hare runs at 5.5 feet per second. How

long until the hare catches the tortoise?

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7.2 Solving Linear Systems by Substitution

Learning Objectives

• Solve systems of equations with two variables by substituting for either variable.

• Manipulate standard form equations to isolate a single variable.

• Solve real-world problems using systems of equations.

• Solve mixture problems using systems of equations.

Introduction

In this lesson, we’ll learn to solve a system of two equations using the method of substitution.

Solving Linear Systems Using Substitution of Variable Expres-

sions

Let’s look again at the problem about Peter and Nadia racing.

Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run

at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How

long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with

Peter?

In that example we came up with two equations:

Nadia’s equation: d = 6t

Peter’s equation: d = 5t + 20

Each equation produced its own line on a graph, and to solve the system we found the point at which the

lines intersected—the point where the values for d and t satisfied both relationships. When the values for

d and t are equal, that means that Peter and Nadia are at the same place at the same time.

But there’s a faster way than graphing to solve this system of equations. Since we want the value of d to

be the same in both equations, we could just set the two right-hand sides of the equations equal to each

other to solve for t. That is, if d = 6t and d = 5t+20, and the two d’s are equal to each other, then by the

transitive property we have 6t = 5t + 20. We can solve this for t:

6t = 5t + 20 subtract 5t f rom both sides :

t = 20 substitute this value f or t into Nadia’s equation :

d = 6 · 20 = 120

Even if the equations weren’t so obvious, we could use simple algebraic manipulation to find an expression

for one variable in terms of the other. If we rearrange Peter’s equation to isolate t:

d = 5t + 20 subtract 20 f rom both sides :

d − 20 = 5t divide by 5 :

d − 20

5

= t

We can now substitute this expression for t into Nadia’s equation (d = 6t) to solve:

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d = 6

(

d − 20

5

)

multiply both sides by 5 :

5d = 6(d − 20) distribute the 6 :

5d = 6d − 120 subtract 6d f rom both sides :

−d = −120 divide by − 1 :

d = 120 substitute value f or d into our expression f or t :

t =

120 − 20

5

=

100

5

= 20

So we find that Nadia and Peter meet 20 seconds after they start racing, at a distance of 120 feet away.

The method we just used is called the Substitution Method. In this lesson you’ll learn several techniques

for isolating variables in a system of equations, and for using those expressions to solve systems of equations

that describe situations like this one.

Example 1

Let’s look at an example where the equations are written in standard form.

Solve the system

2x+ 3y = 6

−4x+ y = 2

Again, we start by looking to isolate one variable in either equation. If you look at the second equation,

you should see that the coefficient of y is 1. So the easiest way to start is to use this equation to solve for

y.

Solve the second equation for y:

−4x+ y = 2 add 4x to both sides :

y = 2 + 4x

Substitute this expression into the first equation:

2x+ 3(2 + 4x) = 6 distribute the 3 :

2x+ 6 + 12x = 6 collect like terms :

14x+ 6 = 6 subtract 6 f rom both sides :

14x = 0 and hence :

x = 0

Substitute back into our expression for y:

y = 2 + 4 · 0 = 2

As you can see, we end up with the same solution (x = 0, y = 2) that we found when we graphed these

functions back in Lesson 7.1. So long as you are careful with the algebra, the substitution method can be

a very efficient way to solve systems.

Next, let’s look at a more complicated example. Here, the values of x and y we end up with aren’t whole

numbers, so they would be difficult to read off a graph!

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Example 2

Solve the system

2x+ 3y = 3

2x − 3y = −1

Again, we start by looking to isolate one variable in either equation. In this case it doesn’t matter which

equation we use—all the variables look about equally easy to solve for.

So let’s solve the first equation for x:

2x+ 3y = 3 subtract 3y f rom both sides :

2x = 3 − 3y divide both sides by 2 :

x =

1

2

(3 − 3y)

Substitute this expression into the second equation:

2 · 12(3 − 3y) − 3y = −1 cancel the f raction and re − write terms :

3 − 3y − 3y = −1 collect like terms :

3 − 6y = −1 subtract 3 f rom both sides :

−6y = −4 divide by − 6 :

y =

2

3

Substitute into the expression we got for x:

x =

1

2

(

3 − 3

(2

3

))

x =

1

2

So our solution is x = 12 , y = 23 . You can see how the graphical solution

(

1

2 ,

2

3

)

might have been difficult to

read accurately off a graph!

Solving Real-World Problems Using Linear Systems

Simultaneous equations can help us solve many real-world problems. We may be considering a pur-

chase—for example, trying to decide whether it’s cheaper to buy an item online where you pay shipping

or at the store where you do not. Or you may wish to join a CD music club, but aren’t sure if you would

really save any money by buying a new CD every month in that way. Or you might be considering two

different phone contracts. Let’s look at an example of that now.

Example 3

Anne is trying to choose between two phone plans. The first plan, with Vendafone, costs $20 per month,

with calls costing an additional 25 cents per minute. The second company, Sellnet, charges $40 per month,

but calls cost only 8 cents per minute. Which should she choose?

You should see that Anne’s choice will depend upon how many minutes of calls she expects to use each

month. We start by writing two equations for the cost in dollars in terms of the minutes used. Since the

number of minutes is the independent variable, it will be our x. Cost is dependent on minutes – the cost

per month is the dependent variable and will be assigned y.

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For Vendafone: y = 0.25x+ 20

For Sellnet: y = 0.08x+ 40

By writing the equations in slope-intercept form (y = mx + b), you can sketch a graph to visualize the

situation:

The line for Vendafone has an intercept of 20 and a slope of 0.25. The Sellnet line has an intercept of 40

and a slope of 0.08 (which is roughly a third of the Vendafone line’s slope). In order to help Anne decide

which to choose, we’ll find where the two lines cross, by solving the two equations as a system.

Since equation 1 gives us an expression for y(0.25x + 20), we can substitute this expression directly into

equation 2:

0.25x+ 20 = 0.08x+ 40 subtract 20 f rom both sides :

0.25x = 0.08x+ 20 subtract 0.08x f rom both sides :

0.17x = 20 divide both sides by 0.17 :

x = 117.65 minutes rounded to 2 decimal places.

So if Anne uses 117.65 minutes a month (although she can’t really do exactly that, because phone plans

only count whole numbers of minutes), the phone plans will cost the same. Now we need to look at the

graph to see which plan is better if she uses more minutes than that, and which plan is better if she uses

fewer. You can see that the Vendafone plan costs more when she uses more minutes, and the Sellnet plan

costs more with fewer minutes.

So, if Anne will use 117 minutes or less every month she should choose Vendafone. If she

plans on using 118 or more minutes she should choose Sellnet.

Mixture Problems

Systems of equations crop up frequently in problems that deal with mixtures of two things—chemicals in

a solution, nuts and raisins, or even the change in your pocket! Let’s look at some examples of these.

Example 4

Janine empties her purse and finds that it contains only nickels (worth 5 cents each) and dimes (worth 10

cents each). If she has a total of 7 coins and they have a combined value of 45 cents, how many of each

coin does she have?

Since we have 2 types of coins, let’s call the number of nickels x and the number of dimes y. We are given

two key pieces of information to make our equations: the number of coins and their value.

# of coins equation: x+ y = 7 (number o f nickels) + (number o f dimes)

value equation: 5x+ 10y = 55 (since nickels are worth 5c and dimes 10c)

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We can quickly rearrange the first equation to isolate x:

x = 7 − y now substitute into equation 2 :

5(7 − y) + 10y = 55 distribute the 5 :

35 − 5y+ 10y = 55 collect like terms :

35 + 5y = 55 subtract 35 f rom both sides :

5y = 20 divide by 5 :

y = 4 substitute back into equation 1 :

x+ 4 = 7 subtract 4 f rom both sides :

x = 3

Janine has 3 nickels and 4 dimes.

Sometimes a question asks you to determine (from concentrations) how much of a particular substance

to use. The substance in question could be something like coins as above, or it could be a chemical in

solution, or even heat. In such a case, you need to know the amount of whatever substance is in each part.

There are several common situations where to get one equation you simply add two given quantities, but

to get the second equation you need to use a product. Three examples are below.

Table 7.5:

Type of mixture First equation Second equation

Coins (items with $ value) total number of items (n1 + n2) total value (item value × no. of

items)

Chemical solutions total solution volume (V1 + V2) amount of solute (vol × concen-

tration)

Density of two substances total amount or volume of mix total mass (volume × density)

For example, when considering mixing chemical solutions, we will most likely need to consider the total

amount of solute in the individual parts and in the final mixture. (A solute is the chemical that is dissolved

in a solution. An example of a solute is salt when added to water to make a brine.) To find the total

amount, simply multiply the amount of the mixture by the fractional concentration. To illustrate, let’s

look at an example where you are given amounts relative to the whole.

Example 5

A chemist needs to prepare 500 ml of copper-sulfate solution with a 15% concentration. She wishes to use

a high concentration solution (60%) and dilute it with a low concentration solution (5%) in order to do

this. How much of each solution should she use?

Solution

To set this problem up, we first need to define our variables. Our unknowns are the amount of concentrated

solution (x) and the amount of dilute solution (y). We will also convert the percentages (60%, 15% and

5%) into decimals (0.6, 0.15 and 0.05). The two pieces of critical information are the final volume (500 ml)

and the final amount of solute (15% of 500 ml = 75 ml). Our equations will look like this:

Volume equation: x+ y = 500

Solute equation: 0.6x+ 0.05y = 75

To isolate a variable for substitution, we can see it’s easier to start with equation 1:

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x+ y = 500 subtract y f rom both sides :

x = 500 − y now substitute into equation 2 :

0.6(500 − y) + 0.05y = 75 distribute the 0.6 :

300 − 0.6y+ 0.05y = 75 collect like terms :

300 − 0.55y = 75 subtract 300 f rom both sides :

− 0.55y = −225 divide both sides by − 0.55 :

y = 409 ml substitute back into equation f or x :

x = 500 − 409 = 91 ml

So the chemist should mix 91 ml of the 60% solution with 409 ml of the 5% solution.

Further Practice

For lots more practice solving linear systems, check out this web page: http://www.algebra.com/algebra/

homework/coordinate/practice-linear-system.epl

After clicking to see the solution to a problem, you can click the back button and then click Try Another

Practice Linear System to see another problem.

Review Questions

1. Solve the system:

x+ 2y = 9

3x+ 5y = 20

2. Solve the system:

x − 3y = 10

2x+ y = 13

3. Solve the system:

2x+ 0.5y = −10

x − y = −10

4. Solve the system:

2x+ 0.5y = 3

x+ 2y = 8.5

5. Solve the system:

3x+ 5y = −1

x+ 2y = −1

6. Solve the system:

3x+ 5y = −3

x+ 2y = −43

7. Solve the system:

x − y = −125

2x+ 5y = −2

8. Of the two non-right angles in a right angled triangle, one measures twice as many degrees as the

other. What are the angles?

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9. The sum of two numbers is 70. They differ by 11. What are the numbers?

10. A number plus half of another number equals 6; twice the first number minus three times the second

number equals 4. What are the numbers?

11. A rectangular field is enclosed by a fence on three sides and a wall on the fourth side. The total length

of the fence is 320 yards. If the field has a total perimeter of 400 yards, what are the dimensions of

the field?

12. A ray cuts a line forming two angles. The difference between the two angles is 18◦. What does each

angle measure?

13. I have $15 and wish to buy five pounds of mixed nuts for a party. Peanuts cost $2.20 per pound.

Cashews cost $4.70 per pound.

(a) How many pounds of each should I buy?

(b) If I suddenly realize I need to set aside $5 to buy chips, can I still buy 5 pounds of nuts with

the remaining $10?

(c) What’s the greatest amount of nuts I can buy?

14. A chemistry experiment calls for one liter of sulfuric acid at a 15% concentration, but the supply

room only stocks sulfuric acid in concentrations of 10% and 35%.

(a) How many liters of each should be mixed to give the acid needed for the experiment?

(b) How many liters should be mixed to give two liters at a 15% concentration?

15. Bachelle wants to know the density of her bracelet, which is a mix of gold and silver. Density is total

mass divided by total volume. The density of gold is 19.3 g/cc and the density of silver is 10.5 g/cc.

The jeweler told her that the volume of silver in the bracelet was 10 cc and the volume of gold was

20 cc. Find the combined density of her bracelet.

16. Jason is five years older than Becky, and the sum of their ages is 23. What are their ages?

17. Tickets to a show cost $10 in advance and $15 at the door. If 120 tickets are sold for a total of $1390,

how many of the tickets were bought in advance?

18. The multiple-choice questions on a test are worth 2 points each, and the short-answer questions are

worth 5 points each.

(a) If the whole test is worth 100 points and has 35 questions, how many of the questions are

multiple-choice and how many are short-answer?

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(b) If Kwan gets 31 questions right and ends up with a score of 86 on the test, how many questions

of each type did she get right? (Assume there is no partial credit.)

(c) If Ashok gets 5 questions wrong and ends up with a score of 87 on the test, how many questions

of each type did he get wrong? (Careful!)

(d) What are two ways you could have set up the equations for part c?

(e) How could you have set up part b differently?

7.3 Solving Linear Systems by Elimination

Learning Objectives

• Solve a linear system of equations using elimination by addition.

• Solve a linear system of equations using elimination by subtraction.

• Solve a linear system of equations by multiplication and then addition or subtraction.

• Compare methods for solving linear systems.

• Solve real-world problems using linear systems by any method.

Introduction

In this lesson, we’ll see how to use simple addition and subtraction to simplify our system of equations

to a single equation involving a single variable. Because we go from two unknowns (x and y) to a single

unknown (either x or y), this method is often referred to by solving by elimination. We eliminate one

variable in order to make our equations solvable! To illustrate this idea, let’s look at the simple example

of buying apples and bananas.

Example 1

If one apple plus one banana costs $1.25 and one apple plus 2 bananas costs $2.00, how much does one

banana cost? One apple?

It shouldn’t take too long to discover that each banana costs $0.75. After all, the second purchase just

contains 1 more banana than the first, and costs $0.75 more, so that one banana must cost $0.75.

Here’s what we get when we describe this situation with algebra:

a+ b = 1.25

a+ 2b = 2.00

Now we can subtract the number of apples and bananas in the first equation from the number in the second

equation, and also subtract the cost in the first equation from the cost in the second equation, to get the

difference in cost that corresponds to the difference in items purchased.

(a+ 2b) − (a+ b) = 2.00 − 1.25→ b = 0.75

That gives us the cost of one banana. To find out how much one apple costs, we subtract $0.75 from the

total cost of one apple and one banana.

a+ 0.75 = 1.25→ a = 1.25 − 0.75→ a = 0.50

So an apple costs 50 cents.

To solve systems using addition and subtraction, we’ll be using exactly this idea – by looking at the sum

or difference of the two equations we can determine a value for one of the unknowns.

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Solving Linear Systems Using Addition of Equations

Often considered the easiest and most powerful method of solving systems of equations, the addition (or

elimination) method lets us combine two equations in such a way that the resulting equation has only one

variable. We can then use simple algebra to solve for that variable. Then, if we need to, we can substitute

the value we get for that variable back into either one of the original equations to solve for the other

variable.

Example 2

Solve this system by addition:

3x+ 2y = 11

5x − 2y = 13

Solution

We will add everything on the left of the equals sign from both equations, and this will be equal to the

sum of everything on the right:

(3x+ 2y) + (5x − 2y) = 11 + 13→ 8x = 24→ x = 3

A simpler way to visualize this is to keep the equations as they appear above, and to add them together

vertically, going down the columns. However, just like when you add units, tens and hundreds, you MUST

be sure to keep the x’s and y’s in their own columns. You may also wish to use terms like “0y” as a

placeholder!

3x+ 2y = 11

+ (5x − 2y) = 13

8x+ 0y = 24

Again we get 8x = 24, or x = 3. To find a value for y, we simply substitute our value for x back in.

Substitute x = 3 into the second equation:

5 · 3 − 2y = 13 since 5 × 3 = 15, we subtract 15 f rom both sides :

−2y = −2 divide by − 2 to get :

y = 1

The reason this method worked is that the y−coefficients of the two equations were opposites of each other:

2 and -2. Because they were opposites, they canceled each other out when we added the two equations

together, so our final equation had no y−term in it and we could just solve it for x.

In a little while we’ll see how to use the addition method when the coefficients are not opposites, but for

now let’s look at another example where they are.

Example 3

Andrew is paddling his canoe down a fast-moving river. Paddling downstream he travels at 7 miles per

hour, relative to the river bank. Paddling upstream, he moves slower, traveling at 1.5 miles per hour. If

he paddles equally hard in both directions, how fast is the current? How fast would Andrew travel in calm

water?

Solution

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First we convert our problem into equations. We have two unknowns to solve for, so we’ll call the speed

that Andrew paddles at x, and the speed of the river y. When traveling downstream, Andrew speed is

boosted by the river current, so his total speed is his paddling speed plus the speed of the river (x + y).

Traveling upstream, the river is working against him, so his total speed is his paddling speed minus the

speed of the river (x − y).

Downstream Equation: x+ y = 7

Upstream Equation: x − y = 1.5

Next we’ll eliminate one of the variables. If you look at the two equations, you can see that the coefficient

of y is +1 in the first equation and -1 in the second. Clearly (+1)+ (−1) = 0, so this is the variable we will

eliminate. To do this we simply add equation 1 to equation 2. We must be careful to collect like terms,

and make sure that everything on the left of the equals sign stays on the left, and everything on the right

stays on the right:

(x+ y) + (x − y) = 7 + 1.5⇒ 2x = 8.5⇒ x = 4.25

Or, using the column method we used in example 2:

x+ y = 7

+ x − y = 1.5

2x+ 0y = 8.5

Again we get 2x = 8.5, or x = 4.25. To find a corresponding value for y, we plug our value for x into either

equation and isolate our unknown. In this example, we’ll plug it into the first equation:

4.25 + y = 7 subtract 4.25 f rom both sides :

y = 2.75

Andrew paddles at 4.25 miles per hour. The river moves at 2.75 miles per hour.

Solving Linear Systems Using Subtraction of Equations

Another, very similar method for solving systems is subtraction. When the x− or y−coefficients in both

equations are the same (including the sign) instead of being opposites, you can subtract one equation

from the other.

If you look again at Example 3, you can see that the coefficient for x in both equations is +1. Instead of

adding the two equations together to get rid of the y’s, you could have subtracted to get rid of the x’s:

(x+ y) − (x − y) = 7 − 1.5⇒ 2y = 5.5⇒ y = 2.75

or...

x+ y = 7

− (x − y) = −1.5

0x+ 2y = 5.5

So again we get y = 2.75, and we can plug that back in to determine x.

The method of subtraction is just as straightforward as addition, so long as you remember the following:

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• Always put the equation you are subtracting in parentheses, and distribute the negative.

• Don’t forget to subtract the numbers on the right-hand side.

• Always remember that subtracting a negative is the same as adding a positive.

Example 4

Peter examines the coins in the fountain at the mall. He counts 107 coins, all of which are either pennies

or nickels. The total value of the coins is $3.47. How many of each coin did he see?

Solution

We have 2 types of coins, so let’s call the number of pennies x and the number of nickels y. The total value

of all the pennies is just x, since they are worth 1c each. The total value of the nickels is 5y. We are given

two key pieces of information to make our equations: the number of coins and their value in cents.

# of coins equation : x+ y = 107 (number o f pennies) + (number o f nickels)

value equation : x+ 5y = 347 pennies are worth 1c, nickels are worth 5c.

We’ll jump straight to subtracting the two equations:

x+ y = 107

− (x+ 5y) = −347

− 4y = −240

y = 60

Substituting this value back into the first equation:

x+ 60 = 107 subtract 60 f rom both sides :

x = 47

So Peter saw 47 pennies (worth 47 cents) and 60 nickels (worth $3.00) making a total of

$3.47.

Solving Linear Systems Using Multiplication

So far, we’ve seen that the elimination method works well when the coefficient of one variable happens to

be the same (or opposite) in the two equations. But what if the two equations don’t have any coefficients

the same?

It turns out that we can still use the elimination method; we just have to make one of the coefficients

match. We can accomplish this by multiplying one or both of the equations by a constant.

Here’s a quick review of how to do that. Consider the following questions:

1. If 10 apples cost $5, how much would 30 apples cost?

2. If 3 bananas plus 2 carrots cost $4, how mush would 6 bananas plus 4 carrots cost?

If you look at the first equation, it should be obvious that each apple costs $0.50. So 30 apples should cost

$15.00.

The second equation is trickier; it isn’t obvious what the individual price for either bananas or carrots is.

Yet we know that the answer to question 2 is $8.00. How?

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If we look again at question 1, we see that we can write an equation: 10a = 5 (a being the cost of 1 apple).

So to find the cost of 30 apples, we could solve for a and then multiply by 30—but we could also just

multiply both sides of the equation by 3. We would get 30a = 15, and that tells us that 30 apples cost $15.

And we can do the same thing with the second question. The equation for this situation is 3b + 2c = 4,

and we can see that we need to solve for (6b+ 4c), which is simply 2 times (3b+ 2c)! So algebraically, we

are simply multiplying the entire equation by 2:

2(3b+ 2c) = 2 · 4 distribute and multiply :

6b+ 4c = 8

So when we multiply an equation, all we are doing is multiplying every term in the equation by a fixed

amount.

Solving a Linear System by Multiplying One Equation

If we can multiply every term in an equation by a fixed number (a scalar), that means we can use the

addition method on a whole new set of linear systems. We can manipulate the equations in a system to

ensure that the coefficients of one of the variables match.

This is easiest to do when the coefficient as a variable in one equation is a multiple of the coefficient in the

other equation.

Example 5

Solve the system:

7x+ 4y = 17

5x − 2y = 11

Solution

You can easily see that if we multiply the second equation by 2, the coefficients of y will be +4 and -4,

allowing us to solve the system by addition:

2 times equation 2:

10x − 4y = 22 now add to equation one :

+ (7x+ 4y) = 17

17x = 34

divide by 17 to get : x = 2

Now simply substitute this value for x back into equation 1:

7 · 2 + 4y = 17 since 7 × 2 = 14, subtract 14 f rom both sides :

4y = 3 divide by 4 :

y = 0.75

Example 6

Anne is rowing her boat along a river. Rowing downstream, it takes her 2 minutes to cover 400 yards.

Rowing upstream, it takes her 8 minutes to travel the same 400 yards. If she was rowing equally hard in

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both directions, calculate, in yards per minute, the speed of the river and the speed Anne would travel in

calm water.

Solution

Step one: first we convert our problem into equations. We know that distance traveled is equal to speed

× time. We have two unknowns, so we’ll call the speed of the river x, and the speed that Anne rows at

y. When traveling downstream, her total speed is her rowing speed plus the speed of the river, or (x+ y).

Going upstream, her speed is hindered by the speed of the river, so her speed upstream is (x − y).

Downstream Equation: 2(x+ y) = 400

Upstream Equation: 8(x − y) = 400

Distributing gives us the following system:

2x+ 2y = 400

8x − 8y = 400

Right now, we can’t use the method of elimination because none of the coefficients match. But if we

multiplied the top equation by 4, the coefficients of y would be +8 and -8. Let’s do that:

8x+ 8y = 1, 600

+ (8x − 8y) = 400

16x = 2, 000

Now we divide by 16 to obtain x = 125.

Substitute this value back into the first equation:

2(125 + y) = 400 divide both sides by 2 :

125 + y = 200 subtract 125 f rom both sides :

y = 75

Anne rows at 125 yards per minute, and the river flows at 75 yards per minute.

Solving a Linear System by Multiplying Both Equations

So what do we do if none of the coefficients match and none of them are simple multiples of each other? We

do the same thing we do when we’re adding fractions whose denominators aren’t simple multiples of each

other. Remember that when we add fractions, we have to find a lowest common denominator—that

is, the lowest common multiple of the two denominators—and sometimes we have to rewrite not just one,

but both fractions to get them to have a common denominator. Similarly, sometimes we have to multiply

both equations by different constants in order to get one of the coefficients to match.

Example 7

Andrew and Anne both use the I-Haul truck rental company to move their belongings from home to the

dorm rooms on the University of Chicago campus. I-Haul has a charge per day and an additional charge

per mile. Andrew travels from San Diego, California, a distance of 2060 miles in five days. Anne travels

880 miles from Norfolk, Virginia, and it takes her three days. If Anne pays $840 and Andrew pays $1845,

what does I-Haul charge

a) per day?

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b) per mile traveled?

Solution

First, we’ll set up our equations. Again we have 2 unknowns: the daily rate (we’ll call this x), and the

per-mile rate (we’ll call this y).

Anne’s equation: 3x+ 880y = 840

Andrew’s Equation: 5x+ 2060y = 1845

We can’t just multiply a single equation by an integer number in order to arrive at matching coefficients.

But if we look at the coefficients of x (as they are easier to deal with than the coefficients of y), we see

that they both have a common multiple of 15 (in fact 15 is the lowest common multiple). So we can

multiply both equations.

Multiply the top equation by 5:

15x+ 4400y = 4200

Multiply the lower equation by 3:

15x+ 6180y = 5535

Subtract:

15x+ 4400y = 4200

− (15x+ 6180y) = 5535

− 1780y = −1335

Divide by − 1780 : y = 0.75

Substitute this back into the top equation:

3x+ 880(0.75) = 840 since 880 × 0.75 = 660, subtract 660 f rom both sides :

3x = 180 divide both sides by 3

x = 60

I-Haul charges $60 per day plus $0.75 per mile.

Comparing Methods for Solving Linear Systems

Now that we’ve covered the major methods for solving linear equations, let’s review them. For simplicity,

we’ll look at them in table form. This should help you decide which method would be best for a given

situation.

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Table 7.6:

Method: Best used when

you...

Advantages: Comment:

Graphing ...don’t need an accu-

rate answer.

Often easier to see

number and quality

of intersections on a

graph. With a graphing

calculator, it can be the

fastest method since

you don’t have to do

any computation.

Can lead to impre-

cise answers with

non-integer solutions.

Substitution ...have an explicit equa-

tion for one variable

(e.g. y = 14x+ 2)

Works on all systems.

Reduces the system to

one variable, making it

easier to solve.

You are not often given

explicit functions in sys-

tems problems, so you

may have to do ex-

tra work to get one of

the equations into that

form.

Elimination by Addi-

tion or Subtraction

...have matching coeffi-

cients for one variable in

both equations.

Easy to combine equa-

tions to eliminate one

variable. Quick to solve.

It is not very likely that

a given system will have

matching coefficients.

Elimination by Multi-

plication and then Addi-

tion and Subtraction

...do not have any vari-

ables defined explicitly

or any matching coeffi-

cients.

Works on all systems.

Makes it possible to

combine equations to

eliminate one variable.

Often more algebraic

manipulation is needed

to prepare the equa-

tions.

The table above is only a guide. You might prefer to use the graphical method for every system in order

to better understand what is happening, or you might prefer to use the multiplication method even when

a substitution would work just as well.

Example 8

Two angles are complementary when the sum of their angles is 90◦. Angles A and B are complementary

angles, and twice the measure of angle A is 9◦ more than three times the measure of angle B. Find the

measure of each angle.

Solution

First we write out our 2 equations. We will use x to be the measure of angle A and y to be the measure of

angle B. We get the following system:

x+ y = 90

2x = 3y+ 9

First, we’ll solve this system with the graphical method. For this, we need to convert the two equations to

y = mx+ b form:

x+ y = 90 ⇒ y = −x+ 90

2x = 3y+ 9 ⇒ y = 2

3

x − 3

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The first line has a slope of -1 and a y−intercept of 90, and the second line has a slope of 23 and a y−intercept

of -3. The graph looks like this:

In the graph, it appears that the lines cross at around x = 55, y = 35, but it is difficult to tell exactly!

Graphing by hand is not the best method in this case!

Next, we’ll try solving by substitution. Let’s look again at the system:

x+ y = 90

2x = 3y+ 9

We’ve already seen that we can start by solving either equation for y, so let’s start with the first one:

y = 90 − x

Substitute into the second equation:

2x = 3(90 − x) + 9 distribute the 3 :

2x = 270 − 3x+ 9 add 3x to both sides :

5x = 270 + 9 = 279 divide by 5 :

x = 55.8◦

Substitute back into our expression for y:

y = 90 − 55.8 = 34.2◦

Angle A measures 55.8◦; angle B measures 34.2◦.

Finally, we’ll try solving by elimination (with multiplication):

Rearrange equation one to standard form:

x+ y = 90 ⇒ 2x+ 2y = 180

Multiply equation two by 2:

2x = 3y+ 9 ⇒ 2x − 3y = 9

Subtract:

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2x+ 2y = 180

− (2x − 3y) = −9

5y = 171

Divide by 5 to obtain y = 34.2◦

Substitute this value into the very first equation:

x+ 34.2 = 90 subtract 34.2 f rom both sides :

x = 55.8◦

Angle A measures 55.8◦; angle B measures 34.2◦.

Even though this system looked ideal for substitution, the method of multiplication worked well too. Once

the equations were rearranged properly, the solution was quick to find. You’ll need to decide yourself which

method to use in each case you see from now on. Try to master all the techniques, and recognize which

one will be most efficient for each system you are asked to solve.

The following Khan Academy video contains three examples of solving systems of equations using addition

and subtraction as well as multiplication (which is the next topic): http://www.youtube.com/watch?v=

nok99JOhcjo (9:57). (Note that the narrator is not always careful about showing his work, and you should

try to be neater in your mathematical writing.)

For even more practice, we have this video. One common type of problem involving systems of equations

(especially on standardized tests) is “age problems.” In the following video the narrator shows two examples

of age problems, one involving a single person and one involving two people. Khan Academy Age Problems

(7:13)

Review Questions

1. Solve the system:

3x+ 4y = 2.5

5x − 4y = 25.5

2. Solve the system:

5x+ 7y = −31

5x − 9y = 17

3. Solve the system:

3y − 4x = −33

5x − 3y = 40.5

4. Nadia and Peter visit the candy store. Nadia buys three candy bars and four fruit roll-ups for $2.84.

Peter also buys three candy bars, but can only afford one additional fruit roll-up. His purchase costs

$1.79. What is the cost of a candy bar and a fruit roll-up individually?

5. A small plane flies from Los Angeles to Denver with a tail wind (the wind blows in the same direction

as the plane) and an air-traffic controller reads its ground-speed (speed measured relative to the

ground) at 275 miles per hour. Another, identical plane, moving in the opposite direction has a

ground-speed of 227 miles per hour. Assuming both planes are flying with identical air-speeds,

calculate the speed of the wind.

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6. An airport taxi firm charges a pick-up fee, plus an additional per-mile fee for any rides taken. If a

12-mile journey costs $14.29 and a 17-mile journey costs $19.91, calculate:

(a) the pick-up fee

(b) the per-mile rate

(c) the cost of a seven mile trip

7. Calls from a call-box are charged per minute at one rate for the first five minutes, then a different

rate for each additional minute. If a 7-minute call costs $4.25 and a 12-minute call costs $5.50, find

each rate.

8. A plumber and a builder were employed to fit a new bath, each working a different number of hours.

The plumber earns $35 per hour, and the builder earns $28 per hour. Together they were paid

$330.75, but the plumber earned $106.75 more than the builder. How many hours did each work?

9. Paul has a part time job selling computers at a local electronics store. He earns a fixed hourly wage,

but can earn a bonus by selling warranties for the computers he sells. He works 20 hours per week.

In his first week, he sold eight warranties and earned $220. In his second week, he managed to sell 13

warranties and earned $280. What is Paul’s hourly rate, and how much extra does he get for selling

each warranty?

Solve the following systems using multiplication.

10. 5x − 10y = 15

3x − 2y = 3

11. 5x − y = 10

3x − 2y = −1

12. 5x+ 7y = 15

7x − 3y = 5

13. 9x+ 5y = 9

12x+ 8y = 12.8

14. 4x − 3y = 1

3x − 4y = 4

15. 7x − 3y = −3

6x+ 4y = 3

Solve the following systems using any method.

16. x = 3y

x − 2y = −3

17. y = 3x+ 2

y = −2x+ 7

18. 5x − 5y = 5

5x+ 5y = 35

19. y = −3x − 3

3x − 2y+ 12 = 0

20. 3x − 4y = 3

4y+ 5x = 10

21. 9x − 2y = −4

2x − 6y = 1

22. Supplementary angles are two angles whose sum is 180◦. Angles A and B are supplementary angles.

The measure of Angle A is 18◦ less than twice the measure of Angle B. Find the measure of each

angle.

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23. A farmer has fertilizer in 5% and 15% solutions. How much of each type should he mix to obtain

100 liters of fertilizer in a 12% solution?

24. A 150-yard pipe is cut to provide drainage for two fields. If the length of one piece is three yards less

that twice the length of the second piece, what are the lengths of the two pieces?

25. Mr. Stein invested a total of $100,000 in two companies for a year. Company A’s stock showed a

13% annual gain, while Company B showed a 3% loss for the year. Mr. Stein made an 8% return on

his investment over the year. How much money did he invest in each company?

26. A baker sells plain cakes for $7 and decorated cakes for $11. On a busy Saturday the baker started

with 120 cakes, and sold all but three. His takings for the day were $991. How many plain cakes did

he sell that day, and how many were decorated before they were sold?

27. Twice John’s age plus five times Claire’s age is 204. Nine times John’s age minus three times Claire’s

age is also 204. How old are John and Claire?

7.4 Special Types of Linear Systems

Learning Objectives

• Identify and understand what is meant by an inconsistent linear system.

• Identify and understand what is meant by a consistent linear system.

• Identify and understand what is meant by a dependent linear system.

Introduction

As we saw in Section 7.1, a system of linear equations is a set of linear equations which must be solved

together. The lines in the system can be graphed together on the same coordinate graph and the solution

to the system is the point at which the two lines intersect.

Or at least that’s what usually happens. But what if the lines turn out to be parallel when we graph them?

If the lines are parallel, they won’t ever intersect. That means that the system of equations they represent

has no solution. A system with no solutions is called an inconsistent system.

And what if the lines turn out to be identical?

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If the two lines are the same, then every point on one line is also on the other line, so every point on the

line is a solution to the system. The system has an infinite number of solutions, and the two equations

are really just different forms of the same equation. Such a system is called a dependent system.

But usually, two lines cross at exactly one point and the system has exactly one solution:

A system with exactly one solution is called a consistent system.

To identify a system as consistent, inconsistent, or dependent, we can graph the two lines on the same

graph and see if they intersect, are parallel, or are the same line. But sometimes it is hard to tell whether

two lines are parallel just by looking at a roughly sketched graph.

Another option is to write each line in slope-intercept form and compare the slopes and y− intercepts of

the two lines. To do this we must remember that:

• Lines with different slopes always intersect.

• Lines with the same slope but different y−intercepts are parallel.

• Lines with the same slope and the same y−intercepts are identical.

Example 1

Determine whether the following system has exactly one solution, no solutions, or an infinite number of

solutions.

2x − 5y = 2

4x+ y = 5

Solution

We must rewrite the equations so they are in slope-intercept form

2x − 5y = 2 − 5y = −2x+ 2 y = 25 x − 25⇒ ⇒

4x+ y = 5 y = −4x+ 5 y = −4x+ 5

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The slopes of the two equations are different; therefore the lines must cross at a single point and the system

has exactly one solution. This is a consistent system.

Example 2

Determine whether the following system has exactly one solution, no solutions, or an infinite number of

solutions.

3x = 5 − 4y

6x+ 8y = 7

Solution

We must rewrite the equations so they are in slope-intercept form

3x = 5 − 4y 4y = −3x+ 5 y = −34 x+ 54⇒ ⇒

6x+ 8y = 7 8y = −6x+ 7 y = −34 x+ 78

The slopes of the two equations are the same but the y−intercepts are different; therefore the lines are

parallel and the system has no solutions. This is an inconsistent system.

Example 3

Determine whether the following system has exactly one solution, no solutions, or an infinite number of

solutions.

x+ y = 3

3x+ 3y = 9

Solution

We must rewrite the equations so they are in slope-intercept form

x+ y = 3 y = −x+ 3 y = −x+ 3

⇒ ⇒

3x+ 3y = 9 3y = −3x+ 9 y = −x+ 3

The lines are identical; therefore the system has an infinite number of solutions. It is a dependent

system.

Determining the Type of System Algebraically

A third option for identifying systems as consistent, inconsistent or dependent is to just solve the system

and use the result as a guide.

Example 4

Solve the following system of equations. Identify the system as consistent, inconsistent or dependent.

10x − 3y = 3

2x+ y = 9

Solution

Let’s solve this system using the substitution method.

Solve the second equation for y:

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2x+ y = 9⇒ y = −2x+ 9

Substitute that expression for y in the first equation:

10x − 3y = 3

10x − 3(−2x+ 9) = 3

10x+ 6x − 27 = 3

16x = 30

x =

15

8

Substitute the value of x back into the second equation and solve for y:

2x+ y = 9⇒ y = −2x+ 9⇒ y = −2 · 15

8

+ 9⇒ y = 21

4

The solution to the system is

(

15

8 ,

21

4

)

. The system is consistent since it has only one solution.

Example 5

Solve the following system of equations. Identify the system as consistent, inconsistent or dependent.

3x − 2y = 4

9x − 6y = 1

Solution

Let’s solve this system by the method of multiplication.

Multiply the first equation by 3:

3(3x − 2y = 4) 9x − 6y = 12

⇒

9x − 6y = 1 9x − 6y = 1

Add the two equations:

9x − 6y = 4

9x − 6y = 1

0 = 13 This statement is not true.

If our solution to a system turns out to be a statement that is not true, then the system doesn’t really

have a solution; it is inconsistent.

Example 6

Solve the following system of equations. Identify the system as consistent, inconsistent or dependent.

4x+ y = 3

12x+ 3y = 9

Solution

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Let’s solve this system by substitution.

Solve the first equation for y:

4x+ y = 3⇒ y = −4x+ 3

Substitute this expression for y in the second equation:

12x+ 3y = 9

12x+ 3(−4x+ 3) = 9

12x − 12x+ 9 = 9

9 = 9

This statement is always true.

If our solution to a system turns out to be a statement that is always true, then the system is dependent.

A second glance at the system in this example reveals that the second equation is three times the first

equation, so the two lines are identical. The system has an infinite number of solutions because they are

really the same equation and trace out the same line.

Let’s clarify this statement. An infinite number of solutions does not mean that any ordered pair (x, y)

satisfies the system of equations. Only ordered pairs that solve the equation in the system (either one of

the equations) are also solutions to the system. There are infinitely many of these solutions to the system

because there are infinitely many points on any one line.

For example, (1, -1) is a solution to the system in this example, and so is (-1, 7). Each of them fits both

the equations because both equations are really the same equation. But (3, 5) doesn’t fit either equation

and is not a solution to the system.

In fact, for every x−value there is just one y−value that fits both equations, and for every y−value there is

exactly one x−value—just as there is for a single line.

Let’s summarize how to determine the type of system we are dealing with algebraically.

• A consistent system will always give exactly one solution.

• An inconsistent system will yield a statement that is always false (like 0 = 13).

• A dependent system will yield a statement that is always true (like 9 = 9).

Applications

In this section, we’ll see how consistent, inconsistent and dependent systems might arise in real life.

Example 7

The movie rental store CineStar offers customers two choices. Customers can pay a yearly membership of

$45 and then rent each movie for $2 or they can choose not to pay the membership fee and rent each movie

for $3.50. How many movies would you have to rent before the membership becomes the cheaper option?

Solution

Let’s translate this problem into algebra. Since there are two different options to consider, we can write

two different equations and form a system.

The choices are “membership” and “no membership.” We’ll call the number of movies you rent x and the

total cost of renting movies for a year y.

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Table 7.7:

flat fee rental fee total

membership $45 2x y = 45 + 2x

no membership $0 3.50x y = 3.5x

The flat fee is the dollar amount you pay per year and the rental fee is the dollar amount you pay when

you rent a movie. For the membership option the rental fee is 2x, since you would pay $2 for each movie

you rented; for the no membership option the rental fee is 3.50x, since you would pay $3.50 for each movie

you rented.

Our system of equations is:

y = 45 + 2x

y = 3.50x

Here’s a graph of the system:

Now we need to find the exact intersection point. Since each equation is already solved for y, we can easily

solve the system with substitution. Substitute the second equation into the first one:

y = 45 + 2x

⇒ 3.50x = 45 + 2x⇒ 1.50x = 45⇒ x = 30 movies

y = 3.50x

You would have to rent 30 movies per year before the membership becomes the better option.

This example shows a real situation where a consistent system of equations is useful in finding a solution.

Remember that for a consistent system, the lines that make up the system intersect at single point. In

other words, the lines are not parallel or the slopes are different.

In this case, the slopes of the lines represent the price of a rental per movie. The lines cross because the

price of rental per movie is different for the two options in the problem

Now let’s look at a situation where the system is inconsistent. From the previous explanation, we can

conclude that the lines will not intersect if the slopes are the same (and the y−intercept is different). Let’s

change the previous problem so that this is the case.

Example 8

Two movie rental stores are in competition. Movie House charges an annual membership of $30 and

charges $3 per movie rental. Flicks for Cheap charges an annual membership of $15 and charges $3 per

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movie rental. After how many movie rentals would Movie House become the better option?

Solution

It should already be clear to see that Movie House will never become the better option, since its membership

is more expensive and it charges the same amount per movie as Flicks for Cheap.

The lines on a graph that describe each option have different y−intercepts—namely 30 for Movie House and

15 for Flicks for Cheap—but the same slope: 3 dollars per movie. This means that the lines are parallel

and so the system is inconsistent.

Now let’s see how this works algebraically. Once again, we’ll call the number of movies you rent x and the

total cost of renting movies for a year y.

Table 7.8:

flat fee rental fee total

Movie House $30 3x y = 30 + 3x

Flicks for Cheap $15 3x y = 15 + 3x

The system of equations that describes this problem is:

y = 30 + 3x

y = 15 + 3x

Let’s solve this system by substituting the second equation into the first equation:

y = 30 + 3x

⇒ 15 + 3x = 30 + 3x⇒ 15 = 30 This statement is always false.

y = 15 + 3x

This means that the system is inconsistent.

Example 9

Peter buys two apples and three bananas for $4. Nadia buys four apples and six bananas for $8 from the

same store. How much does one banana and one apple costs?

Solution

We must write two equations: one for Peter’s purchase and one for Nadia’s purchase.

Let’s say a is the cost of one apple and b is the cost of one banana.

Table 7.9:

cost of apples cost of bananas total cost

Peter 2a 3b 2a+ 3b = 4

Nadia 4a 6b 4a+ 6b = 8

The system of equations that describes this problem is:

2a+ 3b = 4

4a+ 6b = 8

Let’s solve this system by multiplying the first equation by -2 and adding the two equations:

−2(2a+ 3b = 4) − 4a − 6b = −8

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⇒

4a+ 6b = 8 4a+ 6b = 8

0 + 0 = 0

This statement is always true. This means that the system is dependent.

Looking at the problem again, we can see that we were given exactly the same information in both

statements. If Peter buys two apples and three bananas for $4, it makes sense that if Nadia buys twice

as many apples (four apples) and twice as many bananas (six bananas) she will pay twice the price ($8).

Since the second equation doesn’t give us any new information, it doesn’t make it possible to find out the

price of each fruit.

Review Questions

Express each equation in slope-intercept form. Without graphing, state whether the system of equations

is consistent, inconsistent or dependent.

1. 3x − 4y = 13

y = −3x − 7

2. 35 x+ y = 3

1.2x+ 2y = 6

3. 3x − 4y = 13

y = −3x − 7

4. 3x − 3y = 3

x − y = 1

5. 0.5x − y = 30

0.5x − y = −30

6. 4x − 2y = −2

3x+ 2y = −12

7. 3x+ y = 4

y = 5 − 3x

8. x − 2y = 7

4y − 2x = 14

Find the solution of each system of equations using the method of your choice. State if the system is

inconsistent or dependent.

9. 3x+ 2y = 4

− 2x+ 2y = 24

10. 5x − 2y = 3

2x − 3y = 10

11. 3x − 4y = 13

y = −3x − 7

12. 5x − 4y = 1

− 10x+ 8y = −30

13. 4x+ 5y = 0

3x = 6y+ 4.5

14. −2y+ 4x = 8

y − 2x = −4

15. x − 12y = 32

3x+ y = 6

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16. 0.05x+ 0.25y = 6

x+ y = 24

17. x+ 23y = 6

3x+ 2y = 2

18. A movie theater charges $4.50 for children and $8.00 for adults.

(a) On a certain day, 1200 people enter the theater and $8375 is collected. How many children and

how many adults attended?

(b) The next day, the manager announces that she wants to see them take in $10000 in tickets. If

there are 240 seats in the house and only five movie showings planned that day, is it possible to

meet that goal?

(c) At the same theater, a 16-ounce soda costs $3 and a 32-ounce soda costs $5. If the theater sells

12,480 ounces of soda for $2100, how many people bought soda? (Note: Be careful in setting

up this problem!)

19. Jamal placed two orders with an internet clothing store. The first order was for 13 ties and 4 pairs of

suspenders, and totaled $487. The second order was for 6 ties and 2 pairs of suspenders, and totaled

$232. The bill does not list the per-item price, but all ties have the same price and all suspenders

have the same price. What is the cost of one tie and of one pair of suspenders?

20. An airplane took four hours to fly 2400 miles in the direction of the jet-stream. The return trip against

the jet-stream took five hours. What were the airplane’s speed in still air and the jet-stream’s speed?

21. Nadia told Peter that she went to the farmer’s market and bought two apples and one banana, and

that it cost her $2.50. She thought that Peter might like some fruit, so she went back to the seller

and bought four more apples and two more bananas. Peter thanked Nadia, but told her that he did

not like bananas, so he would only pay her for four apples. Nadia told him that the second time she

paid $6.00 for the fruit.

(a) What did Peter find when he tried to figure out the price of four apples?

(b) Nadia then told Peter she had made a mistake, and she actually paid $5.00 on her second trip.

Now what answer did Peter get when he tried to figure out how much to pay her?

(c) Alicia then showed up and told them she had just bought 3 apples and 2 bananas from the same

seller for $4.25. Now how much should Peter pay Nadia for four apples?

7.5 Systems of Linear Inequalities

Learning Objectives

• Graph linear inequalities in two variables.

• Solve systems of linear inequalities.

• Solve optimization problems.

Introduction

In the last chapter you learned how to graph a linear inequality in two variables. To do that, you graphed

the equation of the straight line on the coordinate plane. The line was solid for ≤ or ≥ signs (where the

equals sign is included), and the line was dashed for < or > signs (where the equals sign is not included).

Then you shaded above the line (if the inequality began with y > or y ≥) or below the line (if it began with

y < or y ≤).

In this section, we’ll see how to graph two or more linear inequalities on the same coordinate plane. The

inequalities are graphed separately on the same graph, and the solution for the system is the common

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shaded region between all the inequalities in the system. One linear inequality in two variables divides the

plane into two half-planes. A system of two or more linear inequalities can divide the plane into more

complex shapes.

Let’s start by solving a system of two inequalities.

Graph a System of Two Linear Inequalities

Example 1

Solve the following system:

2x+ 3y ≤ 18

x − 4y ≤ 12

Solution

Solving systems of linear inequalities means graphing and finding the intersections. So we graph each

inequality, and then find the intersection regions of the solution.

First, let’s rewrite each equation in slope-intercept form. (Remember that this form makes it easier to tell

which region of the coordinate plane to shade.) Our system becomes

3y ≤ −2x+ 18 y ≤ −23 x+ 6⇒

− 4y ≤ −x+ 12 y ≥ x4 − 3

Notice that the inequality sign in the second equation changed because we divided by a negative number!

For this first example, we’ll graph each inequality separately and then combine the results.

Here’s the graph of the first inequality:

The line is solid because the equals sign is included in the inequality. Since the inequality is less than or

equal to, we shade below the line.

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And here’s the graph of the second inequality:

The line is solid again because the equals sign is included in the inequality. We now shade above the line

because y is greater than or equal to.

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When we combine the graphs, we see that the blue and red shaded regions overlap. The area where they

overlap is the area where both inequalities are true. Thus that area (shown below in purple) is the solution

of the system.

The kind of solution displayed in this example is called unbounded, because it continues forever in at

least one direction (in this case, forever upward and to the left).

Example 2

There are also situations where a system of inequalities has no solution. For example, let’s solve this

system.

y ≤ 2x − 4

y > 2x+ 6

Solution

We start by graphing the first line. The line will be solid because the equals sign is included in the

inequality. We must shade downwards because y is less than.

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Next we graph the second line on the same coordinate axis. This line will be dashed because the equals

sign is not included in the inequality. We must shade upward because y is greater than.

It doesn’t look like the two shaded regions overlap at all. The two lines have the same slope, so we know

they are parallel; that means that the regions indeed won’t ever overlap since the lines won’t ever cross.

So this system of inequalities has no solution.

But a system of inequalities can sometimes have a solution even if the lines are parallel. For example, what

happens if we swap the directions of the inequality signs in the system we just graphed?

To graph the system

y ≥ 2x − 4

y < 2x+ 6,

we draw the same lines we drew for the previous system, but we shade upward for the first inequality and

downward for the second inequality. Here is the result:

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You can see that this time the shaded regions overlap. The area between the two lines is the solution to

the system.

Graph a System of More Than Two Linear Inequalities

When we solve a system of just two linear inequalities, the solution is always an unbounded region—one

that continues infinitely in at least one direction. But if we put together a system of more than two

inequalities, sometimes we can get a solution that is bounded—a finite region with three or more sides.

Let’s look at a simple example.

Example 3

Find the solution to the following system of inequalities.

3x − y < 4

4y+ 9x < 8

x ≥ 0

y ≥ 0

Solution

Let’s start by writing our inequalities in slope-intercept form.

y > 3x − 4

y < −9

4

x+ 2

x ≥ 0

y ≥ 0

Now we can graph each line and shade appropriately. First we graph y > 3x − 4 :

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Next we graph y < −94 x+ 2 :

Finally we graph x ≥ 0 and y ≥ 0, and we’re left with the region below; this is where all four inequalities

overlap.

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The solution is bounded because there are lines on all sides of the solution region. In other words, the

solution region is a bounded geometric figure, in this case a triangle.

Notice, too, that only three of the lines we graphed actually form the boundaries of the region. Sometimes

when we graph multiple inequalities, it turns out that some of them don’t affect the overall solution; in

this case, the solution would be the same even if we’d left out the inequality y > 3x − 4. That’s because

the solution region of the system formed by the other three inequalities is completely contained within

the solution region of that fourth inequality; in other words, any solution to the other three inequalities is

automatically a solution to that one too, so adding that inequality doesn’t narrow down the solution set

at all.

But that wasn’t obvious until we actually drew the graph!

Solve Real-World Problems Using Systems of Linear Inequalities

A lot of interesting real-world problems can be solved with systems of linear inequalities.

For example, you go to your favorite restaurant and you want to be served by your best friend who happens

to work there. However, your friend only waits tables in a certain region of the restaurant. The restaurant

is also known for its great views, so you want to sit in a certain area of the restaurant that offers a good

view. Solving a system of linear inequalities will allow you to find the area in the restaurant where you

can sit to get the best view and be served by your friend.

Often, systems of linear inequalities deal with problems where you are trying to find the best possible

situation given a set of constraints. Most of these application problems fall in a category called linear

programming problems.

Linear programming is the process of taking various linear inequalities relating to some situation, and

finding the best possible value under those conditions. A typical example would be taking the limitations

of materials and labor at a factory, then determining the best production levels for maximal profits under

those conditions. These kinds of problems are used every day in the organization and allocation of resources.

These real-life systems can have dozens or hundreds of variables, or more. In this section, we’ll only work

with the simple two-variable linear case.

The general process is to:

• Graph the inequalities (called constraints) to form a bounded area on the coordinate plane (called

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the feasibility region).

• Figure out the coordinates of the corners (or vertices) of this feasibility region by solving the system

of equations that applies to each of the intersection points.

• Test these corner points in the formula (called the optimization equation) for which you’re trying

to find the maximum or minimum value.

Example 4

If z = 2x+ 5y, find the maximum and minimum values of z given these constraints:

2x − y ≤ 12

4x+ 3y ≥ 0

x − y ≤ 6

Solution

First, we need to find the solution to this system of linear inequalities by graphing and shading appropri-

ately. To graph the inequalities, we rewrite them in slope-intercept form:

y ≥ 2x − 12

y ≥ −4

3

x

y ≥ x − 6

These three linear inequalities are called the constraints, and here is their graph:

The shaded region in the graph is called the feasibility region. All possible solutions to the system occur

in that region; now we must try to find the maximum and minimum values of the variable z within that

region. In other words, which values of x and y within the feasibility region will give us the greatest and

smallest overall values for the expression 2x+ 5y?

Fortunately, we don’t have to test every point in the region to find that out. It just so happens that the

minimum or maximum value of the optimization equation in a linear system like this will always be found

at one of the vertices (the corners) of the feasibility region; we just have to figure out which vertices. So

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for each vertex—each point where two of the lines on the graph cross—we need to solve the system of just

those two equations, and then find the value of z at that point.

The first system consists of the equations y = 2x − 12 and y = −43 x. We can solve this system by

substitution:

−4

3

x = 2x − 12⇒ −4x = 6x − 36⇒ −10x = −36⇒ x = 3.6

y = 2x − 12⇒ y = 2(3.6) − 12⇒ y = −4.8

The lines intersect at the point (3.6, -4.8).

The second system consists of the equations y = 2x−12 and y = x−6. Solving this system by substitution:

x − 6 = 2x − 12⇒ 6 = x⇒ x = 6

y = x − 6⇒ y = 6 − 6⇒ y = 6

The lines intersect at the point (6, 6).

The third system consists of the equations y = −43 x and y = x − 6. Solving this system by substitution:

x − 6 = −4

3

x⇒ 3x − 18 = −4x⇒ 7x = 18⇒ x = 2.57

y = x − 6⇒ y = 2.57 − 6⇒ y = −3.43

The lines intersect at the point (2.57, -3.43).

So now we have three different points that might give us the maximum and minimum values for z. To

find out which ones actually do give the maximum and minimum values, we can plug the points into the

optimization equation z = 2x+ 5y.

When we plug in (3.6, -4.8), we get z = 2(3.6) + 5(−4.8) = −16.8.

When we plug in (6, 0), we get z = 2(6) + 5(0) = 12.

When we plug in (2.57, -3.43), we get z = 2(2.57) + 5(−3.43) = −12.01.

So we can see that the point (6, 0) gives us the maximum possible value for z and the point

(3.6, –4.8) gives us the minimum value.

In the previous example, we learned how to apply the method of linear programming in the abstract. In

the next example, we’ll look at a real-life application.

Example 5

You have $10,000 to invest, and three different funds to choose from. The municipal bond fund has a 5%

return, the local bank’s CDs have a 7% return, and a high-risk account has an expected 10% return. To

minimize risk, you decide not to invest any more than $1,000 in the high-risk account. For tax reasons,

you need to invest at least three times as much in the municipal bonds as in the bank CDs. What’s the best

way to distribute your money given these constraints?

Solution

Let’s define our variables:

x is the amount of money invested in the municipal bond at 5% return

y is the amount of money invested in the bank’s CD at 7% return

10000 − x − y is the amount of money invested in the high-risk account at 10% return

z is the total interest returned from all the investments, so z = .05x + .07y + .1(10000 − x − y) or z =

1000 − 0.05x − 0.03y. This is the amount that we are trying to maximize. Our goal is to find the values of

x and y that maximizes the value of z.

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Now, let’s write inequalities for the constraints:

You decide not to invest more than $1000 in the high-risk account—that means:

10000 − x − y ≤ 1000

You need to invest at least three times as much in the municipal bonds as in the bank CDs—that means:

3y ≤ x

Also, you can’t invest less than zero dollars in each account, so:

x ≥ 0

y ≥ 0

10000 − x − y ≥ 0

To summarize, we must maximize the expression z = 1000 − .05x − .03y using the constraints:

10000 − x − y ≤ 1000 y ≥ 9000 − x

3y ≤ x y ≤ x

3

x ≥ 0 Or in slope-intercept form: x ≥ 0

y ≥ 0 y ≥ 0

10000 − x − y ≥ 0 y ≤ 10000 − x

Step 1: Find the solution region to the set of inequalities by graphing each line and shading appropriately.

The following figure shows the overlapping region:

The purple region is the feasibility region where all the possible solutions can occur.

Step 2: Next we need to find the corner points of the feasibility region. Notice that there are four corners.

To find their coordinates, we must pair up the relevant equations and solve each resulting system.

System 1:

y = x3

y = 10000 − x

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Substitute the first equation into the second equation:

x

3

= 10000 − x⇒ x = 30000 − 3x⇒ 4x = 30000⇒ x = 7500

y =

x

3

⇒ y = 7500

3

⇒ y = 2500

The intersection point is (7500, 2500).

System 2:

y = x3

y = 9000 − x

Substitute the first equation into the second equation:

x

3

= 9000 − x⇒ x = 27000 − 3x⇒ 4x = 27000⇒ x = 6750

y =

x

3

⇒ y = 6750

3

⇒ y = 2250

The intersection point is (6750, 2250).

System 3:

y = 0

y = 10000 − x.

The intersection point is (10000, 0).

System 4:

y = 0

y = 9000 − x.

The intersection point is (9000, 0).

Step 3: In order to find the maximum value for z, we need to plug all the intersection points into the

equation for z and find which one yields the largest number.

(7500, 2500): z = 1000 − 0.05(7500) − 0.03(2500) = 550

(6750, 2250): z = 1000 − 0.05(6750) − 0.03(2250) = 595

(10000, 0): z = 1000 − 0.05(10000) − 0.03(0) = 500

(9000, 0): z = 1000 − 0.05(9000) − 0.03(0) = 550

The maximum return on the investment of $595 occurs at the point (6750, 2250). This means that:

$6,750 is invested in the municipal bonds.

$2,250 is invested in the bank CDs.

$1,000 is invested in the high-risk account.

Graphing calculators can be very useful for problems that involve this many inequalities. The video at

http://www.youtube.com/watch?v=__wAxkYmhvY shows a real-world linear programming problem worked

through in detail on a graphing calculator, although the methods used there can also be used for pencil-and

paper solving.

Review Questions

1. Consider the system

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y < 3x − 5

y > 3x − 5.

Is it consistent or inconsistent? Why?

2. Consider the system

y ≤ 2x+ 3

y ≥ 2x+ 3.

Is it consistent or inconsistent? Why?

3. Consider the system

y ≤ −x+ 1

y > −x+ 1.

Is it consistent or inconsistent? Why?

4. In example 3 in this lesson, we solved a system of four inequalities and saw that one of the inequalities,

y > 3x − 4, didn’t affect the solution set of the system.

(a) What would happen if we changed that inequality to y < 3x − 4?

(b) What’s another inequality that we could add to the original system without changing it? Show

how by sketching a graph of that inequality along with the rest of the system.

(c) What’s another inequality that we could add to the original system to make it inconsistent?

Show how by sketching a graph of that inequality along with the rest of the system.

5. Recall the compound inequalities in one variable that we worked with back in chapter 6. Compound

inequalities with “and” are simply systems like the ones we are working with here, except with one

variable instead of two.

(a) Graph the inequality x > 3 in two dimensions. What’s another inequality that could be combined

with it to make an inconsistent system?

(b) Graph the inequality x ≤ 4 on a number line. What two-dimensional system would have a graph

that looks just like this one?

Find the solution region of the following systems of inequalities.

6. x − y < −6

2y ≥ 3x+ 17

7. 4y − 5x < 8

− 5x ≥ 16 − 8y

8. 5x − y ≥ 5

2y − x ≥ −10

9. 5x+ 2y ≥ −25

3x − 2y ≤ 17

x − 6y ≥ 27

10. 2x − 3y ≤ 21

x+ 4y ≤ 6

3x+ y ≥ −4

11. 12x − 7y < 120

7x − 8y ≥ 36

5x+ y ≥ 12

Solve the following linear programming problems.

12. Given the following constraints, find the maximum and minimum values of z = −x+ 5y:

x+ 3y ≤ 0

x − y ≥ 0

3x − 7y ≤ 16

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13. Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice elves draw

a wage of five candy canes per hour worked, but can only make four trucks an hour. Senior elves can

make six trucks an hour and are paid eight candy canes per hour. There’s only room for nine elves

in the truck shop, and due to a candy-makers’ strike, Santa Claus can only pay out 480 candy canes

for the whole 8-hour shift.

(a) How many senior elves and how many apprentice elves should work this shift to maximize the

number of trucks that get made?

(b) How many trucks will be made?

(c) Just before the shift begins, the apprentice elves demand a wage increase; they insist on being

paid seven candy canes an hour. Now how many apprentice elves and how many senior elves

should Santa assign to this shift?

(d) How many trucks will now get made, and how many candy canes will Santa have left over?

14. In Adrian’s Furniture Shop, Adrian assembles both bookcases and TV cabinets. Each type of furniture

takes her about the same time to assemble. She figures she has time to make at most 18 pieces of

furniture by this Saturday. The materials for each bookcase cost her $20 and the materials for each

TV stand costs her $45. She has $600 to spend on materials. Adrian makes a profit of $60 on each

bookcase and a profit of $100 on each TV stand.

(a) Set up a system of inequalities. What x− and y−values do you get for the point where Adrian’s

profit is maximized? Does this solution make sense in the real world?

(b) What two possible real-world x−values and what two possible real-world y−values would be

closest to the values in that solution?

(c) With two choices each for x and y, there are four possible combinations of x− and y−values. Of

those four combinations, which ones actually fall within the feasibility region of the problem?

(d) Which one of those feasible combinations seems like it would generate the most profit? Test out

each one to confirm your guess. How much profit will Adrian make with that combination?

(e) Based on Adrian’s previous sales figures, she doesn’t think she can sell more than 8 TV stands.

Now how many of each piece of furniture should she make, and what will her profit be?

(f) Suppose Adrian is confident she can sell all the furniture she can make, but she doesn’t have

room to display more than 7 bookcases in her shop. Now how many of each piece of furniture

should she make, and what will her profit be?

15. Here’s a “linear programming” problem on a line instead of a plane: Given the constraints x ≤ 5 and

x ≥ −2, maximize the value of y where y = x+ 3.

Texas Instruments Resources

In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities

designed to supplement the objectives for some of the lessons in this chapter. See http:

//www.ck12.org/flexr/chapter/9617.

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