Chapter 9 · Worked-Out Solutions Chapter 9 4. Step 1 Draw a right triangle with acute angle θ...

Copyright © Big Ideas Learning, LLC Algebra 2 447All rights reserved. Worked-Out Solutions

Chapter 9

Chapter 9 Maintaining Mathematical Profi ciency (p. 459)

1. ∣ 4 ∣ = 4 ∣ 6 + 4 ∣ = ∣ 10 ∣ = 10

∣ 2 − 9 ∣ = ∣ −7 ∣ = 7 − ∣ 7 ∣ = −7

So, the order is − ∣ 7 ∣ , ∣ 4 ∣ , ∣ 2 − 9 ∣ , and ∣ 6 + 4 ∣ .

2. ∣ 9 − 3 ∣ = ∣ 6 ∣ = 6 ∣ −4 ∣ = 4

∣ 0 ∣ = 0 ∣ −5 ∣

— ∣ 2 ∣

= 5 — 2 = 2.5

So, the order is ∣ 0 ∣ , ∣ −5 ∣

— ∣ 2 ∣

, ∣ −4 ∣ , and ∣ 9 − 3 ∣ .

3.

∣ −83 ∣ = ∣ −512 ∣ = 512 ∣ 9 − 1 ∣ = ∣ 8 ∣ = 8

∣ −2 ⋅ 8 ∣ = ∣ −16 ∣ = 16 ∣ 9 ∣ + ∣ −2 ∣ − ∣ −1 ∣ = 9 + 2 − 1 = 10

So, the order is ∣ 9 − 1 ∣ , ∣ 9 ∣ + ∣ −2 ∣ − ∣ −1 ∣ , ∣ −2 ⋅ 8 ∣ , and ∣ −83 ∣ .

4.

∣ −4 + 20 ∣ = ∣ 16 ∣ = 16 ∣ 5 ∣ − ∣ 3 ⋅ 2 ∣ = 5 − ∣ 6 ∣ = 5 − 6 = −1

− ∣ 42 ∣ = − ∣ 16 ∣ = −16 ∣ −15 ∣ = 15

So, the order is − ∣ 42 ∣ , ∣ 5 ∣ − ∣ 3 ⋅ 2 ∣ , ∣ −15 ∣ , and ∣ −4 + 20 ∣ .

5. a2 + b2 = c2

122 + 52 = c2

144 + 25 = c2

169 = c2

13 = c

So, the length is 13 meters.

6. a2 + b2 = c2

72 + b2 = 252

49 + b2 = 625

b2 = 576

b = 24

So, the length is 24 feet.

7. a2 + b2 = c2

9.62 + 7.22 = c2

92.16 + 51.84 = c2

144 = c2

12 = c

So, the length is 12 millimeters.

8. a2 + b2 = c2

a2 + 212 = 352

a2 + 441 = 1225

a2 = 784

a = 28

So, the length is 28 kilometers.

9. a2 + b2 = c2

a2 + 42 = ( 12 1 —

3 ) 2

a2 + 16 = 152 1 —

9

a2 = 136 1 —

9

a = 11 2 —

3

So, the length is 11 2 —

3 inches.

10. a2 + b2 = c2

( 3 — 10

) 2 + b2 = ( 1 — 2 ) 2

9 —

100 + b2 =

1 —

4

b2 = 4 —

25

b = 2 —

5

So, the length is 2 —

5 yard, or 0.4 yard.

11. Yes, the triangle is a right triangle. The line passing through

the points ( x1, y1 ) and ( x2, y1 ) is horizontal. The line passing

through the points ( x2, y1 ) and ( x2, y2 ) is vertical. Horizontal

and vertical lines are perpendicular, so the triangle formed

by the line segments connecting ( x1, y1 ) , ( x2, y1 ) , and ( x2, y2 )

contains a right angle.

Chapter 9 Mathematical Practices (p. 460)

1. Because θ = 135°, (x, y) lies on the line y = −x.

x2 + y2 = 1

x2 + (−x)2 = 1

2x2 = 1

x2 = 1 — 2

x = − 1 —

√—

2 = −

√—

2 —

2

The coordinates of (x, y) are ( − √

— 2 —

2 ,

√—

2 —

2 ) .

2. Because θ = 315°, (x, y) lies on the line y = −x.

x2 + y2 = 1

x2 + (−x)2 = 1

2x2 = 1

x2 = 1 — 2

x = 1 —

√—

2 =

√—

2 —

2

The coordinates of (x, y) are ( √—

2 —

2 , −

√—

2 —

2 ) .

448 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 9

3. Because θ = 225°, (x, y) lies on the line y = x.

x2 + y2 = 1

x2 + x2 = 1

2x2 = 1

x2 = 1 — 2

x = − 1 —

√—

2 = −

√—

2 —

2

The coordinates of (x, y) are ( − √

— 2 —

2 , −

√—

2 —

2 ) .

9.1 Explorations (p. 461)

1. sin 30° = 1 — 2 cos 30° =

√—

3 —

2 tan 30° =

√—

3 —

3

sin 45° = √

— 2 —

2 cos 45° =

√—

2 —

2 tan 45° = 1

sin 60° = √

— 3 —

2 cos 60° = 1 —

2 tan 60° = √

— 3

2. a. This is true because (90° − θ) represents the other acute

angle in the triangle, cos(90° − θ) = opp —

hyp , which is the

same ratio as sin θ.

b. This is true because (90° − θ) represents the other acute

angle in the triangle, sin(90° − θ) = adj —

hyp , which is the

same ratio as cos θ.

c. This is true by defi nition because cosecant is the

reciprocal of sine.

d. This is true by defi nition because cotangent is the

reciprocal of tangent.

e. (sin θ)2 + (cos θ)2 = ( opp —

hyp ) 2 + ( adj

— hyp

) 2

= opp2

— hyp2 + adj2

— hyp2

= 1 —

hyp2 + (opp2 + adj2)

= 1 —

hyp2 ⋅ hyp2

= 1

So, (sin θ)2 + (cos θ)2 = 1.

f. (sec θ)2 − (tan θ)2 = ( hyp —

adj ) 2 − ( opp

— adj

) 2

= hyp2

— adj2

− opp2

— adj2

= hyp2 − opp2

—— adj2

= adj2 —

adj2

= 1

So, (sec θ)2 − (tan θ)2 = 1.

3. When the side lengths of a right triangle are known,

the ratios of the side lengths can be used to fi nd the

trigonometric function of an acute angle.

4. sin 25° = y —

1 cos 25° =

x —

1

sin 25° = y cos 25° = x

0.42 ≈ y 0.91 = x

So, x ≈ 0.91 and y ≈ 0.42.

9.1 Monitoring Progress (pp. 463–465)

1. From the Pythagorean Theorem, the length of the hypotenuse is

hyp. = √—

42 + 32

= √—

25

= 5.

Using adj. = 4, opp. = 3, and hyp. = 5, the values of the six

trigonometric functions of θ are:

sin θ = 3 — 5 cos θ = 4 — 5 tan θ = 3 — 4

csc θ = 5 — 3 sec θ = 5 — 4 cot θ = 4 — 3

2. From the Pythagorean Theorem, the length of the adjacent

side is

adj. = √—

172 − 152

= √—

64

= 8.

Using adj. = 8, opp. = 15, and hyp. = 17, the values of the

six trigonometric functions of θ are:

sin θ = 15 —

17 cos θ = 8 — 17 tan θ = 15

— 8

csc θ = 17 —

15 sec θ = 17

— 8 cot θ = 8 — 15

3. From the Pythagorean Theorem, the length of the opposite

side is

opp. = √——

( 5 √—

2 ) 2 − 52

= √—

25

= 5.

Using adj. = 5, opp. = 5, and hyp. = 5 √—

2 , the values of the


sin θ = √

— 2 —

2 cos θ =

√—

2 —

2 tan θ = 1

csc θ = √—

2 sec θ = √—

2 cot θ = 1


Chapter 9

4. Step 1 Draw a right triangle with acute angle θ such that

the leg adjacent θ has length 7 and the hypotenuse

has length 10.

opp. = 51

7

10

θ

Step 2 Find the length of the opposite side. By the

Pythagorean Theorem, the length of the other leg is

opp. = √—

102 − 72 = √—

51 .

Step 3 Find the values of the remaining fi ve trigonometric

functions. Because cos θ = 7 — 10

, sec θ = hyp —

adj = 10

— 7 .

The others are:

sin θ = √

— 51 —

10 tan θ =

√—

51 —

7

csc θ = 10 √

— 51 —

51 cot θ =

7 √—

51 —

51

5. Write an equation using a trigonometric function that

involves the ratio of x and 6. Solve the equation for x.

cos 45° = adj —

hyp

√

— 2 —

2 = x —

6

3 √—

2 = x

The length of the side is x = 3 √—

2 ≈ 4.24.

6. Because the triangle is a right triangle, A and B are

complementary angles. So, A = 90° − 45° = 45°. Next,

write two equations using trigonometric functions, one that

involves the ratio of a and 5, and one that involves b and 5.

Solve the fi rst equation for a and the second for b.

sin 45° = opp —

hyp cos 45° = adj

— hyp

sin 45° = a — 5 cos 45° = b —

5

5(sin 45°) = a 5(cos 45°) = b

5 √

— 2 —

2 = a 5 √

— 2 —

2 = b

So, A = 45°, a = 5 √—

2 —

2 , and b = 5 √

— 2 —

2 .


complementary angles. So, B = 90° − 32° = 58°. Next,


involves the ratio of a and 10, and one that involves the ratio

of c and 10.

tan 32° = opp —

adj sec 32° = hyp

— adj

tan 32° = a — 10

sec 32° = c — 10

10 tan 32° = a 10 sec 32° = c

6.25 ≈ a 11.79 ≈ c

So, B = 58°, c ≈ 11.79, and a ≈ 6.25.




involves the ratio of a and 20, and one that involves the ratio

of b and 20.

cos 19° = adj —

hyp sin 19° = opp

— hyp

cos 19° = a — 20

sin 19° = b — 20

20 cos 19° = a 20 sin 19° = b

18.91 ≈ a 6.51 ≈ b

So, B = 19°, a ≈ 18.91, and b ≈ 6.51.




involves the ratio of b and 7, and one that involves the ratio

of c and 7.

cot 30° = adj —

opp sec 30° = hyp

— adj

cot 30° = b — 7 sec 30° = c —

7

7 cot 30° = b 7 sec 30° = c

12.12 ≈ b 14 = c

So, A = 30°, b ≈ 12.12, and c = 14.

10. sec 76° = x — 2

2(sec 76°) = x

8.3 ≈ x

The length is about 8.3 miles.

11.

h72 ft

38°

Write and solve an equation to fi nd the height h.

sin 38° = h — 72

72(sin 38°) = h

44.3 ≈ h

The height of the parasailer above the boat is about 44.3 feet.


Chapter 9

9.1 Exercises (pp. 466–468)

Vocabulary and Core Concept Check

1. In a right triangle, the two trigonometric functions of θ that

are defi ned using the lengths of the hypotenuse and the side

adjacent to θ are cosine and secant.

2. The angle of elevation and angle of depression are always

equal.

3. To solve a right triangle, the missing angles and side lengths

must be found.

4. “What is the ratio of the side opposite θ to the hypotenuse?”

is different. This ratio is 4 —

6 , or

2 —

3 .

csc θ = hyp.

— opp. = 6 —

4 =

3 —

2 ; 1

— sin θ =

1 —

opp.

— hyp.

=

1 —

4 —

6

= 6 —

4 =

3 —

2 ;

= hyp. —

opp. =

6 —

4 =

3 —

2

Monitoring Progress and Modeling with Mathematics

5. From the Pythagorean Theorem, the length of the

hypotenuse is

hyp. = √—

92 + 122

= √—

225

= 15.

Using adj. = 9, opp. = 12, and hyp. = 15, the six


sin θ = opp.

— hyp.

= 12

— 15

= 4 —

5 cos θ =

adj. —

hyp. =

9 —

15 =

3 —

5

tan θ = opp.

— adj.

= 12 —

9 = 4 —

3 csc θ = hyp.

— opp.

= 15 —

12 = 5 —

4

sec θ = hyp. —

adj. = 15

— 9 = 5 —

3 cot θ = adj.

— opp.

= 9 — 12

= 3 — 4

6. From the Pythagorean Theorem, the length of the

hypotenuse is

hyp. = √—

82 + 62

= √—

100

= 10.

Using adj. = 8, opp. = 6, and hyp. = 10, the values of the


sin θ = opp.

— hyp.

= 6 —

10 =

3 —

5 cos θ = adj.

— hyp.

= 8 — 10

= 4 — 5

tan θ = opp. —

adj. = 6 —

8 = 3 —

4 csc θ = hyp.

— opp.

= 10 —

6 = 5 —

3

sec θ = hyp. —

adj. = 10

— 8 = 5 —

4 cot θ = adj.

— opp.

= 8 — 6 = 4 —

3


side to θ is

adj. = √—

72 − 52

= √—

24

= 2 √—

6 .

Using adj. = 2 √—

6 , opp. = 5, and hyp. = 7, the values of the


sin θ = opp.

— hyp.

= 5 —

7 cos θ =

adj. —

hyp. =

2 √—

6 —

7

tan θ = opp. —

adj. =

5 —

2 √—

6 =

5 √—

6 —

12 csc θ =

hyp. — opp. =

7 —

5

sec θ = hyp.

— adj.

= 7 —

2 √—

6 = 7 √

— 6 —

12 cot θ =

adj. — opp. =

2 √—

6 —

5


side to θ is

adj. = √—

92 − 32

= √—

72

= 6 √—

2 .

Using adj. = 6 √—

2 , opp. = 3, and hyp. = 9, the values of the


sin θ = opp.

— hyp.

= 3 —

9 =

1 —

3

cos θ = adj.

— hyp.

= 6 √

— 2 —

9 =

2 √—

2 —

3

tan θ = opp.

— adj.

= 3 —

6 √—

2 =

3 √—

2 —

12 =

√—

2 —

4

csc θ = hyp.

— opp. = 9 —

3 = 3

sec θ = hyp.

— adj.

= 9 —

6 √—

2 =

9 √—

2 —

12 =

3 √—

2 —

12

cot θ = adj.

— opp. = 6 √

— 2 —

3 = 2 √

— 2


Chapter 9


side to θ is

opp. = √—

182 − 102

= √—

224

= 4 √—

14 .

Using adj. = 10, opp. = 4 √—

14 , and hyp. = 18, the values of

the six trigonometric functions of θ are

sin θ = opp.

— hyp.

= 4 √

— 14 —

18 =

2 √—

14 —

9

cos θ =

adj. —

hyp. = 10

— 18

= 5 — 9

tan θ = opp.

— adj.

= 4 √

— 14 —

10 =

2 √—

14 —

5

csc θ = hyp.

— opp. = 18 —

4 √—

14 =

18 √—

14 —

56 =

9 √—

14 —

28

sec θ = hyp.

— adj

= 18 —

10 = 9 —

5

cot θ = adj.

— opp. = 1c —

4 √—

14 =

10 √—

14 —

56 =

5 √—

14 —

28


side to θ is

opp. = √—

262 − 142

= √—

480

= 4 √—

30 .

Using adj. = 14, opp. = 4 √—

30 , and hyp. = 26, the values of

the six trigonometric functions of θ are

sin θ = opp.

— hyp.

= 4 √

— 30 —

26 =

2 √—

30 —

13

cos θ = adj.

— hyp.

= 14

— 26

= 7 —

13

tan θ = opp.

— adj.

= 4 √

— 30 —

14 =

2 √—

30 —

7

csc θ = hyp.

— opp. = 26 —

4 √—

30 =

26 √—

30 —

120 =

13 √—

30 —

60

sec θ = hyp.

— adj.

= 26

— 14

= 13

— 7

cot θ = adj.

— opp. = 14 —

4 √—

30 =

14 √—

30 —

120 =

7 √—

30 —

60

11.

9

497

θ

sin θ = opp.

— hyp.

= 4 —

√—

97 =

4 √—

97 —

97 cot θ =

adj. — opp. =

9 —

4

cos θ = adj.

— hyp.

= 9 —

√—

97 =

9 √—

97 —

97 csc θ =

hyp. — opp. =

√—

97 —

4

12. sin(90° − θ) = cos θ cos(90° − θ) = sin θ tan(90° − θ) = cot θ cot(90° − θ) = tan θ sec(90° − θ) = csc θ csc(90° − θ) = sec θ The expression 90° − θ relates a trigonometric function with

its cofunction.


the leg opposite θ has length 7 and the hypotenuse

has length 11.

Step 2 Find the length of the adjacent side. By the


adj. = √—

112 − 72 = 6 √—

2 .

adj. = 6 2

711

θ


functions. Because sin θ = 7 —

11 , csc θ =

11 —

7 . The

other values are:

cos θ = adj.

— hyp.

= 6 √

— 2 —

11

tan θ = opp.

— adj.

= 7 —

6 √—

2 =

7 √—

2 —

12

sec θ = hyp. —

adj. =

11 —

6 √—

2 =

11 √—

2 —

12

cot θ = adj.

— opp. = 6 √

— 2 —

7



has length 12.



opp. = √—

122 − 32 = √—

119 .

opp. = 119

5

12

θ


Chapter 9


functions. Because cos θ = 5 —

12 , sec θ =

12 —

5 . The

other values are:

sin θ = opp.

— hyp

= √—

119 —

12

tan θ = opp.

— adj.

= √—

119 —

5

csc θ = hyp.

— opp. = 12 —

√—

119 =

12 √—

119 —

119

cot θ = adj.

— opp. = 5 —

√—

119 =

5 √—

119 —

119


the leg adjacent θ has length 6 and the leg opposite

θ has length 7.

Step 2 Find the length of the hypothenuse. By the Pythagorean

Theorem, the length of the hypothenuse is

hyp. = √—

62 + 72 = √—

85 .

hyp. = 85

6

7

θ


functions. Because tan θ = 7 —

6 , cot θ = 6 —

7 . The other

values are:

sin θ = opp.

— hyp.

= 7 —

√—

85 =

7 √—

85 —

85

cos θ = adj.

— hyp.

= 6 —

√—

85 =

6 √—

85 —

85

csc θ = hyp.

— opp = √

— 85 —

7

sec θ = hyp.

— adj.

= √—

85 —

6



has length 15.


Pythagorean Theorem, the length of the adjacent

side is

adj. = √—

152 − 82 = √—

161 .

adj. = 161

815

θ

Step 3 Find the value of the remaining fi ve trigonometric

functions. Because csc θ = 15

— 8 , sin θ =

8 —

15 . The

other values are:

cos θ = adj.

— hyp.

= √—

161 —

15

tan θ = opp.

— adj.

= 8 —

√—

161 =

8 √—

161 —

161

sec θ = hyp.

— adj.

= 15 —

√—

161 =

15 √—

161 —

161

cot θ = adj.

— opp.

= √—

161 —

8



has length 14.


Pythagorean Theorem, the length of the opposite

side is

opp. = √—

142 − 92 = √—

115 .

opp. = 115

9

14

θ


functions. Because sec θ = 14

— 9 , cos θ =

9 —

14 . The

other values are:

sin θ = opp.

— hyp.

= √—

115 —

14

tan θ = opp.

— adj.

= √—

115 —

9

csc θ = hyp.

— opp. = 14 —

√—

115 =

14 √—

115 —

115

cot θ = adj.

— opp. = 9 —

√—

115 =

9 √—

115 —

115


Chapter 9


the leg adjacent θ has length 16 and the leg opposite

θ has length 11.

Step 2 Find the length of the hypotenuse. By the

Pythagorean Theorem, the length of the

hypotenuse is

hyp. = √—

162 + 112 = √—

377 .

hyp. = 377

16

11

θ


functions. Because cot θ = 16

— 11

, tan θ = 11

— 16

. The

other values are:

sin θ = opp.

— hyp.

= 11 —

√—

377 =

11 √—

377 —

377

cos θ = adj.

— hyp.

= 16 —

√—

377 =

16 √—

377 —

377

csc θ = hyp.

— opp. = √—

377 —

11

sec θ = hyp. —

adj. =

√—

377 —

16

19. The adjacent side was used instead of the opposite.

sin θ = opp.

— hyp.

= 8 —

17

20. The reciprocal of csc θ is 1 —

sin θ .

csc θ = 1 —

sin θ =

1 —

6 √

— 2 —

11

= 11 √

— 2 —

12



cos 60° = adj.

— hyp.

1 —

2 = x —

9

9 —

2 = x

The length of the side is x = 9 — 2 = 4.5.



cos 60° = adj.

— hyp.

1 —

2 = x —

6

3 = x

The length of the side is x = 3.



sin 30° = opp.

— hyp.

1 —

2 = x —

12

6 = x




sin 30° = opp.

— hyp.

1 —

2 = x —

13

13

— 2 = x

The length of the side is x = 13 —

2 = 6.5.



tan 45° = opp.

— adj.

1 = x — 8

8 = x




tan 45° = opp.

— adj.

1 = x — 7

7 = x

The length of the side is 7.

27. cos 14° ≈ 0.9703

28. tan 31° ≈ 0.6009

29. csc 59° = 1 —

sin 59° ≈ 1.1666

30. sin 23° ≈ 0.3907

31. cot 6° = 1 —

tan 6° ≈ 9.5144


Chapter 9

32. sec 11° = 1 —

cos 11° ≈ 1.0187




involves the ratio of b and 23, and one that involves c and 23.

Solve the fi rst equation for b and the second for c.

tan 36° = opp.

— adj.

sec 36° = hyp.

— adj.

tan 36° = b — 23

sec 36° = c — 23

23(tan 36°) = b 23 ( 1 —

cos 36° ) = c

16.71 ≈ b 28.43 ≈ c

So, A = 54°, b ≈ 16.71, and c ≈ 28.43.




involves the ratio of a and 9, and one that involves c and 9.

Solve the fi rst equation for a and the second equation for c.

tan 27° = opp.

— adj.

sec 27° = hyp.

— adj.

tan 27° = a — 9 sec 27° = c —

9

9(tan 27°) = a 9 ( 1 —

cos 27° ) = c

4.59 ≈ a 10.10 ≈ c

So, B = 63°, a ≈ 4.59, and c ≈ 10.10.




involves the ratio of b and 17, and one that involves c and 17.

Solve the fi rst equation for b and the second equation for c.

cos 55° = adj.

— hyp.

csc 55° = hyp.

— opp.

cos 55° = b — 17

csc 55° = c — 17

17(cos 55°) = b 17 ( 1 —

sin 55° ) = c

11.90 ≈ b 20.75 ≈ c

So, B = 35°, b ≈ 11.90, and c ≈ 20.75.






cot 16° = adj.

— opp. csc 16° = hyp.

— opp.

cot 16° = a — 14

csc 16° = c — 14

14 ( 1 —

tan 16° ) = a 14 ( 1

— sin 16°

) = c

48.82 ≈ a 50.79 ≈ c

So, A = 74°, a ≈ 48.82, and c ≈ 50.79.






tan 43° = opp.

— adj.

sec 43° = hyp.

— adj.

tan 43° = a — 31

sec 43° = c — 31

31(tan 43°) = a 31 ( 1 —

cos 43° ) = c

28.91 ≈ a 42.39 ≈ c

So, B = 47°, a ≈ 28.91, and c ≈ 42.39.




involves the ratio of b and 23, and one that involves the ratio

of c and 23. Solve the fi rst equation for b and the second

equation for c.

tan 31° = opp.

— adj.

sec 31° = hyp.

— adj.

tan 31° = b — 23

sec 31° = c — 23

23(tan 31°) = b 23 ( 1 —

cos 31° ) = c

13.82 ≈ b 26.83 ≈ c

So, A = 59°, b ≈ 13.82, and c ≈ 26.83.


Chapter 9




involves the ratio of a and 12.8, and one that involves b and

12.8. Solve the fi rst equation for a and the second equation

for b.

cos 72° = adj.

— hyp.

sin 72° = opp.

— hyp.

cos 72° = 9 —

12.8 sin 72° = b

— 12.8

12.8(cos 72°) = a 12.8 (sin 72°) = b

3.96 ≈ a 12.17 ≈ b

So, A = 18°, a ≈ 3.96, and b ≈ 12.17.




involves the ratio of b and 7.4, and one that involves c and

7.4. Solve the fi rst equation for b and the second equation

for c.

cot 64° = adj.

— opp. csc 64° = hyp.

— opp.

cot 64° = b —

7.4 csc 64° = c

— 7.4

7.4 ( 1 —

tan 64° ) = b 7.4 ( 1

— sin 64°

) = c

3.61 ≈ b 8.23 ≈ c

So, B = 26°, b ≈ 3.61, and c ≈ 8.23.

41. tan 79° = w —

100

100(tan 79°) = w

514 ≈ w

The width is about 514 meters.

42. tan 52° = h —

458

458(tan 52°) = h

586 ≈ h

The height is about 586 feet.

43.

h

75

80°

Write and solve an equation to fi nd h.

tan 80° = h — 75

75(tan 80°) = h

425 ≈ h

The height of the building is about 427 meters (including the

person’s eye level above the ground).

44.

h800 ft

30°

Write and solve an equation to fi nd the height h.

sin 30° = h —

800

800(sin 30°) = h

400 ≈ h


45. a. tan 24° = b —

1000

1000(tan 24°) = b

445 ≈ b

The height is about 451 feet (including your eye level of

5.5 feet).

b. The elevation at the top of Mount Rushmore is about

5280 + 451 = 5731 feet.

46. Answers will vary.

47. a. The radius of the Tropic of Cancer is

3960 cos (23.5°) ≈ 3631. So, the circumference is

about 2π ⋅ 3631 ≈ 22,818 miles.

b. The distance between the two points is the length of

the diameter of the Tropic of Cancer. So, it is about

2 ⋅ 3960 cos (23.5°) ≈ 7263 miles.

48. a. The side adjacent θ is x.

b. The side opposite θ is y.

c. Yes, they are equal.

cos θ = adj.

— hyp.

= x — h and sin (90° − θ) =

opp. —

hyp. = x —

h

49. a. sec(90° − 25°) = d —

25,000

25,000 ( 1 ——

cos(90° − 25°) ) = d

59,155 ≈ d

The distance is about 59,155 feet.

b. 25,0002 + x2 ≈ 59,1552

x ≈ √—— 59,1552 − 25,0002

x ≈ 53,613

The distance is about 53,613 feet.

c. tan(90° − 15°) = x + y —

25,000

25,000 tan (90° − 15°) = x + y

25,000 tan (90° − 15°) − 53,613 ≈ y

39,688 ≈ y The distance between the towns is about 39,688 feet.

Use the tangent function to fi nd the horizontal distance,

x + y, from the airplane to the second town and subtract

53,613 feet to fi nd the distance between the two towns.


Chapter 9

50.

h

x50 m32° 53°

tan 32° = h —

50 + x and tan 53° = h —

x

(50 + x) tan 32° = h x tan 53° = h

(50 + x) tan 32° = x tan 53° 50 tan 32° + x tan 32° = x tan 53° 50 tan 32° = x(tan 53° − tan 32°)

50 tan 32° ——

tan 53° − tan 32° = x

44.495 ≈ x

Thus, h = 44.495 tan 53° ≈ 59. So, the height of the

building is about 59 meters.

51. Your friend is correct. The triangle must be a 45-45-90

triangle because both acute angles would be the same and

have the same cosine value.

52. The triangle on the right is equilateral, so all of the angles

must be 60°. Using geometry, it can be shown that θ = 30° and that the two triangles form a larger 30-60-90 triangle.

Because the large triangle is a right triangle, the six

trigonometric ratios can be found. The legs of the triangle are

1 and √—

3 , and the hypotenuse is 2.

sin θ = 1 — 2 cos θ =

√— 3 —

2 tan θ =

√— 3 —

3

csc θ = 2 sec θ = 2 √—

3 —

3 cot θ = √

— 3

53. a. sin 30° = x — 1

sin 30° = x

1 —

2 = x

The perimeter is 6 units.

b. Sample answer: Each side is part of two right triangles

with opposing angles ( 180° — n ) . So, each side length is

2 sin ( 180° — n ) , and there are n sides.

c. 2n sin ( 180° — n ) ≈ 2π

n sin ( 180° — n ) ≈ π

When n = 50, π = 50 sin ( 180° — 50

) ≈ 3.14.

Maintaining Mathematical Profi ciency

54. ( 5 yr —

1 ) ( 365 days

— 1 yr

) ( 24 hr —

1 day ) ( 60 min

— 1 hr

) ( 60 sec —

1 min ) = 157,680,000 sec

55. ( 12 pt —

1 ) ( 1 gal

— 8 pt

) = 1.5 gal

56. ( 5.6 m —

1 ) ( 1000 mm

— 1 m

) = 5600 mm

57. C = 2π(6) ≈ 37.7 cm,

A = π(6)2 ≈ 113.1 cm2

58. C = 2π(11) ≈ 69.1 in.

A = π(11)2 ≈ 380.1 in.2

59. C = π(14) ≈ 44.0 ft,

A = π ( 14 — 2 ) 2 ≈ 153.9 ft2


1. a. Degreemeasure 0° 45° 90° 135° 180° 225° 270° 315° 360°

Radianmeasure

0 π — 4

π — 2

3π — 4 π

5π — 4

3π — 2

7π — 4 2π

b. Degreemeasure 30° 60° 120° 150° 210° 240° 300° 330°

Radianmeasure

π — 6

π — 3

2π — 3

5π — 6

7π — 6

4π — 3

5π — 3

11π — 6

Sample answer: To convert degrees to radians, multiply

degrees by π radians

— 180°

.

2. Radianmeasure

2π — 9

4π —

9

5π —

9

7π —

9

11π —

9

13π —

9

14π —

9

16π —

9

Degreemeasure 40° 80° 100° 140° 220° 260° 280° 320°

Sample answer: To convert radians to degrees, multiply

radians by 180° —

π radians .

3. Sample answer: To convert degrees into radians, multiply

degrees by π radians

— 180°

.

4. 30 radians ⋅ 180° —

π radians ≈ 1719°.

30 radians

— 30 degrees

≈ 1719° — 30°

= 57.3

Sample answer: 1 radian is about 57.3°, so 30 radians is

about 57.3 times greater than 30°, or about 1719°.


Chapter 9


1.

x

y

65°

2. Because 300° is 30° more than 270°, the terminal side is 30° counterclockwise past the negative y-axis.

x

y

300°

3. Because −120° is negative, the terminal side is 30° clockwise from the negative y-axis.

x

y

−120°

4. Because −450° is negative, the terminal side is the negative

y-axis.

x

y

−450°

5. 80° + 360° = 440° 80° − 360° = −280°

6. 230° + 360° = 590° 230° − 360° = −130°

7. 740° − 2 ⋅ 360° = 20° 740° − 3 ⋅ 360° = −340°

8. −135° + 360° = 225° −135° − 360° = −495°

9. 135° = 135 degrees ( π radians —

180 degrees )

= 3π — 4

10. −40° = (−40 degrees) ( π radians —

180 degrees )

= − 2π — 9

11. 5π — 4 = 5π —

4 radians ( 180°

— π radians

) = 225°

12. −6.28 = (−6.28 radians) ( 180° —

π radians )

≈ −359.8°

13. S = rθ = 220 ( π —

2 )

= 110π

≈ 346

The length of the outfi eld fence is about 346 feet. The area of

the fi eld is about 38,013 square feet.



1. An angle is in standard position when its vertex is at the

origin and its initial side lies on the positive x-axis.

2. When the angle is positive, its rotation is counterclockwise

and when the angle is negative, its rotation is clockwise.

3. Sample answer: A radian is a measure of an angle that is

approximately equal to 57.3° and there are 2π radians in a

circle.

4. The expression −90° does not belong because it has a

different terminal side than the other three angles.

A = 1 — 2 r 2θ

= 1 — 2 (220)2 ( π —

2 )

= 12,100π≈ 38,013


Chapter 9


5. Because 110° is 30° more than 90°, the terminal side is 30° counterclockwise past the positive y-axis.

x

y

110°

6. Because 450° is 90° more than 360°, the terminal side is the

positive y-axis.

x

y

450°

7. Because −900° is negative and 180° less than −720°, the

terminal side is the negative x-axis.

x

y

−900°

8. Because −10° is negative, the terminal side is 10° clockwise

from the positive x-axis.

x

y

−10°

9. 70° + 360° = 430° 70° − 360° = −290°

10. 255° + 360° = 615° 255° − 360° = −105°

11. −125° + 360° = 235° −125° − 360° = −485°

12. −800° + 3 ⋅ 360° = 280° −800° + 2 ⋅ 360° = −80°


180 degrees )

= 2π — 9


180 degrees )

= 7π — 4


180 degrees )

= − 13π — 9


180 degrees )

= − 25π — 9

17. π — 9 = ( π —

9 radians ) ( 180° —

π radians )

= 20°

18. 3π — 4 = 3π —

4 radians ( 180° —

π radians )

= 135°

19. −5 = (−5 radians) ( 180° — π radians

) ≈ −286.5°

20. 12 = 12 radians ( 180° — π radians

) ≈ 687.5°

21. A full revolution is 360° or 2π radians. The terminal side

rotates one-sixth of a revolution from the positive x-axis, so

multiply by 1 —

6 to get

1 —

6 ⋅ 360° = 60° and

1 —

6 ⋅ 2π = π —

3 .

22. The angles are 15π —

4 and −

π — 4 .

Sample answer: The angle 315° is equivalent to 7π — 4 radians,

and 7π — 4 + 2π = 15π —

4 and 7π —

4 − 2π = − π —

4 .

23. B; Because 600° is 240° more than 360°, the terminal side

makes one complete rotation 360° counterclockwise plus

240° more.


Chapter 9

24. D; − 9π — 4 = − 9π —

4 radians ( 180° —

π radians ) = −405°;

Because −405° is negative and 45° more than −360°, the

terminal side makes one complete rotation 360° clockwise

plus 45° more.

25. A; 5π — 6 = 5π —

6 radians ( 180° —

π radians ) = 150°; Because 150° is 60°

more than 90°, the terminal side is 60° counterclockwise past

the positive y-axis.

26. C; Because −240° is negative and 60° more than −180°, the

terminal side is 60° clockwise past the negative x-axis.

27. s = rθ

= 10 ( π — 2 )

= 5π ≈ 15.7

The length of the safety rail is about 15.7 yards and the area

of the deck is about 78.5 square yards.

28. a. s = rθ = 21.89 ( 34.92° ⋅

π radians —

180° )

≈ 13.3

The arc length is about 13.3 meters.

b. A = 1 — 2 r 2θ

= 1 — 2 (21.89)2 ( 34.92° ⋅

π radians —

180° )

≈ 146

The area of the sector is about 146 square meters.

29. The wrong conversion was used.

24° = 24 degrees ( π radians —

180 degrees )

= 24π radians —

180

= 0.42 radian

30. The angle was not converted to radians.


180 degrees ) =

2π — 9 radians

A = 1 —

2 (6)2 ( 2π —

9 ) ≈ 12.57 cm2

31. 200 ⋅ 360° = 72,000° 200 ⋅ 2π = 400π The angle is 72,000° or 400π radians.

32.

x

y

5:00

9:00

The hour hand has moved 8 —

12 of a revolution.

8 —

12 ⋅ 360° = 240°

8 —

12 ⋅ 2π = 4π —

3

The angle measure is 240° or 4π — 3 radians.

Sample answer: The minute hand would generate an angle of

2880° or 16π.

33. cos 4π — 3 = − 0.5

34. sin 7π — 8 ≈ 0.383

35. csc 10π — 11

= 1 —

sin 10π — 11

≈ 3.549

36. cot ( − 6π — 5 ) =

1 —

tan ( − 6π — 5 )

≈ −1.376

37. cot (−14) = 1 —

tan (−14) ≈ −0.138

38. cos 6 ≈ 0.960


180 degrees ) = 2 —

3 π radians

A = 1 — 2 r1

2θ − 1 — 2 r2

2θ

= 1 — 2 (25)2 ( 2 —

3 π ) − 1 —

2 (11)2 ( 2 —

3 π )

= 625 —

2 π − 121

— 3 π

= 168π ≈ 528

The area is about 528 square inches.

40. a. 15 revolutions per minute is equivalent to 15

— 60

revolutions

per second.

2π ⋅ 15 —

60 = 1 —

2 π

There are π — 2 radians of rotation per second.

b. s = rθ

= ( 58 —

2 ) ( π —

2 )

≈ 45.6

The arc length is about 45.6 feet.

A = 1 —

2 r2θ

= 1 — 2 (10)2 ( π —

2 )

= 25π≈ 78.5


Chapter 9

41. Four hours is 1 —

6 of a day. So, the rotation is

1 —

6 of Earth, or 60°

or π — 3 radians.

42. Using s = r (π − θ), the arc length of the small sector can be

found to be 1. Therefore, π − θ = 1 and θ = π − 1. So, the

angle is π − 1 radians.

43. Full sector: 1 —

2 ( 6

5 —

8 ) 2 ( 2π —

20 ) ≈ 6.89

Double region: 1 —

2 ( 6

5 —

8 ) 2 ( 2π —

20 ) −

1 —

2 ( 6

5 —

8 −

3 —

8 ) ( 2π —

20 ) ≈ 0.76

Triple region: 1 —

2 ( 3

3 —

4 +

3 —

8 ) 2 ( 2π —

20 ) −

1 —

2 ( 3

3 —

4 ) 2 ( 2π —

20 ) ≈ 0.46

The area of the entire sector is about 6.89 square inches, the

area of the double region is about 0.76 square inches, and the

area of the triple region is about 0.46 square inches.

44. Sample answer: This continued fraction (which is irrational)

gives rise to a sequence of rational approximations for π.

When the next fraction is added, the value gets closer to the

value of π = 3.1415926535 . . ., as shown.

3 = 3

3 + 1 —

7 =

22 —

7 = 3.142857143 . . .

3 + 1 —

7 + 1 —

15

= 333

— 106

= 3.141509434 . . .

3 + 1 ——

7 + 1 —

15 + 1 —

1

= 355

— 113

= 3.141592920 . . .

45. Your friend is correct. When the arc length is equal to the

radius, the equation s = rθ shows that θ = 1 and A = 1 —

2 r 2θ

is equivalent to A = s2

— 2 for r = s and θ = 1.

46. a. s = rθ = 42 ⋅ π — 8 ≈ 16.49

The length of the arc is about 16.49 inches.

b. The angle you would rotate through is 15π —

8 radians.

c. A = 1 —

2 r 2θ =

1 —

2 (42)2 ( π —

8 ) = 441π —

4 ≈ 346.4

Each step has 441π — 4 square inches of area, so

15 ⋅ 441π — 4 ≈ 5195.4 square inches of carpeting is needed.

47. a. 0.55 ⋅ 60 = 33

So, the measure is 70°33′. b. The measure is 110.76° because

110 + 45

— 60

+ 36 —

3600 = 110.76°.


48. d = √——

(x2 − x1)2 + (y2 − y1)

2

= √——

(3 − 1)2 + (6 − 4)2

= √—

22 + 22

= √—

8

≈ 2.83

49. d = √——

(x2 − x1)2 + (y2 − y1)

2

= √———

[10 − (−7)]2 + [8 − (−13)]

2

= √—

172 + 212

= √—

730

≈ 27.02

50. d = √——

(x2 − x1)2 + (y2 − y1)

2

= √———

[ −3 − (−3) ] 2 + (16 − 9)2

= √—

02 + 72

= √—

49

= 7

51. d = √——

(x2 − x1)2 + (y2 − y1)

2

= √——

(8 − 2)2 + (−5 − 12)2

= √——

62 + (−17)2

= √—

325

≈ 18.03

52. d = √——

(x2 − x1)2 + (y2 − y1)

2

= √————

[ −20 − (−14) ] 2 + [ −32 − (−22) ] 2

= √——

(−6)2 + (−10)2

= √—

136

≈ 11.66

53. d = √——

(x2 − x1)2 + (y2 − y1)

2

= √——

(−1 − 4)2 + (34 − 16)2

= √——

(−5)2 + 182

= √—

349

≈ 18.68


1. a. sin θ = √—

3 —

2 cos θ = − 1 —

2 tan θ = − √

— 3

b. sin θ = √

— 2 —

2 cos θ = −

√—

2 —

2 tan θ = −1

c. sin θ = −1 cos θ = 0 tan θ is undefi ned.

d. sin θ = − √—

3 —

2 cos θ =

1 —

2 tan θ = − √

— 3


Chapter 9

e. sin θ = − √

— 2 —

2 cos θ =

√—

2 —

2 tan θ = −1

f. sin θ = 0 cos θ = −1 tan θ = 0

2. Sample answer: The coordinates of the point on the unit

circle can be used to fi nd the ratios of the six trigonometric

functions where sin θ = y — r , cos θ =

x — r , tan θ =

y — x , csc θ =

r — y ,

sec θ = r — x , and cot θ =

x — y .

3. a. tan θ = y —

x is undefi ned when x = 0, which occurs when

θ = π — 2 + nπ, where n is an integer.

b. cot θ = x —

y is undefi ned when y = 0, which occurs when

θ = nπ, where n is an integer.

c. sec θ = r —

x is undefi ned when x = 0, which occurs when

θ = π — 2 + nπ, where n is an integer.

d. csc θ = r —

y is undefi ned when y = 0, which occurs when

θ = nπ, where n is an integer.


1. Use the Pythagorean Theorem to fi nd the length of r.

r = √—

x 2 + y 2

= √—

32 + (−3)2

= √—

18

= 3 √—

2

Using x = 3, y = −3, and r = 3 √—

2 , the values of the six


sin θ = y —

r = −

√—

2 —

2 csc θ =

r —

y = − √

— 2

cos θ = x —

r =

√—

2 —

2 sec θ =

r —

x = √

— 2

tan θ = y —

x = −1 cot θ =

x —

y = −1


r = √—

x 2 + y 2

= √——

(−8)2 + 152

= √—

289

= 17

Using x = −8, y = 15, and r = 17, the values of the six


sin θ = y —

r =

15 —

17 csc θ =

r —

y =

17 —

15

cos θ = x —

r = − 8 —

17 sec θ =

r —

x = − 17

— 8

tan θ = y —

x = − 15

— 8 cot θ =

x —

y = − 8 —

15


r = √—

x 2 + y 2

= √——

(−5)2 + (−12)2

= √—

169

= 13

Using x = −5, y = −12, and r = 13, the values of the six


sin θ = y —

r = − 12

— 13

csc θ = r —

y = − 13

— 12

cos θ = x —

r = − 5 —

13 sec θ =

r —

x = − 13

— 5

tan θ = y —

x =

12 —

5 cot θ =

x —

y =

5 —

12

4. Draw a unit circle with the angle θ = 180° in standard

position.

Identify the point where the terminal side of θ intersects the

unit circle. The terminal side of θ intersects the unit circle at

(−1, 0).

x

y

(−1, 0)

θ

Find the values of the six trigonometric functions. Let

x = −1 and y = 0 to evaluate the trigonometric functions.

sin θ = y —

r = 0 csc θ =

r —

y =

1 —

0 undefi ned

cos θ = x —

r = −1 sec θ =

r —

x = −1

tan θ = y —

x = 0 cot θ =

x —

y =

−1 —

0 undefi ned

5.

x

y

210°

30°

The terminal side lies in Quadrant III. So, the reference angle

is 210° − 180° = 30°.


Chapter 9

6.

x

y

−260°

80°

The terminal side lies in Quadrant II. So, the reference angle

is 180° − 100° = 80°.

7.

x

y

−79

π29π


is 11π —

9 − π = 2π —

9 .

8.

x

y

154π

4π

The terminal side lies in Quadrant IV. So, the reference angle

is 2π − 7π — 4 = π —

4 .

9. The angle −210° is coterminal with 150°. The reference

angle is 180° − 150° = 30°. The cosine function is negative

in Quadrant II, so cos (−210°) = −cos (30°) = − √—

3 —

2 .

10. The angle 11π —

4 is coterminal with

3π — 4 . The reference angle is

π − 3π — 4 =

π — 4 . The secant function is negative in Quadrant II, so

sec 11π —

4 = −sec

π — 4 = − √

— 2 .

11. d = v 2

— 32

sin 2θ

= 272

— 32

sin(2 ⋅ 20°)

≈ 14.6

The horizontal distance traveled is about 14.6 feet.



1. A quadrantal angle is an angle in standard position whose

terminal side lies on an axis.

2. After fi nding the reference angle θ′, evaluate − cosθ′ (cosine

is negative in Quadrant III).



r = √—

x 2 + y 2

= √—

42 + (−3)2

= √—

25

= 5

Using x = 4, y = −3, and r = 5, the values of the six


sin θ = y —

r = − 3 —

5 csc θ =

r —

y = − 5 —

3

cos θ = x —

r =

4 —

5 sec θ =

r —

x = 5 —

4

tan θ = y —

x = − 3 —

4 cot θ =

x —

y = − 4 —

3


r = √—

x 2 + y 2

= √——

52 + (−12)2

= √—

169

=13



sin θ = y —

r = − 12

— 13

csc θ = r —

y = − 13

— 12

cos θ = x —

r = 5 —

13 sec θ =

r —

x = 13

— 5

tan θ = y —

x = − 12

— 5 cot θ =

x —

y = − 5 —

12


r = √—

x 2 + y 2

= √——

(−6)2 + (−8)2

= √—

100

= 10



sin θ = y —

r = − 4 —

5 csc θ =

r —

y = − 5 —

4

cos θ = x —

r = − 3 —

5 sec θ =

r —

x = − 5 —

3

tan θ = y —

x =

4 —

3 cot θ =

x —

y =

3 —

4


Chapter 9


r = √—

x 2 + y 2

= √—

32 + 12

= √—

10

Using x = 3, y = 1, and r = √—



sin θ = y —

r =

√—

10 —

10 csc θ =

r —

y =

√—

10 —

1 = √

— 10

cos θ = x —

r =

3 —

√—

10 =

3 √—

10 —

10 sec θ =

r —

x =

√—

10 —

3

tan θ = y —

x =

1 —

3 cot θ =

x —

y =

3 —

1 = 3


r = √—

x 2 + y 2

= √——

(−12)2 + (−9)2

= √—

225

= 15



sin θ = y —

r = − 3 —

5 csc θ =

r —

y = − 5 —

3

cos θ = x —

r = − 4 —

5 sec θ =

r —

x = − 5 —

4

tan θ = y —

x =

3 —

4 cot θ =

x —

y =

4 —

3


r = √— x 2 + y 2

= √— 12 + (−2)2

= √—

5

Using x = 1, y = −2, and r = √—



sin θ = y —

r =

−2 —

√—

5 = − 2 √

— 5 —

5 csc θ =

r —

y = −

√—

5 —

2

cos θ = x —

r =

1 —

√—

5 =

√—

5 —

5 sec θ =

r —

x =

√—

5 —

1 = √

— 5

tan θ = y —

x =

−2 —

1 = −2 cot θ =

x —

y = −

1 —

2

9. Draw a unit circle with the angle θ = 0° in standard position.


unit circle. The terminal side of θ intersects the unit circle

at (1, 0).

x

y

(1, 0)

Find the values of the six trigonometric functions. Let x = 1

and y = 0 to evaluate the trigonometric functions.

sin θ = y —

r =

0 —

1 = 0 csc θ =

r —

y =

1 —

0 undefi ned

cos θ = x —

r =

1 —

1 = 1 sec θ =

r —

x =

1 —

1 = 1

tan θ = y —

x =

0 —

1 = 0 cot θ =

x —

y =

1 —

0 undefi ned

10. Draw a unit circle with the angle θ = 540° in standard

position.


unit circle. The terminal side of θ intersects the unit circle

at (−1, 0).

x

y

(−1, 0)θ

Find the values of the six trigonometric functions. Let x = −1


sin θ = y —

r =

0 —

1 = 0 csc θ =

r —

y =

1 —

0 undefi ned

cos θ = x —

r =

−1 —

1 = −1 sec θ =

r —

x =

1 —

−1 = −1

tan θ = y —

x =

0 —

−1 = 0 cot θ =

x —

y =

−1 —

0 undefi ned


Chapter 9

11. Draw a unit circle with the angle θ = π — 2 in standard position.

Identify the point where the terminal side of θ intersects

the unit circle. The terminal side of θ intersects the unit

circle at (0, 1).

x

y(0, 1)

θ



sin θ = y —

r =

1 —

1 = 1 csc θ =

r —

y =

1 —

1 = 1

cos θ = x —

r =

0 —

1 = 0 sec θ =

r —

x =

1 —

0 undefi ned

tan θ = y —

x = 1 —

0 undefi ned cot θ =

x —

y =

0 —

1 = 0

12. Draw a unit circle with the angle θ = 7π — 2 in standard

position.



circle at (0, −1).

x

y

(0, −1)

θ


and y = −1 to evaluate the trigonometric functions.

sin θ = y —

r =

−1 —

1 = −1 csc θ =

r —

y =

1 —

−1 = −1

cos θ = x —

r =

0 —

1 = 0 sec θ =

r —

x =

1 —

0 undefi ned

tan θ = y —

x = −1

— 0 undefi ned cot θ =

x —

y =

0 —

−1 = 0

13. Draw a unit circle with the angle θ = −270° in standard

position.



circle at (0, 1).

x

y(0, 1)

θ



sin θ = y —

r =

1 —

1 = 1 csc θ =

r —

y =

1 —

1 = 1

cos θ = x —

r =

0 —

1 = 0 sec θ =

r —

x =

1 —

0 undefi ned

tan θ = y —

x =

1 —

0 undefi ned cot θ =

x —

y =

0 —

1 = 0

14. Draw a unit circle with the angle θ = −2π in standard

position.



circle at (1, 0).

x

y

(1, 0)θ



sin θ = y —

r =

0 —

1 = 0 csc θ =

r —

y =

1 —

0 undefi ned

cos θ = x —

r =

1 —

1 = 1 sec θ =

r —

x =

1 —

1 = 1

tan θ = y —

x =

0 —

1 = 0 cot θ =

x —

y =

1 —

0 undefi ned

15.

x

y

−100°80°


is 260° − 180° = 80°.


Chapter 9

16.

x

y

150°30°


is 180° − 150° = 30°.

17.

x

y

320°

40°


is 360° − 320° = 40°.

18.

x

y

−370°

10°


is 360° − 350° = 10°.

19.

x

y

154π

4π


is 2π − 7π — 4 =

π — 4 .

20.

x

y

83π

3π


is π − 2π — 3 =

π — 3 .

21.

x

y

−56

π6π


is 7π — 6 − π =

π — 6 .

22.

x

y

−136

π

6π


is 2π − 11π — 6 =

π — 6 .

23. The equation for tangent is tan θ = y —

x .

tan θ = y —

x = − 2 —

3

24. The angle found is the angle between the terminal side and

the y-axis, instead of the x-axis.

θ is coterminal with 290°, whose terminal side lies in

Quadrant IV. So, θ′ = 360° − 290° = 70°.

25. The angle 135° has the reference angle 180° − 135° = 45°.

The secant function is negative in Quadrant II, so

sec 135° = −sec 45° = − √—

2 .


Chapter 9

26. The angle 240° has reference angle 240° − 180° = 60°.

The tangent function is positive in Quadrant III, so

tan 240° = tan 60° = √—

3 .


angle is 210° − 180° = 30°. The sine function is negative in

Quadrant III, so sin(−150°) = −sin 30° = − 1 — 2 .

28. The angle −420° is coterminal with 300°. The reference angle

is 360° − 300° = 60°. The cosecant function is negative in

Quadrant IV, so csc (−420°) = −csc 60° = − 2 √

— 3 —

3 .

29. The angle −3π —


5π — 4 . The reference angle

is 5π — 4 − π =

π — 4 . The tangent function is positive in

Quadrant III, so tan ( −3π — 4 ) = tan

π — 4 = 1.

30. The angle −8π —


4π — 3 . The reference angle

is 4π — 3 − π =

π — 3 . The cotangent function is positive in

Quadrant III, so cot ( −8π — 3 ) = cot

π — 3 =

√— 3 —

3 .

31. The angle 7π — 4 has reference angle 2π −

7π — 4 =

π — 4 . The

cosine function is positive in Quadrant IV, so

cos 7π — 4 = cos

π — 4 =

√—

2 —

2 .


6 has reference angle 2π −

11π — 6 =

π — 6 .

The secant function is positive in Quadrant IV, so

sec 11π —

6 = sec π —

6 =

2 √—

3 —

3 .

33. d = v2

— 32

sin 2θ

= 492

— 32

sin(2.60°)

≈ 65

The horizontal distance traveled is about 65 feet.

34. d = v2

— 32

sin 2θ

= 142

— 32

sin(2.45°)

≈ 6.1

The distance of a jump is about 6.1 feet.

35. d = v2

— 32

sin 2θ

5 = v2

— 32

sin(2 ⋅ 18°)

160 —

sin(36°) = v2

v ≈ 16.5

The initial speed needs to be about 16.5 feet per second.

36. You win the competition because your javelin travels

712

— 32

sin(2 ⋅ 40°) ≈ 155.1 feet.

37. First, determine the distance the top of the treadmill is above

the midpoint. The reference angle is 180° − 110° = 70°,

and by using the sine function, sin 70° = opp.

— hyp.

= opp.

— 5 , or

opp. = 5 sin 70° ≈ 4.7. So, the top of the treadmill is about

4.7 feet above the midpoint. This puts the top of the treadmill

about 4.7 + 6 = 10.7 feet above the ground.

38. Use the following diagram.

x

y

10 ft

75 ft

75 ft

255°15°

First, determine the height above the center of the Ferris wheel.

This gives sin 15° = opp.

— 75

or opp. = 75 sin 15° ≈ 19 feet. So,

the total height is about 19 + 75 + 10 = 104 feet. If the radius

is doubled the height is not doubled.

Sample answer: The initial height that the Ferris wheel is

above the ground is not doubled so the entire height is not

doubled.

39. a. Angle of sprinkler, θ

Horizontal distance water travels, d

30° 16.9

35° 18.4

40° 19.2

45° 19.5

50° 19.2

55° 18.4

60° 16.9

b. The maximum distance is 45°. Because v2

— 32

is constant in this

situation, the maximum distance traveled will occur when

sin 2θ is as large as possible. The maximum value of

sin 2θ occurs when 2θ = 90°, that is, when θ = 45°.

c. The distances are the same.

40. Determine the x-value as described in the diagram,

x = 50 cos 300° = 25. So, the stopping position is

50 − 25 = 25 feet farther from the goal line than the

starting position. So, the stopping position is 125 feet from

the goal line.


Chapter 9

41.

x

y√22

(1, 0)

(0, 1)

(−0, 1)

(−1, 0)

,( )12

√32

,( )√22

( )12

, √32√2

2

, )12

√32

,( )√22

)12

, √32−

√22

,( )√22

−√22

,( )√22

−−

(−

, )12

√32( −, )1

2√32( −−

(−

)12

, √32( −)1

2, √3

2( −−

0°

360°

45°135°

225° 315°

180°

90°

270°

30°

60°

150°

120°

240° 300°

330°210°

42. a. This is not possible. If sin θ > 0 and cos θ < 0, then

tan θ < 0.

b. If sin θ > 0, cos θ < 0, and tan θ < 0, then the terminal

side of the angle in standard position lies in Quadrant II.

So, 90° < θ < 180°.

43. tan θ = y —

x =

y —

r —

x —

r =

sin θ — cos θ

sin 90° = 1 and cos 90° = 0, so tan 90° is undefi ned because

you cannot divide by 0, but cot 90° = 0 —

1 = 0.

44. Sine and cosecant are negative because the y-coordinate

is negative in Quadrant IV. Cosine and secant are positive

because the x-coordinate is positive in Quadrant IV. Tangent

and cotangent are negative because the y-coordinate is

negative and the x-coordinate is positive.

45. A point on the line will have the form (x, y). Using this and

the origin, the slope of the line is m = y —

x . Because tan θ =

y —

x ,

tan θ = m.

46. Your friend is incorrect. θ = 240° is also a solution and any

angle coterminal to 60° and 240° are also solutions.

47. a. The reference angle for 117° is 180° − 117° = 63°. So,

sin 63° = opp.

— hyp.

= opp.

— 128

and cos 63° = adj.

— hyp.

= adj.

— 128

, or

opp. = 128 sin 63° ≈ 114 and adj. = 128 cos 63° ≈ 58.1.

Because the point is in Quadrant II, the point is

(−58.1, 114).

b. d = √—— (x2 − x1)

2 + (y2 − y1)2

≈ √——— [128 − (−58.1)]2 + (0 − 114)2

= √—— 186.12 + (−114)2

≈ 218

The distance is about 218 picometers.

48. When C is the circumference of Earth, then the equation is

1 — 2 C = C sin P. So,

1 —

2 = sin P. Because the angle is between

0° and 90°, this happens when P = 60°.


49. Find the rational zeros of f. Because f is a polynomial

function, the possible rational zeros are ±1, ±2, ±3, ±4,

±6, and ±12. Use synthetic division.

−3 1 2 1 8 −12

−3 3 −12 12

1 −1 4 −4 0

1 1 2 1 8 −12

1 3 4 12

1 3 4 12 0

So, −3 and 1 are zeros.

So, f (x) = (x − 1)(x + 3)(x2 + 4) and x = −3 and x = 1 are

the only real zeros.

50. Find the rational zeros of f. Because f is a polynomial

function, the possible rational zeros are ±1, ±2, ±3, ±6,

and ±18. Use synthetic division.

−6 1 4 −14 −14 −15 −18

−6 12 12 12 18

1 −2 −2 −2 −3 0

−1 1 4 −14 −14 −15 −18

−1 −3 17 −3 18

1 3 −17 3 −18 0

3 1 4 −14 −14 −15 −18

3 21 21 21 18

1 7 7 7 6 0

So, −6, −1 and 3 are zeros.

So, f (x) = (x − 3)(x + 1)(x + 6)(x2 + 1) and x = 6,

x = −1, and x = 3 are the only real zeros.


Chapter 9

51. Step 1 Plot the x-intercepts. Because −3 and 1 are zeros

of f, plot (−3, 0) and (−1, 0).

Step 2 Plot points between and beyond the x-intercepts.

x −4 −2 0 2

y −10 −6 −18 50

Step 3 Determine the end behavior. Because f has three

factors of the form x − k and a constant factor

of 2, it is a cubic function with positive leading

coeffi cient. So, f(x) → −∞ as x → −∞ and

f(x) → +∞ as x → +∞.

Step 4 Draw the graph so that it passes through the plotted

points and has the appropriate end behavior.

x

y

4

2

6

−8

−6

−10

−20

−4

2 3 4 5 6−4 −1−5−6−7

52. Step 1 Plot the x-intercepts. Because 4, 5, and −9 are zeros

of f, plot (4, 0), (5, 0), and (−9, 0).


x −10 0 3 9 — 2

y − 70 — 3 −60 −32 171

— 8


factors of the form x − k and a constant factor

of 1 —

3 , it is a cubic function with positive leading

coeffi cient. So, f(x) → −∞ as x → −∞ and

f(x) → +∞ as x → +∞.



x

y

10

20

−20

−60

−70

82 6−6 −2−10

53. Step 1 Plot the x-intercepts. Because 0, −1, and 2 are zeros

of f, plot (0, 0), (−1, 0), and (2, 0).


x −2 − 1 — 2 1 3

y 16 − 5 — 16 −2 36


factors of the form x − k, it is a quartic function

with positive leading coeffi cient. So, f(x) → +∞

as x → −∞ and f (x) → +∞ as x → +∞.



x

y

8

4

2

6

10

12

−8

−6

−10

−12

−14

−16

−18

−20

−4

1 3 4−1−2−3−4


1. a. x −2π −

7π — 4 −

3π — 2 −

5π — 4 −π

y = sin x 0 √

— 2 —

2 1

√—

2 —

2 0

x π — 4

π — 2

3π — 4 π

5π — 4

y = sin x √

— 2 —

2 1

√—

2 —

2 0 −

√—

2 —

2

x −

3π — 4 −

π — 2 −

π — 4 0

y = sin x − √

— 2 —

2 −1 −

√—

2 —

2 0

x 3π — 2

7π — 4 2π

9π — 4

y = sin x −1 − √

— 2 —

2 0

√—

2 —

2


Chapter 9

b.

x

y1

π−2

y = sin x

c. The x-intercepts are x = −2π, π, 0, π, 2π. The local

maximums are at x = − 3π — 2 ,

π — 2 . The local minimums

are at x = − π — 2 ,

3π — 2 . The function is increasing when

−2π < x < − 3π — 2 , −

π — 2 < x <

π — 2 , and

3π — 2 < x < 2π and

decreasing when − 3π — 2 < x < −

π — 2 and

π — 2 < x <

3π — 2 . The

sine function is an odd function.

2. a. x −2π −

7π — 4 −

3π — 2 −

5π — 4 −π

y = cos x 1 √

— 2 —

2 0

− √—

2 —

2 −1

x π — 4

π — 2

3π — 4 π

5π — 4

y = cos x √

— 2 —

2 0 −

√—

2 —

2 −1 −

√—

2 —

2

x −

3π — 4 −

π — 2 −

π — 4 0

y = cos x − √

— 2 —

2 0

√—

2 —

2 1

x 3π — 2

7π — 4 2π

9π — 4

y = cos x 0 √

— 2 —

2 1

√—

2 —

2

b.

x

y1

π π2

−1

π−

y = cos x

c. The x-intercepts are x = − 3π — 2 , −

π — 2 , π —

2 , 3π —

2 . The local

maximums are at x = −2π, 0, 2π. The local minimums

are at x = −π, π. The function is increasing when

−π < x < 0 and π < x < 2π and decreasing when

−2π < x < −π and −π < x < π. The cosine function is

an even function.

3. Sample answer: The minimum value of each function is −1

and the maximum value is 1. Both functions have a repeating

pattern. The x-intercepts for y = sin x occur when x = 0,

±π, ±2π, ±3π . . ., and the x-intercepts for y = cos x occur

when x = ± π — 2 , ±

3π — 2 , ±

5π — 2 . . .

4. As x increases (or decreases) without bound, the graph

of y = sin x does not approach a single value. The graph

oscillates between y = 1 and y = −1.

9.4 Monitoring Progress (pp. 488−490)

1. The function is of the form g(x) = a sin bx, where a = 1 —

4 and

b = 1. So, the amplitude is a = 1 —

4 and the period is

2π — b =

2π — 1 = 2π.

Intercepts: (0, 0); ( 1 — 2 ∙ 2π, 0 ) = (π, 0); (2π, 0)

Maximum: ( 1 — 4 ∙ 2π,

1 —

4 ) = ( π —

2 ,

1 —

4 )

Minimum: ( 3 — 4 ∙ 2π, −

1 —

4 ) = ( 3π —

2 , −

1 —

4 )

x

y

π− ππ−2

1

−1

g(x) = sin x14

The graph of g is a vertical shrink by a factor of 1 —

4 of the

graph of f (x) = sin x.

2. The function is of the form g(x) = a cos bx, where a = 1 and

b = 2. So, the amplitude is a = 1 and the period is

2π — b =

2π — 2 = π.

Intercepts: ( π — 4 , 0 ) ; ( 3π —

4 , 0 ) ; ( 5π —

4 , 0 )

Maximum: (0, 1); (π, 1)

Minimum: ( π — 2 , −1 ) ; ( 3π —

2 , −1 )

x

y1

π

−1

g(x) = cos (2x)

The graph of g is a horizontal shrink by a factor of 1 —

2 of the

graph of f (x) = cos x.


Chapter 9

3. The function is of the form g(x) = a sin bx, where a = 2

and b = π. So, the amplitude is a = 2 and the period is

2π — b =

2π — π

= 2.

Intercepts: (0, 0); (1, 0); (2, 0)

Maximum: ( 1 — 2 , 2 )

Minimum: ( 3 — 2 , −2 )

x

y2

1

0.5 1 1.5

−2

−1

g(x) = 2sin xπ

The graph of g is a horizontal shrink by a factor of 1 — π

and a vertical stretch by a factor of 2 of the graph of

f (x) = sin x.

4. The function is of the form g(x) = a cos bx, where a = 1 —

3

and b = 1 —

2 . So, the amplitude is a =

1 —

3 and the period is

2π — b =

2π — 1 —

2

= 4π.

Intercepts: (π, 0); (3π, 0)

Maximum: ( 0, 1 —

3 ) ; ( 4π,

1 —

3 )

Minimum: ( 2π, − 1 —

3 )

x

y

−

π2

13

13

g(x) = cos x13

12

The graph of g is a horizontal stretch by a factor of 2

followed by a vertical shrink by a factor of 1 —

3 of the graph

of f (x) = cos x.

5. Step 1 Identify the amplitude, period, horizontal shift, and

vertical shift.

Amplitude: a = 1 Horizontal shift: h = 0

Period: 2π — b =

2π — 1 = 2π Vertical shift: k = 4

Step 2 Draw the midline of the graph, y = 4.

Step 3 Find the fi ve key points.

On y = k: ( π — 2 , 4 ) ; ( 3π —

2 , 4 )

Maximum: (0, 5); (2π, 5)

Minimum: (π, 3)

Step 4 Draw the graph through the key points.

x

y

1

2

3

4

π


vertical shift.

Amplitude: a = 1 —

2 Horizontal shift: h =

π — 2

Period: 2π — b =




On y = k: ( π — 2 , 0 ) ; ( 3π —

2 , 0 )

Maximum: ( π, 1 —

2 )

Minimum: ( 0, − 1 —

2 ) ; ( 2π, −

1 —

2 )


x

y1

π

−1


Chapter 9


vertical shift.

Amplitude: a = 1 Horizontal shift: h = −π

Period: 2π — b =

2π — 1 = 2π Vertical shift: k = −1

Step 2 Draw the midline of the graph, y = −1.


On y = k: (0, −1); (π, −1)

Maximum: ( − π — 2 , 0 ) ; ( 3π —

2 , 0 )

Minimum: ( π — 2 , −2 )


x

y

π

−1

−2


vertical shift.

Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = − π — 2

Period: 2π — b =


Step 2 Draw the midline of the graph. Because k = 0, the

midline is the x-axis.

Step 3 Find the fi ve key points of f (x) = ∣ −1 ∣ cos ( x − π — 2 ) .

On y = k: (−π, 0); (0, 0); (π, 0)

Maximum: ( − π — 2 , 1 )

Minimum: ( π — 2 , −1 )

Step 4 Refl ect the graph. Because a < 0, the graph is

refl ected in the midline y = 0. So, ( − π — 2 , 1 ) becomes

( − π — 2 , −1 ) and ( π —

2 , −1 ) becomes ( π —

2 , 1 ) .


x

y1

−1


vertical shift.

Amplitude: ∣ a ∣ = ∣ −3 ∣ = 3 Horizontal shift: h = 0

Period: 2π — b

= 2π — 1 —

2

= 4π Vertical shift: k = 2


midline is y = 2.

Step 3 Find the fi ve key points of g(x) = −3 sin 1 —

2 x + 2.

On y = k: (0, 2); (2π, 2)

Maximum: (−π, 5); (3π, 5)

Minimum: (π, −1)


x

y

1

2

4

5

π π2 π3


vertical shift.


Period: 2π — b =

2π — 4 =

π — 2 Vertical shift: k = −1

Step 2 Draw the midline of the graph. Because k = −1, the

midline is y = −1.

Step 3 Find the fi ve key points of g(x) = −2 cos 4x − 1.

On y = −1: ( π — 8 , −1 ) ; ( 3π —

8 , −1 )

Maximum: ( π — 4 , 1 )

Minimum: (0, −3); ( π — 2 , −3 )


x

y1

−3

9.4 Exercises (pp. 491−494)


1. The shortest repeating portion of the graph of a periodic

function is called a cycle.

2. The amplitude of the fi rst function is 1 —

2 and the amplitude of

the second function is 3. The period of the fi rst function is 2π

and the period of the second function is π.


Chapter 9

3. A phase shift is a horizontal translation of a periodic

function.

Sample answer: y = sin ( x − π — 2 ) is a phase shift of y = sin x.

4. The midline of y = 2 sin 3(x + 1) − 2 is y = −2.

5. The function is a periodic function with a period of 2.

6. The function is a periodic function with a period of 2π.

7. The function is not a periodic function.

8. The function is not a periodic function.

9. The amplitude is 1 and the period is 6π.

10. The amplitude is 1 —

2 and the period is 2.

11. The amplitude is 4 and the period is π.



and b = 1. So, the amplitude is a = 3 and the period is

2π — b =

2π — 1 = 2π.

Intercepts: (0, 0); (π, 0); (2π, 0)

Maximum: ( π — 2 , 3 )

Minimum: ( 3π — 2 , −3 )

x

y

3

4

2

1

π

−3

−4

The graph of g is a vertical stretch by a factor of 3 of the




2π — b =

2π — 1 = 2π.

Intercepts: (0, 0); (π, 0); (2π, 0)

Maximum: ( π — 2 , 2 )

Minimum: ( 3π — 2 , −2 )

x

y2

1

−1

−2



15. The function is of the form g(x) = a cos bx, where a = 1


2π — b =

2π — 3 .

Intercepts: ( π — 6 , 0 ) ; ( π —

2 , 0 )

Maximum: (0, 1); ( 2π — 3 , 1 )

Minimum: ( π — 3 , −1 )

x

y1

π

−1

π3

2π3


3 of the


16. The function is of the form g(x) = a cos bx, where a = 1


2π — b =

2π — 4 =

π — 2 .

Intercepts: ( π — 8 , 0 ) ; ( 3π —

8 , 0 )

Maximum: (0, 1); ( π — 2 , 1 )

Minimum: ( π — 4 , −1 )

x

y1

π

−1


4 of the



Chapter 9


and b = 2π. So, the amplitude is a = 1 and the period is

2π — b =

2π — 2π

= 1.

Intercepts: (0, 0); (0.5, 0); (1, 0)

Maximum: (0.25, 1)

Minimum: (0.75, −1)

x

y1

0.5

−1

0.8 1.20.4 1.6


2π of the




2π — b =

2π — 2 = π.

Intercepts: (0, 0); ( π — 2 , 0 ) ; (π, 0)

Maximum: ( π — 4 , 3 )

Minimum: ( π — 4 , −3 )

x

y

1

2

3

4

π

−3

−4

−2

−1


2 and a

vertical stretch by a factor of 3 of the graph of f (x) = sin x.


3

and b = 4. So, the amplitude is a = 1 —

3 and the period is

2π — b =

2π — 4 =

π — 2 .

Intercepts: ( π — 4 , 0 ) ; ( 3π —

4 , 0 )

Maximums: ( 0, 1 —

3 ) ; ( π,

1 —

3 )

Minimum: ( π — 2 , −

1 —

3 )

x

y

π

−

13

13


4 and a

vertical shrink by a factor of 1 —

3 of the graph of f (x) = cos x.


2

and b = 4π. So, the amplitude is a = 1 —

2 and the period is

2π — b =

2π — 4π

= 1 —

2 .

Intercepts: ( 1 — 8 , 0 ) ; ( 3 —

8 , 0 )


2 ) ; ( 1 —

2 ,

1 —

2 )

Minimum: ( 1 — 4 , −

1 —

2 )

x

y

−

12

12

0.25 0.5 0.75 1


4π and a


2 of the graph of f (x) = cos x.

21. B; D; The function y = −4 sin πx is of the form

g(x) = a sin bx, where a = −4 and b = π. So, the amplitude

is ∣ a ∣ = ∣ −4 ∣ = 4 and the period is 2π — b =

2π — π

= 2. The

function y = 4 cos πx is of the form g(x) = a cos bx, where

a = 4 and b = π. So, the amplitude is a = 4 and the period

is 2π — b =

2π — π

= 2.


Chapter 9

22. a. The period is 5, so 2π — b = 5 or b =

2π — 5 . The equation is

y = sin 2π — 5 x.

b. The period is 4, so 2π — b = 4 or b =

π — 2 . The equation is

y = 10 sin π — 2 x.

c. The period is 2π, so 2π — b = 2π or b = 1. The equation is

y = 2 sin x.

d. The period is 3π, so 2π — b = 3π or b =

2 —

3 . The equation is

y = 1 —

2 sin

2 —

3 x.

23. The period is 1 —

4 and represents the amount of time, in

seconds, that it takes for the pendulum to go back and forth

and return to the same position. The amplitude is 4 and

represents the maximum distance, in inches, the pendulum

will be from its resting position.

t

d

0.25 0.5 0.75 1

2

3

4

1

−1

−2

−3

−4

24. The period is 6 and represents

t

y

1 2 3 4 5 6

1.75

−1.75

the amount of time, in seconds,

that it takes for the buoy to bob

up and down and return to the same

position. The amplitude is 1.75 and

represents the maximum distance,

in feet, the buoy will be from

its midline.


vertical shift.


Period: 2π — b =




On y = k: (0, 2); (π, 2); (2π, 2)

Maximum: ( π — 2 , 3 )

Minimum: ( 5π — 2 , 1 )


x

y

2

3

1

π π2

−1


vertical shift.


Period: 2π — b =




On y = −4: ( π — 2 , −4 ) ; ( 3π —

2 , −4 )

Maximums: (0, −3); (2π, −3)

Minimum: (π, −5)


x

y1

π π2−1

−2

−3

−4

−5


vertical shift.

Amplitude: a = 1 Horizontal shift: h = π — 2

Period: 2π — b =





On y = k: (0, 0); (π, 0); (2π, 0)

Maximum: ( π — 2 , 1 )

Minimum: ( 3π — 2 , −1 )


x

y1

−1

π2


Chapter 9


vertical shift.

Amplitude: a = 1 Horizontal shift: h = − π — 4

Period: 2π — b =





On y = k: ( 3π — 4 , 0 ) ; ( 7π —

4 , 0 ) ; ( − π —

4 , 0 )

Maximum: ( π — 4 , 1 )

Minimum: ( 5π — 4 , 1 )


x

y1

π

−1

π2


vertical shift.


Period: 2π — b =




On y = k: ( π — 2 , −1 ) ; ( 3π —

2 , −1 )

Maximums: (0, 1); (2π, 1)

Minimums: (π, −3)


x

y1

π π2

−1

−2

−3


vertical shift.


Period: 2π — b =




On y = k: (0, 1); (π, 1); (2π, 1)

Maximum: ( π — 2 , 4 )

Minimum: ( 3π — 2 , −2 )


x

y

1

2

3

4

π π2−1

−2


vertical shift.

Amplitude: a = 1 Horizontal shift: h = −2π

Period: 2π — b =

2π — 2 = π Vertical shift: k = 0




On y = k: (0, 0); ( π — 2 , 0 ) ; (π, 0)

Maximum: ( π — 4 , 1 )

Minimum: ( 3π — 4 , −1 )


x

y1

−1

π5 4


Chapter 9


vertical shift.

Amplitude: a = 1 Horizontal shift: h = 2π

Period: 2π — b =

2π — 2 = π Vertical shift: k = 0




On y = k: ( π — 4 , 0 ) ; ( 3π —

4 , 0 )

Maximums: (0, 1); (π, 1)

Minimum: ( π — 2 , −1 )


x

y1

−1

π π2


vertical shift.

Amplitude: a = 1 Horizontal shift: k = −π

Period: 2π — b =

2π — 1 —

2




On y = k: (0, 3); (2π, 3); (4π, 3)

Maximum: (3π, 4)

Minimum: (π, 2)


x

y

1

2

3

4

π π4π3π2−1


vertical shift.

Amplitude: a = 1 Horizontal shift: h = 3π — 2

Period: 2π — b =

2π — 1 —

2

= 4π Vertical shift: k = −5



On y = k: (0, −5); (2π, −5); (4π, −5)

Maximum: (3π, −4)

Minimum: (π, −6)


x

y1

π π4π3π2−1

−2

−3

−4

−5

−6

−7

35. To fi nd the period, use the expression 2π — ∣ b ∣

.

Period: 2π — ∣ b ∣

= 2π —

∣ 2 — 3 ∣ = 3π

36. π — 2 should be added to the x-coordinate.

Maximum: ( ( 1 — 4 ⋅ 2π ) +

π — 2 , 2 ) = ( π —

2 +

π — 2 , 2 ) = (π, 2)

37. The graph of g is a vertical stretch by a factor of 2 followed

by a translation π — 2 units right and 1 unit up of the graph of f.

38. The graph of g is a vertical stretch by a factor of 3 followed

by a translation π — 4 units left and 2 units down of the graph

of f.

39. The graph of g is a horizontal shrink by a factor of 1 —

3

followed by a translation 3π units left and 5 units down of

the graph of f.

40. The graph of g is a horizontal shrink by a factor of 1 —

6

followed by a translation π units right and 9 units up of the

graph of f.


Chapter 9


vertical shift.


Period: 2π — b =

2π — 1




On y = k: ( π — 2 , 3 ) ; ( 3π —

2 , 3 )

Maximum: (π, 4)

Minimums: (0, 2); (2π , 2)


x

y

1

2

3

4

π π2−1


vertical shift.


Period: 2π — b =

2π — 1




On y = k: (0, −5); (π, −5); (2π, −5)

Maximum: ( 3π — 2 , −4 )

Minimum: ( π — 2 , −6 )


x

y1

π π2−1

−2

−3

−4

−5

−6

−7


vertical shift.


Period: 2π — b =

2π — 1 —

2




On y = k: (0, −2); (2π, −2); (4π, −2)

Maximum: (3π, −1)

Minimum: (π, −3)


x

y1

π π4π3π2−1

−2

−3


vertical shift.


Period: 2π — b =

2π — 2

= π Vertical shift: k = 1



On y = k: ( π — 4 , 1 ) ; ( 5π —

4 , 1 )

Maximum: ( π — 2 , 2 )

Minimums: (0, 0); (π, 0)


x

y

1

2

3

4

π−1


vertical shift.

Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = π

Period: 2π — b =

2π — 1


Step 2 Draw the midline through the graph, y = 4.


On y = k: (0, 4); (π, 4); (2π, 0)

Maximum: ( π — 2 , 5 )

Minimum: ( 3π — 2 , 3 )


x

y

1

2

3

4

5

6

π π2−1


Chapter 9


vertical shift.

Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = −π

Period: 2π — b =

2π — 1


Step 2 Draw the midline through the graph, y = −2.


On y = k: ( π — 2 , −2 ) ; ( 3π —

2 , −2 )

Maximums: (0, −1); (2π, −1)

Minimum: (π, −3)


x

y1

π π2−1

−2

−3

−4


vertical shift.

Amplitude: ∣ a ∣ = ∣ −4 ∣ = 4 Horizontal shift: h = − π — 2

Period: 2π — b =


Step 2 Draw the midline through the graph, y = −1.


On y = k: ( π — 4 , −1 ) ; ( 5π —

4 , −1 )

Maximums: ( 3π — 4 , 3 ) ; ( 11π —

4 , 3 )

Minimum: ( 7π — 4 −5 )


x

y

1

2

3

4

π π2−1

−2

−3

−4

−5

−6


vertical shift.

Amplitude: ∣ a ∣ = ∣ −5 ∣ = 5 Horizontal shift: h = π — 2

Period: 2π — b =

2π — 1


Step 2 Draw the midline through the graph, y = 3.


On y = k: ( π — 2 , 3 ) ; ( 3π —

2 , 3 )

Maximums: (0, 8); (2π, 8)

Minimum: (π, −2)


x

y

1

2

3

4

5

6

7

8

π π2−1

−2

49. A; The amplitude is ∣ a ∣ = ∣ −4 ∣ = 4. The period is

2π — b =

2π — 1

= 2π. The horizontal shift is h = π — 2 . Because the

graph is shifted π — 2 units horizontally, a maximum value

occurs at ( − π — 2 , 4 ) .

50. a. B; The graph of sine has been translated 3 units up.

b. C; The graph of cosine has been translated 3 units down.

c. A; The graph of sine has been shrunk horizontally by a

factor of 1 —

2 then translated

π — 2 units right.

d. D; The graph of cosine has been shrunk horizontally by a

factor of 1 —

2 then translated


51. The transformation is g(x) = 3 sin(x − π) + 2.

52. The transformation is g(x) = cos 2π(x + 3) − 4.

53. The transformation is g(x) = – 1 — 3 cos π x − 1.

54. The transformation is g(x) =– 1 —

2 sin 6(x − 1) – 3 — 2 .

55.

0

1

2

3

4

5

6

7

8

0 10 20 30 40 50 60 70 80 90 θ

h

Angle (degrees)

Hei

gh

t (f

eet)

When θ = 45°, h = −8 cos 45° + 10 ≈ 4.3 feet.


Chapter 9

56. a.

t H L H — L

0 45 11.5 45

— 11.5

2.5 22.5 18 27.5

— 18

5 10 11.5 10

— 11.5

7.5 27.5 5 27.5

— 5

b. When the number of lynx is at the midline and increasing,

the number of hares decreases. Once the number of lynx

reaches a maximum and begins to decrease, the number of

hares decreases to its minimum. At this point, the number

of hares begins to increase to its midline and the number

of lynx starts to decrease until it reaches its minimum. The

number of lynx then starts to increase as the number of

hares reaches its maximum.

57. The days on which the average speed is 10 miles per hour are

days 205 and 328. When the function is graphed with the line

y = 10, the two points of intersection are about (205.5, 10)

and (328.7, 10).

Time (days)

Win

d s

pee

d (

mile

s p

er h

ou

r)

t

s

8

12

16

4

0200 3001000

58. The times when the depth is 7 feet are at midnight and about

12:24 p.m.. When the function is graphed with the line

y = 7, the two points of intersection are (0, 7) and (12.4, 7).

Time (hours)

Dep

th (

feet

)

t

d

30

45

60

15

012 1860

59. a. 2cos π − 2cos 0 ——

π − 0 =

−2 − 2 —

π

≈ −1.27

The average rate of change is about −1.27.

b. 1 − (−1) —

π − 0 =

2 — π

≈ 0.64

The average rate of change is about 0.64.

c. 1 − (−1) —

π − 0 ≈

2 —

π

≈ 0.64

The average rate of change is about 0.64.

60. a. x −2π − 3π —

2 −π − π —

2 0

y = sin(−x) 0 −1 0 1 0

y = cos(−x) 1 0 −1 0 1

x π — 2 π 3π —

2 2π

y = sin(−x) −1 0 1 0

y = cos(−x) 0 −1 0 1

b.

x

y1

π2

−1

y = sin(−x) y = cos(−x)

c. Both graphs are refl ections across the y-axis. Because

cosine is an even function, it is symmetrical across the

y-axis and the graph of y = cos(−x) is the same as the

graph of y = cos (x).

61. a.

0

20

40

60

80

100

120

140

160

180

0 40 80 120 160 t

h

Time (seconds)

Hei

gh

t (f

eet)

b. The Ferris wheel goes through 4.5 cycles in 180 seconds.

c. The maximum height is 175 feet and the minimum height

is 5 feet.


Chapter 9

62. a. The graph represents a function of the form

f(x) = a cos bx because the y-coordinate is 5 when x = 0.

b. The maximum value is 5, the minimum value is −5, the

period is π, and the amplitude is 5.

63. Sample answer: The x-intercepts can be represented by

the expression (2n + 1) π — 4 , where n is an integer. So, the

x-intercepts occur when x = ± π — 4 , ±

3π — 4 , ± 5π —

4 , . . . .

64. Your friend is incorrect. The value of a indicates a vertical

stretch or a vertical shrink and changes the amplitude of the

graph. It does not affect the x-intercepts of the function. The

value of b indicates a horizontal stretch or a horizontal shrink

and changes the period of the graph, which is the horizontal

length of each cycle. So, only the value of b affects the

x-intercepts of the function.

65. The graph of g(x) = cos x is a translation π — 2 units to the right

of the graph of f(x) = sin x.

66. The form is y = sin x + cos 2x, where b1 = 1 and b2 = 2.

67. The period is 2π — b =

2π — 8π — 3

= 3 —

4 , so there is one heartbeat every

3 —

4 of a second. Thus, the pulse rate is

60 —

3 —

4

= 80 heartbeats

per minute.

68. a. The initial displacement is when t = 0. So, the initial

displacement is y = 0.2 cos (6.0) = 0.2 feet. The period

is 2π — b =

2π — 6 =

π — 3 .

t

y

0.1

0.2

−0.1

−0.2

3π

6π

b.

t

y

0.1

0.2

−0.1

−0.2

3π

4π

12π

The motion becomes imperceptible as the displacement

rapidly decreases toward 0.


69. x2 + x − 6

— x + 3

= (x + 3)(x − 2)

—— x + 3

= x − 2 , x ≠ −3

70. x3 − 2x2 − 24x —— x2 – 2x − 24

= x (x + 4)(x − 6)

—— (x + 4)(x − 6)

= x , x ≠ −4, 6

71. x2 − 4x − 5

— x2 + 4x − 5

= (x − 5)(x + 1)

—— (x + 5)(x − 1)

72. x2 − 16

— x2 + x − 20

= (x + 4)(x − 4)

—— (x + 5)(x − 4)

= x + 4

— x + 5

, x ≠ 4

73. The least common multiple of 2x and 2(x − 5) is 2x (x − 5).

74. The least common multiple of (x2 − 4) = (x + 2)(x − 2) and

(x + 2) is (x + 2)(x − 2).

75. The least common multiple of x2 + 8x + 12 = (x + 6)(x + 2)

and x + 6 is (x + 6)(x + 2).

9.1−9.4 What Did You Learn? (p. 495)

1. Sample answer: The water from the sprinkler will travel the

maximum horizontal distance when the angle is 45°. If the

sprinkler is tilted up n degrees, the water will travel the same

horizontal distance when the sprinkler is tilted down n degrees.

2. Sample answer: The hare is the prey of the lynx. When the

hare population fl ourishes, there is more food for the lynx

and that population fl ourishes as well. As the lynx population

increases, the hare population will begin to decrease because

of the increase in predators and vice versa.

9.1–9.4 Quiz (p. 496)



has length 7.

adj. = 3 5

27

θ


Pythagorean Theorem, the length of the other side is

adj. = √—

72 − 22 = 3 √—

5 .


functions. Because sin θ = 2 —

7 , csc θ =

hyp. —

opp. =

7 —

2 .

The other values are:

cos θ = adj. —

hyp. =

3 √—

5 —

7 tan θ =

opp. —

adj. =

2 √—

5 —

15

sec θ = hyp.

— adj.

= 7 √

— 5 —

15 cot θ =

adj. —

opp. =

3 √—

5 —

2


Chapter 9

2 . Write an equation using a trigonometric function that


sec 60° = hyp.

— adj.

2 = x —

8

16 = x




tan 30° = opp.

— adj.

√

— 3 —

3 =

x —

12

4 √—

3 = x

The length of the side is x = 4 √—

3 .



cos 49° = adj. —

hyp.

cos 49° = x —

27

27 cos 49° = x

17.7 ≈ x

The length of the side is x ≈ 17.7.

5. The terminal side is 40° counterclockwise from the positive

x-axis.

x

y

40°

40° + 360° = 400°

40° − 360° = −320°

6. Because 5π — 6 is

π — 3 more than

π — 2 , the terminal side is

π — 3

counterclockwise past the positive y-axis.

x

y

56π

5π — 6 + 2π =

17π — 6

5π — 6 − 2π = − 7π —

6

7. Because –960° is negative and 60° less than –900°, the

terminal side is 60° clockwise past the negative x-axis.

x

y

−960°

−960° + 3 ⋅ 360° = 120°

−960° + 2 ⋅ 360° = −240°

8. 3π — 10

= 3π — 10

radians ( 180 degrees —

π radians ) = 54°

9. −60° = −60 degrees ( π radians —

180 degrees )

= − π — 3


180 degrees )

= 2π — 5


Chapter 9


r = √—

x2 + y2

= √——

(−2)2 + (−6)2

= √—

40

= 2 √—

10

Using x = −2, y = −6, and r = 2 √—



sin θ = y —

r =

−6 —

2 √—

10 = − 6 √

— 10 —

20 = − 3 √

— 10 —

10

cos θ = x —

r =

−2 —

2 √—

10 = − 2 √

— 10 —

20 = − √

— 10 —

10

tan θ = y —

x = −6

— −2 = 3

csc θ = r —

y = 2 √

— 10 — −6

= − √—

10 —

3

sec θ = r —

x = 2 √

— 10 — −2

= − √—

10

cot θ = x —

y = −2

— −6 = 1 —

3

12. sin θ = 1 csc θ = 1

cos θ = 0 sec θ = undefi ned

tan θ = undefi ned cot θ = 0

13. sin θ = √

— 3 —

2 csc θ =

2 √—

3 —

2

cos θ = − 1 — 2 sec θ = −2

tan θ = − √—

3 cot θ = − √—

3 —

3


and b = 1. So, the amplitude is a = 3 and the period

2π — b =

2π — 1

= 2π.

Intercepts: (0, 0); (π, 0); (2π, 0)

Maximum: ( π — 2 , 3 )

Minimum: ( 3π — 2 , −3 )

x

y

1

2

3

−1

−2

−3

3π2

π2


graph of f.


vertical shift.


Period: 2π — b =

2π — 5π

= 2 —

5 Vertical shift: k = 3



On y = k: ( 1 — 10

, 3 ) ; ( 3 — 10

, 3 ) Maximums: (0, 4), ( 2 —

5 , 4 )

Minimum: ( 1 — 5 , 2 )


x

y

1

2

3

4

−115

25

The graph of g is a horizontal shrink by a factor

of 1 —

5π followed by a translation 3 units up of the

graph of f.

16. a. sin 70° = opp.

— 400

400 sin 70° = opp.

376 ≈ opp.

The total height of the kite is about 380 feet, including the

reel’s height above the ground.

b. Make the following diagram, using the height from part (a).

70°85°

376 ft

x y

tan 70° = 376

— x tan 85° =

376 —

y

x = 376 —

tan 70° y =

376 —

tan 85°

x ≈ 137 y ≈ 33

You and your friend are standing x + y feet apart, or about

137 + 33 = 170 feet.


Chapter 9

17. The time from 7:00 p.m. to 8:55 p.m. is 1 hour and

55 minutes or 23

— 12

hours. So, there are 23

— 12

revolutions,

which is 23

— 12

⋅ 2π = 23π —

6 radians. While you are seated, you

have rotated a distance of about (47.25) ( 23π — 6 ) ≈ 569 feet.

A person 5 feet from the window will have rotated a distance

of about (47.25 − 5) ( 23π — 6 ) ≈ 509 feet.


1. a. x – π —

2 – π —

3 – π —

4 – π —

6 0

y = tan x undefi ned – √—

3 –1 – √—

3 —

3 0

x 2π — 3

3π — 4

5π — 6 π

7π — 6

y = tan x – √—

3 –1 – √—

3 —

3 0

√—

3 —

3

x π —

6 π —

4 π —

3 π —

2

y = tan x √

— 3 —

3 1 √

— 3 undefi ned

x 5π — 4

4π — 3

3π — 2

5π — 3

y = tan x 1 √—

3 undefi ned – √—

3

b.

x

y

4

2

1

3

5

6

−4

−3

−5

−6

−2

22−π π 74ππ

c. The asymptotes are at x = – π — 2 ,

π — 2 , –

3π — 2 . The x-intercepts

are x = 0, π. The function is increasing when – π — 2 < x <

π — 2

and π — 2 < x <

3π — 2 . The function is an odd function.

2. Sample answer: The tangent function does not have

maximum or minimum values and has a repeating pattern.

The asymptotes occur when x = ± π — 2 , ± 3π —

2 , ±

5π — 2 , . . . .

The x-intercepts occur when x = 0, ± π, ± 2π, ± 3π, . . . .

3. Because the function f (x) = cot x is undefi ned when x = 0

and x = π, the graph of y = cot x has asymptotes at x = 0

and x = π.


1. The function is of the form g(x) = a tan bx, where a = 1 and

b = 2. So, the period is π —

∣ b ∣ =

π — 2 .

Intercept: (0, 0)

Asymptotes: x = π —

2 ∣ b ∣ =

π — 2(2)

, or x = π — 4 ; x = –

π — 2 ∣ b ∣

= – π —

2(2) ,

or x = – π — 4

Halfway points: ( π — 4b

, a ) = ( π — 4(2)

, 1 ) = ( π — 8 , 1 ) ;

( – π — 4b

, –a ) = ( – π — 4(2)

, –1 ) = ( – π — 8 , –1 )

x

y4

2

1

3

−4

−3

−2

4−π4π

8π

−1


2 of the

graph of f (x) = tan x.

2. The function is of the form g(x) = a cot bx, where a = 1 —

3 and


∣ b ∣ =

π — 1 = π.

Intercept: ( π — 2b

, 0 ) = ( π — 2 , 0 )

Asymptotes: x = 0; x = π —

∣ b ∣ =

π — 1 = π


, a ) = ( π — 4(1)

, 1 —

3 ) = ( π —

4 ,

1 —

3 ) ; ( 3π —

4b , –a )

= ( 3π — 4(1)

, − 1 —

3 ) = ( 3π —

4 , – 1 —

3 )

x

y

1

2

3

4

π−1

−2

−3

−4

2π


3 of the

graph of f (x) = cot x.


Chapter 9

3. The function is of the form g(x) = a cot bx, where a = 2 and

b = 4. So, the period is π — b =

π — 4 .


, 0 ) = ( π — 8 , 0 )


∣ b ∣ =

π — 4


, a ) = ( π — 4(4)

, 2 ) = ( π — 16

, 2 ) ; ( 3π — 4b

, –a ) = ( 3π —

4(4) , –2 ) = 3π —

16 , –2

x

y

1

2

3

4

5

−2

−3

−4

−5

−6

4π

8π

−1

The graph of g is a vertical stretch by a factor of 2 and a

horizontal shrink by a factor of 1 —

4 of the graph of f (x) = cot x.


b = π. So, the period is π —

∣ b ∣ = π — π

= 1.

Intercept: (0, 0)


2 ∣ b ∣ =

π — (2π)

, or x = 1 —

2 ; x = –

π —

2 ∣ b ∣ = – π —

(2π) ,

or x = – 1 — 2


, a ) = ( π — 4(π)

, 5 ) = ( 1 — 4 , 5 ) ;

( – π — 4b

, –a ) = ( – π — 4(π)

, –5 ) = ( – 1 —

4 , –5 )

x

y

4

2

6

−6

−4

0.5−0.5

The graph of g is a vertical stretch by a factor of 5 and a

horizontal shrink by a factor of 1 — π of the graph of f (x) = tan x.

5. Step 1 Graph the function y = sin 3x. The period is 2π — 3 .

Step 2 Graph asymptotes of g. Because the asymptotes of g

occur when sin 3x = 0, graph x = 0, x = π — 3 , and

x = 2π — 3 .

Step 3 Plot points on g, such as ( π — 6 , 1 ) and ( π —

2 , –1 ) .

Then use the asymptotes to sketch the curve.

x

y

1

2

3

4

5

−1

−2

−3

−4

−5

−6

23π

3π


3 of the

graph of f (x) = csc x.

6. Step 1 Graph the function y = 1 —

2 cos x. The period is

2π — 1 = 2π.


occur when 1 —

2 cos x = 0, graph x = – π —

2 , x = π —

2 , and

x = 3π — 2 .

Step 3 Plot points on g, such as ( 0, 1 —

2 ) and ( π,

1 —

2 ) .


x

y

1

2

3

4

−2

−3

−4

2−π 32π

2π

−1


2 of the

graph of f (x) = sec x.


Chapter 9

7. Step 1 Graph the function y = 2 sin 2x. The period is 2π — 2 = π.


occur when 2 sin 2x = 0, graph x = 0, x = π — 2 , and

x = π.

Step 3 Plot points on g, such as ( π — 4 , 2 ) and ( 3π —

4 , –2 ) .


2π x

y

2

4

π−2

−4

−6


2 and a

vertical stretch by a factor of 2 of the graph of f (x) = csc x.

8. Step 1 Graph the function y = 2 cos πx. The period is

2π — π = 2.


occur when 2 cos πx = 0, graph x = – 1 — 2 , x =

1 —

2 , and

x = 3 —

2 .

Step 3 Plot points on g, such as (0, 2) and (1, −2). Then

use the asymptotes to sketch the curve.

x

y

4

6

8

−4

−6

−8

−20.5 1 1.5−.5

The graph of g is a horizontal shrink by a factor of 1 — π and a

vertical stretch by a factor of 2 of the graph of f (x) = sec x.



1. The graphs of the tangent, cotangent, secant, and cosecant

functions have no amplitude because the graphs do not have

minimum and maximum values.

2. The cosecant and cotangent functions are undefi ned for

x-values at which sin x = 0.

3. The period of the function y = sec x is 2π, and the period of

y = cot x is π.

4. To graph y = a sec bx, fi rst graph y = a cos bx. Use the

asymptotes and several points of y = a sec bx to graph the

function.




∣ b ∣ =

π — 1 = π.

Intercept: (0, 0)


2 ∣ b ∣ =

π — 2(1)

, or x = π — 2 ;

x = − π —

2 ∣ b ∣ = −

π — 2(1)

, or x = − π — 2


, a ) = ( π — 4(1)

, 2 ) = ( π — 4 , 2 ) ;

( − π — 4b

, −a ) = ( − π —

4(1) , −2 ) = ( −

π — 4 , −2 )

2π

4π

2−π4−π x

y4

2

1

3

−4

−3

−2





∣ b ∣ =

π — 1 = π.

Intercept: (0, 0)


2 ∣ b ∣ =

π — 2(1)

, or x = π — 2 ;

x = − π —

2 ∣ b ∣ = −

π — 2(1)

, or x = − π — 2


, a ) = ( π — 4(1)

, 3 ) = ( π — 4

, 3 ) ;

( − π — 4b

, −a ) = ( − π —

4(1) , −3 ) = ( −

π — 4 , −3 )

4π

4−π2π

2−π x

y4

2

1

3

−4

−3

−2




Chapter 9



∣ b ∣ =

π — 3 .


, 0 ) = ( π — 2(3)

, 0 ) = ( π — 6 , 0 )


∣ b ∣ =

π — 3


, a ) = ( π — 4(3)

, 1 ) = ( π — 12

, 1 ) ;

( 3π — 4b

, –a ) = ( 3π — 4(3)

, –1 ) = ( π — 4 , –1 )

3π x

y

1

2

3

−1

−2

−3

−4


3 of the


8. The function is of the form g (x) = a cot bx, where a = 1 and


∣ b ∣ =

π — 2 .


, 0 ) = ( π — 2(2)

, 0 ) = ( π — 4 , 0 )

Asymptotes: x = 0; x = π — b =

π — 2


, a ) = ( π — 4(2)

, 1 ) = ( π — 8 , 1 ) ;

( 3π — 4b

, −a ) = ( 3π — 4(2)

, −1 ) = ( 3π — 8 , −1 )

4π

2π x

y

1

2

3

−1

−2

−3

−4


2 of the



b = 1 —

4 . So, the period is

π — ∣ b ∣

= π — 1 —

4

= 4π.


, 0 ) = ( π —

2 ( 1 — 4 ) , 0 ) = (2π, 0)


∣ b ∣ =

π — 1 —

4

= 4π


, a ) = ( π —

4 ( 1 — 4 ) , 3 ) = (π, 3);

( 3π — 4b

, −a ) = ( 3π —

4 ( 1 — 4 ) , −3 ) = (3π,−3)

32π

2π

x

y

2

4

6

8

10

−2

−4

−6

−8

−10

−12

π π2 π3 72π π4

The graph of g is a horizontal stretch by a factor of 4 and a

vertical stretch by a factor of 3 of the graph of f (x) = cot x.

10. The function is of the form g(x) = a cot bx, where a = 4

and b = 1 —


π — ∣ b ∣

= π — 1 —

2

= 2π.


, 0 ) = ( π —

2 ( 1 — 2 ) , 0 ) = ( π, 0 )


∣ b ∣ =

π — 1 —

2

, or x = 2π


, a ) = ( π —

4 ( 1 — 2 ) , 4 ) = ( π —

2 , 4 ) ;

( 3π — 4b

, −a ) = ( 3π —

4 ( 1 — 2 ) , −4 ) = ( 3π —

2 , −4 )

32π

2π x

y

4

8

−4

−8

−12

π π2

The graph of g is a horizontal stretch by a factor of 2 and a

vertical stretch by a factor of 4 of the graph of f (x) = cot x.


Chapter 9

11. The function is of the form g(x) = a tan bx, where a = 1 —

2 and

b = π. So, the period is π —

∣ b ∣ =

π — π

= 1.

Intercept: (0, 0)


2 ∣ b ∣ =

π — 2(π)

, or x = 1 —

2 ;

x = − π —

2 ∣ b ∣ = − π —

2(π) , or x = − 1 —

2


, a ) = ( π — 4(π)

, 1 —

2 ) = ( 1 —

4 ,

1 —

2 ) ;

( − π — 4b

, −a ) = ( − π — 4(π)

, − 1 — 2 ) = ( − 1 —

4 , − 1 —

2 )

4−π2π

2−π x

y4

2

1

3



2 of the graph of f (x) = tan x.

12. The function is of the form g(x) = a tan bx, where a = 1 —

3 and

b = 2π. So, the period is π —

∣ b ∣ =

π — 2π

= 1 —

2 .

Intercept: (0, 0)


2 ∣ b ∣ =

π — 2(2π)

, or x = 1 —

4 ;

x = − π —

2 ∣ b ∣ = − π —

2(2π) , or x = −

1 —

4


, a ) = ( π — 4(2π)

, 1 —

3 ) = ( 1 —

8 ,

1 —

3 )

( − π — 4b

, −a ) = ( − π — 4(2π)

, − 1 — 3 ) = ( − 1 —

8 , − 1 —

3 )

x

y

2

1

−2

−1

−0.25 0.25


2π and a


3 of the graph of f (x) = tan x.

13. To fi nd the period, use the expression π —

∣ b ∣ .

Period: π —

∣ b ∣ =

π — 3

14. The horizontal and vertical shrink factors are switched.

A vertical stretch by a factor of 2 and a horizontal shrink

by a factor of 1 —

5 .

15. a.

x

y

4

2

1

5

6

−4

−3

−5

−6

−2

4−π2π

4π

f(x) = 3 sec 2x

y = 3 cos 2x

b.

x

y

2

4

6

−2

−4

−6

−8

6π

3π

2π

f(x) = 4 csc 3x

y = 4 sin 3x

16. A; The asymptotes of the graph are x = π —

2 ∣ b ∣ =

π — 8 and

x = − π —

2 ∣ b ∣ = −

1 —

8 .

17. Step 1 Graph the function y = 3 sin x. The period is

2π — 1 = 2π.

Step 2 Graph asymptotes of g. Because the asymptotes of

g occur when 3 sin x = 0, graph x = 0, x = π, and

x = 2π.


2 , −3 ) .


x

y

2

4

6

8

10

π π2−2

−4

−6

−8

−10

−12

32π

2π




Chapter 9

18. Step 1 Graph the function y = 2 sin x. The period is 2π — 1 = 2π.


g occur when 2 sin x = 0, graph x = 0, x = π, and

x = 2x.


2 , −2 ) .


x

y

2

4

6

8

π π2−2

−4

−6

−8

−10

32π

2π



19. Step 1 Graph the function y = cos 4x. The period is 2π — 4 =

π — 2 .


g occur when cos 4x = 0, graph x = − π — 8 , x =

π — 8 ,

and x = 3π — 8 .

Step 3 Plot points on g, such as (0, 1) and ( π — 9 , −1 ) .


8−π 38π

8π

4π x

y4

2

3

−4

−3

−2

−1


4 of the


20. Step 1 Graph the function y = cos 3x. The period is 2π — 3 .


occur when cos 3x = 0, graph x = − π — 6 , x =

π — 6 , and

x = π — 2 .

Step 3 Plot points on g, such as (0, 1) and ( π — 3 , −1 ) .


6−π6π

3π

2π x

y4

2

3

−4

−3

−2

−1


3 of the



2 cos π x. The period

is 2π — π

= 2.


g occur when 1 —

2 cos π x = 0, graph x = − 1 —

2 , x =

1 —

2 ,

and x = 3 —

2 .


2 ) and ( 1, − 1 —

2 ) .


x

y4

2

3

−4

−3

−2

−10.5 1.5−0.5


π and a


2 of the graph of f (x) = sec x.


Chapter 9


4 cos 2π x. The period is

2π — 2π

= 1.


occur when 1 —

4 cos 2 π x = 0, graph x = − 1 —

4 , x =

1 —

4 ,

and x = 3 —

4 .


4 ) and ( 1 —

2 , − 1 —

4 ) .


x

y

2

1

−2

−1

0.25 0.75−0.25


2π and a


4 of the graph of f (x) = sec x.

23. Step 1 Graph the function y = sin π — 2 x. The period is

2π — π — 2 = 4.


g occur when sin π — 2 x = 0, graph x = 0, x = 2, and

x = 4.



x

y

2

1

3

−4

−3

−2

−121 3 4

The graph of g is a horizontal stretch by a factor of 2 — π of the


24. Step 1 Graph the function y = sin π — 4 x. The period is

2π — π — 4 = 8.

Step 2 Graph the asymptotes of g. Because the asymptotes

of g occur when sin π — 4 x = 0, graph x = 0, x = 4,

and x = 8.



x

y

2

1

3

−4

−3

−2

−1421 3 5 6 87

The graph of g is a horizontal stretch by a factor of 4 — π of the


25. The period is π, so b = 1, and 6 = a tan π — 4 , so a = 6. Thus,

y = 6 tan x.

26. The period is π, so b = 1, and 1 —

2 = a tan

π — 4

, so a = 1 —

2 . Thus,

y = 1 —

2 tan x.

27. The period is 1, so b = π, and 2 = a tan ( π ⋅ 1 —

4 ) , so a = 2.

Thus, y = 2 tan π x.

28. The period is π — 2 , so b = 2, and 5 = a tan ( 2 ⋅

π — 8 ) , so a = 5.

Thus, y = 5 tan 2x.

29. B; The parent function is the tangent function and the graph

has an asymptote at x = π — 2 .

30. C; The parent function is the cotangent function and the

graph has an asymptote at x = 0.

31. D; The parent function is the cosecant function and the graph

has an asymptote at x = 1.

32. F; The parent function is the secant function and the graph

has an asymptote at x = 1 —

2 .

33. A; The parent function is the secant function and the graph


34. E; The parent function is the cosecant function and the graph


35. The tangent function that passes through the origin and has

asymptotes at x = π and x = −π can be stretched or shrunk

vertically to create more tangent functions with the same

characteristics.


Chapter 9

36. a.

x

y

2

1

3

5

6

7

8

−4

−3

−2

4−π2−π

4π

2π 3

4π 5

4π 3

2ππ

The graph of g is a translation 3 units up of the graph of

f (x) = sec x.

b.

54π 7

4π3

2π x

y

1

2

3

π π2−1

−2

−3

−4

−5

−6

The graph of g is a translation 2 units down of the graph

of f (x) = csc x.

c.

x

y

1

2

3

4

π−1

−2

−3

−4

4π

2π

The graph of g is a translation π units right of the graph of

f (x) = cot x.

d.

x

y4

2

1

3

−4

−3

−2

−12−π4−π

2π

The graph of g is a refl ection across the x-axis of the


37. The rule for the transformation is g (x) = cot ( 2x + π — 2 ) + 3.

38. The rule for the transformation is g(x) = 2 tan (3x − π).

39. The rule for the transformation is g(x) = −5 sec (x − π) + 2.

40. The rule for the transformation is g(x) = −8 csc x.

41. Function B has a local maximum value of −5 so Function

A’s local maximum value of − 1 — 4 is greater. Function A has a

local minimum of 1 —

4 so Function B’s local minimum value of

5 is greater.

42. Graph Average Rate of Change

A 1 − (−1)

— π — 4 − ( −

π — 4 )

= 4 —

π

B −1 −1

— π — 4 − ( −

π — 4 )

= − 4 — π

C

1 —

2 − ( −

1 —

2 ) —

π — 4 − ( −

π — 4 )

= 1 —

π

D −2 − 2 —

π — 4 − ( − π —

4 )

= − 8 — π

The order is D, B, C, and A.

43.

0

100

200

300

400

500

600

700

800

0 10 20 30 40 50 60 70 80 90 θ

d

Angle (degrees)

Dis

tan

ce (

ft)

As d increases, θ increases because as the car gets farther

away, the angle required to see the car gets larger.

44.

150

0

320

The viewing window is 0 < t < 15 and 0 < h < 320. The

Statue of Liberty is approximately 305 feet tall so it would

take almost 15 seconds to span the statue.


Chapter 9

45. a. tan θ = 260 − d

— 120

120 tan θ = 260 − d d = 260 − 120 tan θ b.

0

40

80

120

160

200

240

280

0 10 20 30 40 50 60 70 80 90 θ

d

The graph shows a negative correlation meaning that as

the angle gets larger, the distance from your friend to the

top of the building gets smaller. As the angle gets smaller,

the distance from your friend to the top of the building

gets larger.

46. a. tan θ = 200 − d

— 300

300 tan θ = 200 − d

d = 200 − 300 tan θ b.

0

50

100

150

200

250

300

0 10 20 30 40 θ

d

Angle (degrees)

Dis

tan

ce f

rom

to

p

of

clif

f (f

eet)

c. Using the trace feature, your friend is halfway down the

cliff, 100 feet from the top, at about 18.4°.

47. Your friend is incorrect. The graph of cosecant can be

translated π — 2 units right to create the same graph as y = sec x.

48. a. The period is 4.

b. The range is y > 2 and y < −2.

c. The function is of the form y = a csc bx because the

cosecant function has an asymptote at x = 0.

49. a sec bx = a —

cos bx

Because the cosine function is at most 1, y = a cos bx will

produce a maximum when cos bx = 1 and y = a sec bx will

produce a minimum. When cos bx = −1, y = a cos bx will

produce a minimum and y = a sec bx will produce

a maximum.

50.

2

−3

0

3

π

Graphing the function produces the same graph as the

cosecant function with asymptotes at 0, ±π, ±2π, . . .

So, csc x = 1 —

2 ( tan

x —

2 + cot

x —

2 ) .

51. Sample answer: y = 5 tan ( 1 — 2 x −

3π — 4 )


52. Step 1 Use the three x-intercepts to write the function in

factored form. y = a (x + 1)(x − 1)(x − 3)

Step 2 Find the value of a by substituting the coordinates of

the point (0, 3).

3 = a (0 + 1)(x − 1)(x − 3)

3 = 3a

1 = a

The function is y = 3(x + 1)(x − 1)(x − 3)

= x 3 − 3x 2 − x + 3.


factored form.

y = a (x + 2)(x − 1)(x − 3)


the point (0, −6).

−6 = a (0 + 2)(0 − 1)(0 − 3)

−6 = 6a

−1 = a

The function is y = −(x + 2)(x − 1)(x − 3)

= −x 3 + 2 x 2 + 5x − 6.


factored form.

y = a (x + 1)(x − 2)(x − 3)


the point (1, −2).

−2 = a (1 + 1)(1 − 2)(1 − 3)

−2 = 4a

− 1 —

2 = a

The function is y = − 1 —

2 (x + 1)(x − 2)(x − 3)

= − 1 — 2 x 3 + 2 x 2 −

1 —

2 x − 3.


factored form.

y = a (x + 3)(x + 1)(x − 3)


the point (−2, 1).

1 = a (−2 + 3)(−2 + 1)(−2 − 3)

1 = 5a

1 —

5 = a

The function is y = 1 —

5 (x + 3)(x + 1)(x − 3)

= 1 —

5 x 3 +

1 —

5 x 2 −

9 —

5 x −

9 —

5 .


57. The amplitude is 3 and the period is π.


Chapter 9



1. a. The function y = 5 sin 1.38π(x + 0.5) can model the

electric current shown. The amplitude is 5 and the period

is about 1.45.

b. The function y = 20 sin π(x + 0.4) can model the electric

current shown. The amplitude is 20 and the period is 2.

c. The function y = 15 sin 2π(x – 0.4) can model the electric


d. The function y = 15 sin 1 —

2 π(x – 1.2) can model the


is 4.

e. The function y = 12 sin 1.46π (x + 0.2) can model the


is about 1.37.

f. The function y = 3 sin π(x + 3) can model the electric


2. Sample answer: The problems involve oscillating motion or

patterns that repeat in cycles.

3. Answers will vary.


1. Because period and frequency are inversely related to each

other, the period would increase because the frequency is

decreased. The new function would be P = 2 sin 2000π t.

2. Sample answer:

Step 1 Find the maximum and minimum values. From the

graph, the maximum value is 2 and the minimum

value is –2.

Step 2 Identify the vertical shift, k. The value of k is the

mean of the maximum and minimum values.

k = (maximum value) + (minimum value)

——— 2 =

2 + (–2) —

2 =

0 —

2 = 0

Step 3 Decide whether the graph should be modeled by a

sine or cosine function. Because the graph crosses

the y-axis at a maximum value, the graph is a cosine

curve with no horizontal shift. So, h = 0.

Step 4 Find the amplitude and period. The period is

2π — 3 =

2π — b → b = 3.

The amplitude is

∣ a ∣ = (maximum value) – (minimum value) ———

2 =

2 – (–2) —

2 =

4 —

2 = 2.

The graph is not a refl ection, so a > 0. Therefore,

a = 2. The function is y = 2 cos 3x.

3. Sample answer:

Step 1 Find the maximum and minimum values. From the


value is –3.




——— 2 =

1 + (–3) —

2 =

−2 —

2 = −1



the midline y = –1 on the y-axis, the graph is a sine



2 = 2π — b → b = π.

The amplitude is

∣ a ∣ = (maximum value) – (minimum value)

——— 2 =

1 – (–3) —

2 =

4 —

2 = 2.

The graph is not a refl ection, so a > 0. Therefore,

a = 2. The function is y = 2 sin π x – 1.

4. The amplitude changes to 32.5 and the vertical shift becomes

37.5, but the period is not affected.

5. Enter the data in a graphing calculator and make a scatter

plot. The scatter plot appears sinusoidal. So, perform a

sinusoidal regression. Graph the data and the model in the

same viewing window.

120

0

80

The model appears to be a good fi t. So, a model for the data

is T = 21.8 sin (0.514m – 2.18) + 51.3. The period of the

graph represents the amount of time it takes for the weather

to repeat its cycle, which is about 12 months.



1. Graphs of sine and cosine functions are called sinusoids.

2. From the graph, it can be determined that the period is 1 —

3 .

The frequency is the reciprocal of the period, so 3.


Chapter 9


3. The period is 2π.

frequency = 1 —

period

= 1 —

2π

4. The period is 2π — 3 .

frequency = 1 —

period

= 1 —

2π — 3

= 3 —

2π

5. The period is π — 2 .

frequency = 1 —

period

= 1 —

π — 2

= 2 —

π

6. The period is π.

frequency = 1 —

period

= 1 —

π

7. The period is 2 —

3 .

frequency = 1 —

period

= 1

— 2 —

3

= 3 —

2

8. The period is 8.

frequency = 1 —

period

= 1 —

8

9. The period is 8π — 3 .

frequency = 1 —

period

= 1

— 8π — 3

= 3 —

8π

10. The period is 10π.

frequency = 1 —

period

= 1 —

10π

11. Step 1 Find the values of a and b in the model P = a sin bt. The maximum pressure is 0.02, so a = 0.02. Use the

frequency to fi nd b.

frequency = 1 —

period

20 = b —

2π

40π = b

The pressure P as a function of time t is given by

P = 0.02 sin 40π t. Step 2 Graph the model. The amplitude is a = 0.02 and the

period is

1 —

f =

1 —

20 .

The key points are:

Intercepts: (0, 0); (0.025, 0); (0.05, 0)

Maximum: (0.0125, 0.02)

Minimum: (0.0375, –0.02)

x

y

0.01

0.02

−0.01

−0.02

0.01 0.02 0.03 0.04


Chapter 9

12. Step 1 Find the values of a and b in the model P = a sin bt. The maximum pressure is 5, so a = 5. Use the

frequency to fi nd b.

frequency = 1 —

period

440 = b —

2π

880π = b

The pressure P as a function of time t is given by

P = 5 sin 880π t. Step 2 Graph the model. The amplitude is a = 5 and the

period is 1 —

f =

1 —

440 .

The key points are:

Intercepts: (0, 0); ( 1 —

880 , 0 ) ; ( 1

— 440

, 0 ) Maximum: ( 1

— 1760

, 5 ) Minimum: ( 3

— 1760

, −5 )

x

y

4

2

1

3

5

−4

−3

−5

−2

12200

11100

32200

1550

13. Step 1 Find the maximum and minimum values. From the


value is −3.




———— 2

= 3 + (−3)

— 2 =

0 —

2 = 0



the midline y = 0 on the y-axis, the graph is a sine



π = 2π — b → b = 2.

The amplitude is

∣ a ∣ = (maximum value) − (minimum value)

———— 2

= 3 + (−3)

— 2 =

6 —

2 = 3.

The graph is not a refl ection, so a > 0. Therefore, a = 3.

The function is y = 3 sin 2x.



value is −5.




———— 2

= 5 + (−5)

— 2 =

0 —

2 = 0



the y-axis at a maximum value, the graph is a cosine



π — 2 =

2π — b → b = 4.

The amplitude is

∣ a ∣ = (maximum value) – (minimum value)

———— 2

= 5 – (−5)

— 2 =

10 —

2 = 5.


The function is y = 5 cos 4x.



value is −2.

Step 2 Identify the vertical shift, k. The value k is the mean

of the maximum and minimum values.


———— 2

= 2 + (−2)

— 2 =

0 —

2 = 0.

Step 3 Decide whether the graph should be modeled by

a sine or cosine function. Because the graph has a

minimum value on the y-axis, the graph is a cosine

curve with a horizontal shift 4 units left.

So, h = −4.


4 = 2π — b → b =

π — 2 .

The amplitude is


———— 2

= 2 − (−2)

— 2 =

4 —

2 = 2.

The graph is a refl ection, so a < 0. Therefore, a = −2.

The function is y = −2 cos π — 2 (x + 4).


Chapter 9


graph, the maximum value is −1 and the minimum

value is −3.

Step 2 Identify the vertical shift, k. The value k is the mean

of the maximum and minimum values.


———— 2

= −1 + (−3)

— 2 =

−4 —

2 = −2



the midline y = −2 on the y-axis, the graph is a sine



2 = 2π — b → b = π.

The amplitude is


———— 2

= −1 − (−3)

— 2 =

2 —

2 = 1.


The function is y = −sinπ x − 2.

17. To fi nd the amplitude, take half of the difference between the

maximum and the minimum.

10 − (−6)

— 2 =

16 —

2 = 8

18. To fi nd the vertical shift, use the y-coordinates of the points.

−2 + (−8)

— 2 =

−10 —

2 = −5

19. Step 1 Identify the maximum and minimum values. The

maximum height is 9 feet. The minimum height is

4 feet.

Step 2 Identify the vertical shift, k.


———— 2

= 9 + 4

— 2 =

13 —

2 = 6.5

Step 3 Decide whether the height should be modeled by a

sine or cosine function. When t = 0, the height is at

its minimum. So, use a cosine function whose graph

is a refl ection in the x-axis with no horizontal shift

(h = 0).

Step 4 Find the amplitude and period.

The amplitude is


———— 2

= 9 − 4

— 2 = 2.5.

Because the graph is a refl ection in the x-axis, a < 0.

So, a = −2.5. Because the fl ywheel is rotating at a rate of

1 revolution per 2 seconds, one revolution is completed in

2 seconds. So, the period is 2π — b = 2, and b = π.

A model for the height of the handle is

h = −2.5 cos π t + 6.5.

20. Step 1 Identify the maximum and minimum values.

The maximum height of the bucket is 70.5 feet.

The minimum height is −2 feet.



———— 2

= 70.5 + (−2)

—— 2 = 34.25


sine or cosine function. When t = 0, the height is at its

minimum. So, use a cosine function whose graph is a

refl ection in the x-axis with no horizontal shift (h = 0).


The amplitude is


———— 2

= 70.5 – (−2)

— 2 = 36.25.

Because the graph is a refl ection in the x-axis, a < 0.

So, a = −36.25.

Because the bucket is rotating at a rate of 1 revolution per

24 seconds, one revolution is completed in 24 seconds.

So, the period is 2π — b = 24, and b =

π — 12

.

A model for the height of the bucket is

h = −36.25 cos π —

12t + 34.25.

21. Enter the data in a graphing calculator. Make a scatter plot.

The scatter plot appears sinusoidal. So, perform a sinusoidal

regression. Graph the data and the model in the same

viewing window.

130

0

110


is D = 19.81 sin (0.549t − 2.40) + 79.8. The period of the


to repeat its cycle, which is about 11.4 months.



regression. Graph the data and model in the same viewing

window.

130

0

100


is D = 7.38 sin (0.498t − 2.05) + 78.6. The period of the


to repeat its cycle, which is about 12.6 months.


Chapter 9



value is −100.




———— 2

= 100 + (−100)

—— 2 =

0 —

2 = 0


sine or cosine curve. Because the graph crosses the

midline y = 0 on the y-axis, the graph is a sine curve

with no horizontal shift. So, h = 0.


1 —

2 =

2π —

b → b = 4π.

The amplitude is


———— 2

= 100 − (−100)

—— 2 =

200 —

2 = 100.


The function is V = 100 sin 4π t.

24. Louisville has the greater average daily temperature. The

graph of the average daily temperature for Louisville is

always higher than the one for Lexington.

t

y

40

20

10

30

50

60

70

80

421 3 5 6 7 8 9 10 11

Louisville

Lexington

25. a. Using sinusoidal regression on a calculator, the data can

be modeled by N = 3.68 sin (0.776t − 0.70) + 20.4.

b. Using the model, when t = 12,

N = 3.68 sin (0.776(12) − 0.70) + 20.4 ≈ 23.1. So, there

are about 23,100 employees in the 12th year.

26.

x

y2

−2

−1

4 631−1−3−6

(−2π, 1)

(−π, −1) (π, −1)

(2π, 1)(0, 1)

( , 0)π2π(− , 0)π

2π

( , 0)32π(− , 0)3

2π

The slope of the graph of y = sin x is given by the function

y = cos x.

27. a. and b. Use a cosine function because it does not require

determining a horizontal shift.

c. Use a sine function because it does not require

determining a horizontal shift.

28. The frequency is 2 because the graph completes 2 full cycles

in 1 unit of x.

29. The vertical shift is k = 8 + 3

— 2 =

11 —

2 = 5.5. The period is

π — 2 =

2π — b ⇒ b = 4. The amplitude is ∣ a ∣ =

8 − 3 —

2 =

5 —

2 = 2.5.

The midline is 5.5. For the sine function, the graph is

shifted π — 8 units right, and for the cosine function, the graph is

a refl ection. So, the sine function is y = 2.5 sin 4 ( x − π — 8 ) + 5.5

and the cosine function is y = −2.5 cos 4x + 5.5.

30. Your friend is incorrect. The period is the reciprocal of

the frequency. The reciprocal of 1 —

2 , 2, is greater than the

reciprocal of 2, 1 —

2 .

31. a. The vertical shift is 16.5 + 3.5

— 2 = 10, so k = 10.

The period is 12, so b = π — 6 . The amplitude is

16.5 – 3.5

— 2 = 6.5, so a = −6.5. Therefore, the model is

d = −6.5 cos π — 6 t + 10.

b.

Time (hours)

Dep

th (

feet

)

t

d

8

12

16

4

012 1860

Low tide occurs at 12:00 a.m. (t = 0) and 12:00 p.m.

(t = 12), and high tide occurs at 6:00 a.m. (t = 6) and

6:00 p.m. (t = 18).

c. The graphs are related by a horizontal shift to the left by

3 units.


32. 17

— √

— 2 =

17 —

√—

2 ⋅

√—

2 —

√—

2

= 17 √

— 2 —

( √—

2 ) 2

= 17 √

— 2 —

2


Chapter 9

33. 3 —

√—

6 − 2 =

3 —

√—

6 − 2 ⋅

√—

6 + 2 —

√—

6 + 2

= 3 ( √

— 6 + 2 ) —

√—

6 2 − 22

= 6 + 3 √

— 6 —

2

34. 8 —

√—

10 + 3 =

8 —

√—

10 + 3 ⋅

√—

10 − 3 —

√—

10 − 3

= 8 ( √

— 10 − 3 ) —

√—

10 2 − 32

= −24 + 8 √—

10

35. 13 —

√—

3 + √—

11 =

13 —

√—

3 + √—

11 ⋅

√—

3 − √—

11 —

√—

3 − √—

11

= 13 ( √

— 3 − √

— 11 ) ——

√—

3 2 − √—

11 2

= 13 √

— 11 − 13 √

— 3 ——

8

36. log8 x —

7 = log8 x − log8 7

37. ln 2x = ln 2 + ln x

38. log3 5x3 = log3 5 + log3 x3

= log3 5 + 3 log3 x

39. ln 4x6

— y = ln 4x6 − ln y

= ln 4 + ln x6 − ln y

= ln 4 + 6 ln x − ln y


1. a. By the Pythagorean Theorem, a2 + b2 = c2.

b. The expressions are sin θ = opp.

— hyp.

= b —

c and

cos θ = adj.

— hyp.

= a —

c .

c. a2 + b2 = c2

a2

— c2

+ b2

— c2

= c2

— c2

sin2 θ + cos2 θ = 1

d. Sample answer:

θ sin2 θ cos2 θ sin2 θ + cos2 θ

QI π — 3

3 —

4

1 —

4 1

QII 3π — 4

1 —

2

1 —

2 1

QIII 7π — 6

1 —

4

3 —

4 1

QIV 7π — 4

1 —

2

1 —

2 1

2. a. sin2 θ + cos2 θ = 1

sin2 θ — cos2 θ

+ cos2 θ — cos2 θ =

1 —

cos2 θ

1 + tan2 θ = sec2 θ

b. sin2 θ + cos2 θ = 1

sin2 θ — sin2 θ

+ cos2 θ — sin2 θ

= 1 —

sin2 θ

cot2 θ + 1 = csc2 θ

3. Sample answer: To verify a trigonometric identity,

manipulate one side of an equation until both sides are equal.

4. no; sin θ = cos θ is not true for all values of θ. For example,

sin π — 3 =

√—

3 —

2 and cos

π — 3 =

1 —

2 .

5. Sample answer: From Lesson 9.1, you know that

csc θ = 1 —

sin θ and sec θ = 1 —

cos θ . These are examples of

trigonometric identities.


1. Find sin θ.

sin2 θ + csc2 θ = 1

sin2 θ + ( 1 — 6 ) 2 = 1

sin2 θ = 1 − ( 1 — 6 ) 2

sin2 θ = 35

— 36

sin θ = ± √—

35 —

6

sin θ = √

— 35 —

6

Find the values of the other four trigonometric functions of θ

using values of sin θ and cos θ.

tan θ = sin θ — csc θ

= √

— 35

—

6 —

1 —

6

= √

— 35 cot θ =

cos θ — sin θ

= 1 — 6 —

√

— 35 —

6 =

√—

35 —

35

csc θ = 1 —

sin θ =

1 — √

— 35

— 6 =

6 √—

35 —

35 sec θ =

1 —

cos θ = 1 —

1 — 6 = 6

2. sin x cot x sec x = ( sin x ) ( cos x —

sin x ) ( 1

— cos x

) = 1

3. cos θ − cos θ sin2 θ = cos θ (1 − sin2 θ)

= cos θ (cos2 θ)

= cos3 θ


Chapter 9

4. tan x csc x — sec x

=

( sin x —

cos x ) ( 1

— sin x

) ——

1 —

cos x

=

1 —

cos x —

1 —

cos x

= 1

5. cot(−θ) = 1 —

tan(−θ)

= 1 —

−tan θ

= −cot θ

6. csc2 x(1 − sin2 x) = csc2 x(cos2 x)

= ( 1 —

sin2x ) cos2 x

= cos2 x

— sin2 x

= cot2 x

7. cos x csc x tan x = cos x ⋅ 1 —

sin x ⋅

sin x —

cos x

= 1

8. (tan2 x + 1)(cos2 x − 1) = (sec2 x)(−sin2 x)

= ( 1 —

cos2x ) −sin2x

= − sin2 x

— cos2 x

= −tan2 x



1. A trigonometric equation is true for some values of a

variable but a trigonometric identity is true for all values of

the variable for which both sides of the equation are defi ned.

2. The reciprocal identity for secant can be used to write an

expression in terms of cosine. The negative angle identity can be

used to simplify the expression and then the reciprocal identity

can again be used to write the expression in terms of cosine.


3. Find cos θ.


( 1 — 3 ) 2 + cos2 θ = 1

cos2 θ = 1 − ( 1 — 3 ) 2

cos2 θ = 8 —

9

cos θ = ± 2 √

— 2 —

3

cos θ = 2 √

— 2 —

3


using the values of sine and cosine.

tan θ = sin θ — cos θ

=

1 —

3

—

2 √

— 2 —

3

= 2 —

√—

4 cot θ =

cos θ — sin θ

=

2 √

— 2 —

3

— 1 —

3

= 2 √—

2

csc θ = 1 —

sin θ = 1 —

1 —

3 = 3 sec θ =

1 —

cos θ =

1 —

2 √

— 2 —

3

= 3 √

— 2 —

4

4. Find cos θ.


( −7 —

10 ) 2 + cos2 θ = 1

cos2 θ = 1 − ( −7 —

10 ) 2

cos2 θ = 51

— 100

cos θ = ± √

— 51 —

10

cos θ = − √—

51 — 10




=

− 7 —

10

—

− √

— 51 —

10

= 7 √

— 51 —

51

cot θ = cos θ — sin θ

=

− √

— 51 —

10

—

− 7 —

10

= √

— 51 —

7

csc θ = 1 — sin θ

= 1 —

− 7 —

10

= − 10

— 7

sec θ = 1 —

csc θ =

1 —

− √

— 51 —

10

= − 10 √

— 51 —

51


Chapter 9

5. Find sec θ.


1 + ( − 3 —

7 ) 2 = sec2 θ

58 — 49

= sec2 θ

± √

— 58 —

7 = sec θ

− √—

58 — 7 = sec θ


using the values of tangent and secant.

sin θ = tan θ — sec θ

=

− 3 —

7

—

− √

— 58 —

7

= 3 √

— 58 —

58

cos θ = 1 —

sec θ =

1 —

− √

— 58 —

7

= − 7 √—

58 —

58

cot θ = 1 —

tan θ =

1 —

− 3 —

7

= − 7 — 3

csc θ = sec θ — tan θ

=

− √

— 58 —

7

—

− 3 —

7

= √

— 58 —

3

6. Find csc θ.

1 + cot2 θ = csc2 θ

1 + ( − 2 —

5 ) 2 = csc2 θ

29 — 25

= csc2 θ

± √

— 29 —

5 = csc θ

√—

29 — 5 = csc θ


using the values of cosecant and cotangent.

sin θ = 1 —

csc θ =

1 —

√

— 29 —

5

= 5 √

— 29 —

29

cos θ = cot θ — csc θ

=

− 2 —

5

—

√

— 29 —

5

= − 2 √

— 29 —

29

tan θ = 1 —

cot θ =

1 —

− 2 —

5

= − 5 —

2

sec θ = csc θ — cot θ

=

√

— 29 —

5

—

− 2 —

5

= − √

— 29 —

2

7. Find sin θ.


sin2 θ + ( − 5 —

6 ) 2 = 1

sin2 θ = 1 − ( − 5 —

6 ) 2

sin2 θ = 11

— 36

sin θ = ± √

— 11 —

6

sin θ = − √

— 11 —

6




=

− √

— 11 —

6

—

− 5 —

6

= √

— 11 —

5

cot θ = cos θ — sin θ

=

− 5 —

6

—

− √

— 11 —

6

= 5 √

— 11 —

11

sec θ = 1 —

cos θ =

1 —

− 5 —

6

= − 6 —

5

csc θ = 1 —

sin θ =

1 —

− √

— 11 —

6

= − 6 √

— 11 —

11

8. Find tan θ.


1 + tan2 θ = ( 9 — 4 ) 2

tan2 θ = ( 9 — 4 ) 2 −1

tan2 θ = 65

— 16

tan θ = ± √

— 65 —

4

tan θ = − √—

65 —

4


using the values of tangent and secant.

sin θ = tan θ — sec θ

=

− √

— 65 —

4

— 9 —

4

= − √

— 65 —

9

cos θ = 1 —

sec θ =

1 —

9 —

4

= 4 —

9

cot θ = 1 —

tan θ =

1 —

− √

— 65 —

4

= − 4 √

— 65 —

65

csc θ = sec θ — tan θ

=

9 —

4

—

− √

— 65 —

4

= − 9 √—

65 —

65


Chapter 9

9. Find csc θ.

1 + cot2 θ = csc2 θ 1 + (−3)2 = csc2 θ 10 = csc2 θ ± √

— 10 = csc θ

− √—

10 = csc θ


using the values of cosecant and cotangent.

sin θ = 1 —

csc θ =

1 —

− √—

10 = −

√—

10 —

10

cos θ = cot θ — csc θ

= −3

— − √

— 10 =

3 √—

10 —

10

tan θ = 1 —

cot θ =

1 —

−3 = −

1 —

3

sec θ = csc θ — cot θ

= − √

— 10 —

−3 =

√—

10 —

3

10. Find cot θ.

1 + cot2 θ = csc2 θ

1 + cot2 θ = ( − 5 —

3 ) 2

cot2 θ = ( − 5 —

3 ) 2 −1

cot2 θ = 16

— 9

cot θ = 4 —

3


using the values of cosecant and tangent.

sin θ = 1 —

csc θ =

1 —

− 5 —

3

= − 3 — 5 cos θ =

cot θ — csc θ

=

4 —

3

— −

5 —

3

= − 4 — 5

tan θ = 1 —

cot θ =

1 —

4 —

3

= 3 —

4 sec θ =

csc θ — cot θ

=

− 5 —

3

— 4 —

3

= − 5 — 4

11. sin x cot x = (sin x) ( cos x —

sin x )

= cos x

12. cos θ (1 + tan2 θ) = cos θ(sec2 θ)

= cos θ ( 1 —

cos2 θ )

= 1 —

cos θ

= sec θ

13. sin(−θ) —

cos(−θ) =

−sin θ — cos θ

= −tan θ

14. cos2 x — cot2 x

= cos2 x

—

( cos2 x —

sin2 x )

= sin2 x

15. cos ( π —

2 − x ) —

csc x =

sin x —

1 —

sin x

= sin2 x

16. sin ( π — 2 − θ ) sec θ = cos θ ( 1

— cos θ )

= 1

17. csc2 x − cot2 x —— sin(−x) cot(x)

= 1 ——

sin(−x) cot x

= 1 —

−sin x cot x

= 1 ——

−sin x ( cos x —

sin x )

= − 1 —

cos x

= −sec x

18. cos2 x tan2(−x) − 1 ——

cos2 x =

cos2 x tan2 x − 1 ——

cos2 x

=

cos2 x ( sin2 x —

cos2 x ) − 1 ——

cos2 x

= sin2 x − 1

— cos2 x

= −cos2 x

— cos2 x

= −1


Chapter 9

19. cos ( π —

2 − θ ) —

csc θ + cos2 θ =

sin θ —

1 —

sin θ + cos2 θ

= sin2 θ + cos2 θ

= 1

20. sec x sin x + cos ( π —

2 − x ) ———

1 + sec x =

sec x sin x + sin x ——

1 + sec x

= sin x(sec x + 1)

—— 1 + sec x

= sin x

21. (1 + cos2 θ) was substituted for sin2 θ instead of (1 − cos2 θ).

1 − sin2 θ = 1 − (1 − cos2 θ)

= 1 − 1 + cos2 θ = cos2 θ

22. The identity for tan x involving sin x and cos x is

tan x = sin x

— cos x

.

tan x csc x = sin x

— cos x

⋅ 1 —

sin x

= 1 —

cos x

= sec x

23. sin x csc x = sin x ⋅ 1 —

sin x

= 1

24. tan θ csc θ cos θ = sin θ — cos θ

⋅ 1 —

sin θ ⋅ cos θ

= 1

25. cos ( π — 2 − x ) cot x = sin x ⋅

cos x —

sin x

= cos x

26. sin ( π — 2 − x ) tan x = cos x ⋅

sin x —

cos x

= sin x

27. cos ( π —

2 − θ ) + 1

—— 1 − sin(−θ)

= sin θ + 1

—— 1 − sin(− θ)

= sin θ + 1

—— 1 − (−sin θ)

= sin θ + 1

— 1 + sin θ

= 1

28. sin2(−x) —

tan2 x =

(−sin x)2

— tan2 x

= sin2 x

— tan2 x

= sin2 x ⋅ cot2 x

= sin2 x ⋅ cos2 x

— sin2 x

= cos2 x

29. 1 + cos x — sin x

+ sin x —

1 + cos x =

1 + cos x —

sin x +

sin x(1 − cos x) ——

(1 + cos x)(1 − cos x)

= 1 + cos x

— sin x

+ sin x(1 − cos x)

—— 1 − cos2 x

= 1 + cos x

— sin x


—— sin2 x

= sin x(1 + cos x)

—— sin2 x


—— sin2 x

= sin x(1 + cos x) + sin x(1 − cos x)

——— sin2 x

= sin x(1 + cos x + 1 − cos x)

——— sin2 x

= sin x(2)

— sin2 x

= 2 —

sin x

= 2 csc x

30. sin x —— 1 − cos(−x)

= sin x —

1 − cos x

= sin x(1 + cos x)

—— (1 − cos x)(1 + cos x)

= sin x(1 + cos x)

—— 1 − cos2 x

= sin x(1 + cos x)

—— sin2 x

= 1 + cos x

— sin x

= 1 —

sin x +

cos x —

sin x

= csc x + cot x


Chapter 9

31.

x

y

2

1

3

−2π

( , 1)π2π

(− , −1)π2π

y = csc x

y = sin x

x

y

2

1

3

( , 1)π4π

(− , −1)π4π

y = tan x

y = cot x

−2π

x

y

3

−1

−3

−2

2π−2π

(− , −1)π ( , −1)π

y = sec xy = cos x

The functions sin x, csc x, tan x, and cot x are odd functions,

and cos x and sec x are even functions.

sin(−θ) = −sin θ

csc(−θ) = 1 —

sin(− θ) =

1 —

−sin θ = −csc θ

tan(−θ) = −tan θ cot(−θ) =

1 —

tan(− θ) =

1 —

−tan θ = −cot θ

cos(−θ) = cos θ

sec(−θ) = 1 —

cos(− θ) =

1 —

cos θ = sec θ

32. Just as 1 —

x decreases as x increases, sec θ =

1 —

cos θ decreases as

cos θ increases. This happens on the intervals

π + 2nπ < θ < 3π — 2 + 2nπ and

3π — 2 + 2nπ < θ < 2π + 2nπ,

where n is an integer.

33. Your answers are equivalent.

sec x tan x − sin x = 1 —

cos x ⋅

sin x —

cos x − sin x

= sin x

— cos2 x

− sin x

= sec2 x sin x − sin x

= sin x(sec2 x −1)

= sin x tan x

34. a. The function sin θ is positive because y is positive and

both cos θ and tan θ are negative because x is negative.

b. The angle −θ lies in Quadrant III.

c. The function sin(−θ) is negative because y is negative,

cos (−θ) is negative because x is negative, and tan (−θ) is

positive because x and y are negative.

35. s = h sin (90° − θ)

—— sin θ

s = h cos θ — sin θ

s = h cot θ

36. Sample answer: Multiply both sides by 1 —

cos x , so

sin x —

cos x = 1.

Then tan x = 1 because sin x

— cos x

= tan x.

37. a. u Wcos θ = Wsin θ u cos θ = sin θ

u = sin θ — cos θ

u = tan θ b. As θ increases from 0° to 90°, u starts at 0 and increases

without bound.

38. a. n1 ——

√— cot2 θ 1 + 1 =

n2 ——

√— cot2 θ 2 + 1

n1 —

√— csc2 θ 1 =

n2 —

√—

csc2 θ 2

n1 —

csc θ 1 =

n2 — csc θ 2

n1 sin θ1 = n2 sin θ2

b. When θ1 = 55°, θ2 = 35°, and n2 = 2:

n1 sin 55° = 2 sin 35°

n1 = 2 sin 35°

— sin 55°

n1 ≈ 1.4

c. It would be the case that n1 = n2. This situation could

occur when the mediums have the same composition.

39. You can obtain the graph of y = cos x by refl ecting the

graph of f (x) = sin x in the y-axis and translating it


40. a. ln ∣ sec θ ∣ = ln 1 —

∣ cos θ ∣

= ln ∣ (cos θ)−1 ∣ = −ln ∣ cos θ ∣ b. ln ∣ tan θ ∣ = ln ∣ sin θ —

cos θ ∣

= ln ∣ sin θ ∣ − ln ∣ cos θ ∣


Chapter 9



involves the ratio of x and 11.

cot 45° = adj.

— opp.

1 = x —

11

11 = x



involves the ratio of x and 13

sin 60° = opp.

— hyp.

√

— 3 —

2 =

x —

13

13 √

— 3 —

2 = x

The length of the side is x = 13 √

— 3 —

2 .


involves the ratio of x and 7.

cos 30° = adj.

— hyp.

√

— 3 —

2 =

7 —

x

x √—

3 = 14

x = 14

— √

— 3

x = 14 √

— 3 —

3

The length of the side is x = 14 √

— 3 —

3 .


1. a. Sample answer: For each triangle shown, d is a side that is

opposite the angle a − b and the other two sides are 1 unit

in length. So, the triangles are congruent by the SAS

Congruence Theorem.

b. d = √———

(cos b − cos a) 2 + (sin b − sin a) 2

c. d = √————

(cos(a − b) − 1) 2 + (sin(a − b) − 0) 2

d. √———


= √————

(cos(a − b) − 1) 2 + (sin(a − b) − 0) 2


= (cos(a − b) − 1) 2 + (sin(a − b) − 0) 2

cos2 b − 2 cos a cos b + cos2 a + sin2 b

− 2 sin a sin b + sin2 a = cos2(a − b) − 2 cos(a − b) + 1 + sin2(a − b)

−2 cos a cos b − 2 sin a sin b + 2

= −2 cos(a − b) + 2

cos a cos b + sin a sin b

= cos(a − b)

2. cos(a + b) = cos[a − (−b)]

= cos a cos(−b) + sin a sin(−b)

= cos a cos b + sin a(−sin b)

= cos a cos b − sin a sin b

3. sin(a − b) = cos [ π — 2 − (a − b) ]

= cos [ ( π — 2 − a ) + b ]

= cos ( π — 2 − a ) cos b − sin ( π —

2 − a ) sin b

= sin a cos b − cos a sin b

sin(a + b) = cos [ π — 2 − (a + b) ]

= cos [ ( π — 2 − a ) − b ]

= cos ( π — 2 − a ) cos b + sin ( π —

2 − a ) sin b

= sin a cos b + cos a sin b

4. Sample answer: Rewrite the formula using a sum or

difference formula and simplify.

5. a. sin 75° can be written as sin(45° + 30°) and cos 75° can

be written as cos(45° + 30°) so that the sum formulas can

be applied.

sin 75° = sin(45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

= √

— 6 + √

— 2 —

4

cos 75° = cos(45° + 30°)

= cos 45° cos 30° − sin 45° sin 30°

= √—

6 − √—

2 —

4

b. sin 75° = sin(120° − 45°)

= sin 120° cos 45° − cos 120° sin 45°

= √

— 6 + √

— 2 —

4

cos 75° = cos(120° − 45°)

= cos 120° cos 45° + sin 120° sin 45°

= √

— 6 − √

— 2 —

4

The values use the same.


Chapter 9

9.8 Monitoring Progress (pp. 521–522) 1. sin 105° = sin(60° + 45°)

= sin 60° cos 45° + cos 60° sin 45°

= √

— 3 —

2 ( √

— 2 —

2 ) + 1 —

2 ( √

— 2 —

2 )

= √

— 6 + √

— 2 —

4

2. cos 15° = cos(45° − 30°)

= cos 45° cos 30° + sin 45° sin 30°

= √

— 2 —

2 ( √

— 3 —

2 ) +

√—

2 —

2 ( 1 —

2 )

= √

— 6 + √

— 2 —

4

3. tan 5π — 12

= tan ( π — 6 +

π — 4 )

=

tan π — 6 + tan

π — 4

—— 1 − tan

π — 6 tan

π — 4

=

√

— 3 —

3 + 1

—

1 − √

— 3 —

3 (1)

= 2 + √—

3

4. cos π — 12

= cos ( π — 3 −

π — 4 )

= cos π — 3 cos

π — 4 + sin

π — 3 sin

π — 4

= 1 —

2 ( √

— 2 —

2 ) +

√—

3 —

2 ( √

— 2 —

2 )

= √

— 6 + √

— 2 —

4

5. Find cos a and sin b. Because sin a = 8 —

17 and a is in

Quadrant I, cos a = 15

— 17

, as shown in the fi gure.

178

172 − 82 = 15

ax

y

Because cos b = − 24

— 25 and b is in Quadrant III, sin b = − 7 — 25 ,

as shown in the fi gure.

24

25252 − 242 = 7 b x

y

Use the difference formula for sine to fi nd sin(a − b).

sin(a − b) = sin a cos b − cos a sin b

= 8 — 17

( − 24

— 25 ) − 15

— 17

( − 7 — 25 )

= − 87

— 425

6. sin(x + π) = sin x cos π + cos x sin π = (sin x)(−1) + (cos x)(0)

= −sin x

7. cos(x − 2π) = cos x cos 2π + sin x sin 2π = (cos x)(1) + (sin x)(0)

= cos x

8. tan(x − π) = tan x − tan π ——

1 + tan x tan π

= tan x − 0 ——

1 + (tan x)(0)

= tan x

9. sin ( π — 4 − x ) − sin ( x +

π — 4 ) = 1

sin π — 4 cos x − cos

π — 4 sin x − ( sin x cos

π — 4 + cos x sin

π — 4 ) = 1

√

— 2 —

2 cos x −

√—

2 —

2 sin x −

√—

2 —

2 sin x −

√—

2 —

2 cos x = 1

− √—

2 sin x = 1

sin x = − √

— 2 —

2

In the interval 0 ≤ x < 2π, the solutions are x = 5π — 4 and x =

7π — 4 .



1. The expression is cos 170°.

2. First break 75° into the sum or difference of two angles

whose tangent values are known such as 45° + 30°. Rewrite

the expression using the corresponding sum or difference

formula and evaluate.


3. tan(−15°) = tan(30° − 45°)

= tan 30° − tan 45° ——

1 + tan 30° tan 45°

=

√

— 3 —

3 − 1

—

1 + √

— 3 —

3 (1)

= √—

3 − 2

4. tan 195° = tan(150° + 45°)

= tan 150° + tan 45°

—— 1 − tan 150° tan 45°

= −

√—

3 —

3 + 1

——

1 − ( − √

— 3 —

3 ) (1)

= 2 − √—

3


Chapter 9

5. sin 23π — 12

= sin ( π — 4 +

5π — 3 )

= sin π — 4 cos

5π — 3 + cos

π — 4 sin

5π — 3

= √

— 2 —

2 ( 1 —

2 ) +

√—

2 —

2 ( −

√—

3 —

2 )

= √

— 2 − √

— 6 —

4

6. sin(−165°) = sin(−120° + (−45°))

= sin(−120°) cos(−45°) + cos(−120°) sin(−45°)

= − √

— 3 —

2 ( √

— 2 —

2 ) + ( −

1 —

2 ) ( −

√—

2 —

2 )

= √

— 2 − √

— 6 —

4

7. cos 105° = cos(60° + 45°)

= cos 60° cos 45° − sin 60° sin 45°

= 1 —

2 ( √

— 2 —

2 ) −

√—

3 —

2 ( √

— 2 —

2 )

= √

— 2 − √

— 6 —

4

8. cos 11π — 12

= cos ( π — 4 +

2π — 3 )

= cos π — 4 cos

2π — 3 − sin

π — 4 sin

2π — 3

= √

— 2 —

2 ( −

1 —

2 ) −

√—

2 —

2 ( √

— 3 —

2 )

= − √—

2 − √—

6 —

4

9. tan 17π — 12

= tan ( 3π — 4 +

2π — 3 )

=

tan 3π — 4 + tan

2π — 3

—— 1 − tan

3π — 4 tan

2π — 3

= −1 + ( − √

— 3 ) ——

1 − (−1) ( − √—

3 )

= √—

3 + 2

10. sin ( − 7π — 12

) = sin ( − π — 4 −

π — 3 )

= sin ( − π — 4 ) cos

π — 3 − cos ( −

π — 4 ) sin

π — 3

= ( − √—

2 —

2 ) ( 1 —

2 ) − ( √

— 2 —

2 ) ( √

— 3 —

2 )

= − √—

2 − √—

6 —

4

11. Find sin a and cos b.

Because cos a = 4 —

5 and a is in Quadrant I, sin a =

3 —

5 as

shown in the fi gure.

5

4

52 − 42 = 3

ax

y

Because sin b = − 15

— 17

and b is in Quadrant IV, cos b = 8 —

17 as


1715

172 − 152 = 8b x

y

Use the sum formula for sine to fi nd sin(a + b).

sin(a + b) = sin a cos b + cos a + sin b

= 3 —

5 ( 8 —

17 ) +

4 —

5 ( −

15 —

17 )

= −36

— 85



5 and a is in Quadrant I, sin a = 3 — 5 as shown

in the fi gure.

5

4

52 − 42 = 3

ax

y


— 17 and b is in Quadrant IV, cos b = 8 —

17 as


1715

172 − 152 = 8b x

y

Use the difference formula for sine to fi nd sin(a – b).

sin(a − b) = sin a cos b − cos a sin b

= 3 —

5 ( 8 —

17 ) − ( 4 —

5 ) ( −

15 — 17 )

= 84

— 85


Chapter 9




3 —

5 as shown

in the fi gure.

5

4

52 − 42 = 3

ax

y


— 17 and b is in Quadrant IV, cos b = 8 — 17 as


1715

172 − 152 = 8b x

y

Use the difference formula for cosine to fi nd cos(a − b).

cos(a − b) = cos a cos b + sin a sin b

= 4 — 5 ( 8 —

17 ) +

3 —

5 ( −

15 — 17 )

= − 13

— 85




3 —

5 as shown

in the fi gure.

5

4

52 − 42 = 3

ax

y


— 17 and b is in Quadrant IV, cos b = 8 —

17 as


1715

172 − 152 = 8b x

y

Use the sum formula for cosine to fi nd cos(a + b).

cos(a + b) = cos a cos b − sin a sin b

= ( 4 — 5 ) ( 8 —

17 ) − ( 3 —

5 ) ( −

15 — 17 )

= 77 — 85

15. Find tan a and tan b.


5 and a is in Quadrant I, tan a =

3 —

4 as shown

in the fi gure.

5

4

52 − 42 = 3

ax

y


— 17 and b is in Quadrant IV, tan b = − 15

— 8 as


1715

172 − 152 = 8b x

y

Use the sum formula for tangent to fi nd tan(a + b).

tan(a + b) = tan a + tan b

—— 1 − tan a tan b

= 3 —

4 + ( −

15 —

8 ) ——

1 − ( 3 — 4 ) ( − 15

— 8 )

= − 36

— 77


Chapter 9

16. Find tan a and tan b.


5 and a is in Quadrant I, tan a =

3 —

4 as


5

4

52 − 42 = 3

ax

y


— 17

and b is in Quadrant IV, tan b = − 15

— 8

as shown in the fi gure.

1715

172 − 152 = 8b x

y

Use the difference formula for tangent to fi nd tan(a − b).

tan(a − b) = tan a − tan b

—— 1 + tan a tan b

=

3 —

4 − ( −

15 —

8 ) ——

1 + ( 3 — 4 ) ( −

15 —

8 )

= − 84

— 13

17. tan(x + π) = tan x + tan π ——

1 − tan x tan π

= tan x + 0

—— 1 − (tan x)(0)

= tan x

18. cos ( x − π — 2 ) = cos x cos

π — 2 + sin x sin

π — 2

= (cos x)(0) + (sin x)(1)

= sin x

19. cos(x + 2π) = cos x cos 2π − sin x sin 2π = (cos x)(1) − (sin x)(0)

= cos x

20. tan(x − 2π) = tan x − tan 2π ——

1 + tan x tan 2π

= tan x − 0

—— 1 + (tan x)(0)

= tan x

21. sin ( x − 3π — 2

) = sin x cos 3π — 2 − cos x sin

3π — 2

= (sin x)(0) − (cos x)(−1)

= cos x

22. tan ( x + π — 2 ) =

sin ( x + π — 2 ) —

cos ( x + π — 2 )

=

sin x cos π — 2 + cos x sin

π — 2

——— cos x cos

π — 2 − sin x sin

π — 2

= sin x(0) + cos x (1)

—— cos x (0) − sin x (1)

= cos x

— −sin x

= −cot x

23. The sign in the denominator should be negative when using

the sum formula.

tan ( x + π — 4 ) =

tan x + tan π — 4

—— 1 − tan x tan

π — 4

= tan x + 1 —

1 − tan x

24. The a and b were reversed when the difference formula

was used.

sin ( x − π — 4 ) = sin x cos

π — 4 − cos x sin

π — 4

= (sin x) ( √—

2 —

2 ) − (cos x) ( √

— 2 —

2 )

= √

— 2 —

2 (sin x − cos x)

25. B, D; 2 sin π — 6 − 1 = 0 and 2 sin

5π — 6 − 1 = 0

26. B, D; tan 3π — 4 + 1 = 0 and tan

7π — 4 + 1 = 0

27. sin ( x + π — 2 ) =

1 —

2


π — 2 =

1 —

2

0 sin x + 1 cos x = 1 —

2

cos x = 1 —

2

In the interval 0 ≤ x < 2π, the solutions are x = π — 3 and x = 5π —

3 .

28. tan ( x − π — 4 ) = 0

tan x − tan π — 4

—— 1 + tan x tan

π — 4 = 0

tan x − 1

— 1 + tan x

= 0

tan x − 1 = 0 tan x = 1

In the interval 0 ≤ x < 2π, the solutions are x = π — 4 and x = 5π —

4 .


Chapter 9

29. cos ( x + π — 6 ) − cos ( x − π —

6 ) = 1

cos x cos π — 6 − sin x sin

π — 6 − ( cos x cos

π — 6 + sin x sin

π — 6 ) = 1

√

— 3 —

2 cos x − 1 —

2 sin x − √

— 3 —

2 cos x − 1 —

2 sin x = 1

− sin x = 1 sin x = −1

In the interval 0 ≤ x < 2π, the solution is x = 3π — 2 .

30. sin ( x + π — 4 ) + sin ( x − π —

4 ) = 0


π — 4 + sin x cos

π — 4 − cos x sin

π — 4 = 0

√

— 2 —

2 sin x +

√—

2 —

2 cos x +

√—

2 —

2 sin x −

√—

2 —

2 cos x = 0

√—

2 sin x = 0 sin x = 0

In the interval 0 ≤ x < 2π, the solutions are x = 0 and x = π.

31. tan(x + π) − tan(π − x) = 0

tan x + tan π —— 1− tan x tan π

− tan π − tan x ——

1+ tan π tan x = 0

tan x − (− tan x) = 0 2 tan x = 0 tan x = 0


32. sin(x + π) + cos(x + π) = 0

sin x cos π + cos x sin π + cos x cos π − sin x sin π = 0 −sin x − cos x = 0 −sin x = cos x

In the interval 0 ≤ x < 2π, the solutions are x = 3π — 4 and x =

7π — 4 .

33. sin ( π — 2 − θ ) = sin

π — 2 cos θ − cos

π — 2 sin θ

= (1) cos θ − (0) sin θ = cos θ

34. Your friend is incorrect. The difference formula for tan ( π — 2 − θ )

would require fi nding tan π — 2 , which is undefi ned.

35. When t = 45°:

WQ — NA

= 35 tan(θ − 45°) + 35 tan 45° ———

h tan θ

= 35 ( tan θ − tan 45°

—— 1 + tan θ tan 45°

) + 35

——— h tan θ

= 35 ( tan θ − 1

— 1 + tan θ

) + 35

—— h tan θ

= 35(tan θ − 1) + 35(1 + tan θ) ———

h tan θ(1 + tan θ)

= 35 tan θ − 35 + 35 + 35 tan θ ——— h tan θ(1 + tan θ)

= 70 tan θ —— h tan θ(1 + tan θ)

= 70 ——

h(1 + tan θ)

36. When t = 1:

y = A cos ( 2π — 3 − 2π x

— 5 ) + A cos ( 2π —

3 + 2π x

— 5 )

= A cos 2π — 3 cos 2π x

— 5 + A sin

2π — 3 sin 2π x

— 5

+ A cos 2π — 3 cos

2π x —

5 − A sin

2π — 3 sin 2π x

— 5

= A cos 2π — 3 cos 2π x

— 5 + A cos

2π — 3 cos 2π x

— 5

= 2A cos 2π — 3 cos 2π x

— 5

= 2A ( − 1 —

2 ) cos 2π x

— 5

= −A cos 2π x —

5

37. y1 + y2 = cos 960πt + cos 1240πt

= cos(1100πt − 140πt) + cos(1100πt + 140πt)

= cos 1100πt cos 140πt + sin 1100πt sin 140πt

+ cos 1100πt cos 140πt − sin 1100πt sin 140πt

= cos 1100πt cos 140πt + cos 1100πt cos 140πt

= 2 cos 1100πt cos 140πt

38. Any point where the two graphs intersect is a solution

because if f(x) = g(x), then f(x) − g(x) = 0.


Chapter 9

39. a. First note that tan θ1 = m1 and tan θ2 = m2.

tan(θ2 − θ1) = tan θ2 − tan θ1

—— 1 + tan θ2 tan θ1

= m2 – m1

— 1 + m2 m1

b. The slopes of the lines are m1 = 1 and m2 = 1 —

√—

3 − 2 .

tan(θ2 − θ1) = m2 − m1 — 1 + m2 m1

tan(θ2 − θ1) =

1 —

√—

3 −2 − 1

——

1 + ( 1 —

√—

3 −2 ) (1)

tan(θ2 − θ1) = 1 − ( √

— 3 −2 ) ——

( √—

3 −2 ) + 1

tan(θ2 − θ1) = 3 − √

— 3 —

√—

3 − 1

tan(θ2 − θ1) = ( 3 − √

— 3 ) ( √

— 3 + 1 ) ——

√—

3 2 − 12

tan(θ2 − θ1) = 2 √—

3 —

2

tan(θ2 − θ1) = √—

3

θ2 − θ1 = 60°

So, the acute angle is 60°.

40. a. sin 3x = sin(2x + x)

= sin 2x cos x + cos 2x sin x

= sin(x + x) cos x + cos(x + x) sin x

= (sin x cos x + cos x sin x) cos x + (cos x cos x −

sin x sin x) sin x

= sin x cos2 x + cos2 x sin 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 − 3 sin3 x − sin3 x

= 3 sin x − 4 sin3 x

b. cos 3x = cos(2x + x)

= cos 2x cos x − sin 2x sin x

= cos(x + x) cos x − sin(x + x) sin x

= (cos x cos x − sin x sin x) cos x −

(sin x cos x + cos x sin x) sin x

= (cos2 x − sin2 x) cos x − 2 sin x cos x sin x

= (cos2 x − sin2 x) cos x − 2 sin2 x cos x

= (cos2 x − (1 − cos2 x)) cos x − 2(1 − cos2 x)

cos x

= (2 cos2 x − 1) cos x − 2 (cos x − cos3 x)

= 2 cos3 x − cos x − 2 cos x + 2 cos3 x

= 4 cos3 x − 3 cos x

c. tan 3x = sin 3x —

cos 3x

= 3 sin x − 4 sin3x ——

4 cos3x − 3 cos x

= 3 sin x − 4 sin3x ——

4 cos3x − 3 cos x ⋅

1 —

cos3 x —

1 —

cos3 x

= 3 sin x

— cos3x

− 4 sin3x

— cos3x

——

4cos3x

— cos3x

− 3cos x

— cos3x

= 3 tan x

— cos2x

− 4 tan3x ——

4 − 3 —

cos2x

= 3 sec2x tan x − 4 tan3x

—— 4 − 3 sec2x

= 3(1 + tan2x) tan x − 4 tan3x ———

4 − 3(1 + tan2x)

= 3 tan x + 3 tan3x − 4 tan3x ———

4 − 3 − 3 tan2x

= 3 tan x − tan3x

—— 1 − 3 tan2x


41. 1 − 9 —

x − 2 = −

7 —

2 Check x = 4:

− 9 —

x − 2 = −

9 —

2 1 −

9 —

4 − 2 =

? −

7 —

2

9 —

x − 2 =

9 —

2 1 −

9 —

2 =

? −

7 —

2

2 ⋅ 9 = 9(x − 2) − 7 —

2 = −

7 —

2 ✓

2 = x − 2

4 = x

The solution is x = 4.


Chapter 9

42. 12

— x +

3 —

4 =

8 —

x

4x ( 12 —

x +

3 —

4 ) = 4x ( 8 —

x )

4 ⋅ 12 + 3x = 4 ⋅ 8

48 + 3x = 32

3x = −16

x = − 16

— 3

The solution is x = − 16

— 3 .

43. 2x − 3

— x + 1

= 10 —

x2 − 1 + 5

2x − 3

— x + 1

= 10 ——

( x + 1 ) ( x − 1 ) + 5

(x + 1)(x − 1) ( 2x − 3 —

x + 1 ) = (x + 1)(x − 1) ( 10

—— (x + 1)(x − 1)

+ 5 ) (x − 1)(2x − 3) = 10 + 5 (x + 1)(x − 1)

2x2 − 5x + 3 = 10 + 5x2 − 5

2x2 − 5x + 3 = 5x2 + 5

0 = 3x2 + 5x + 2

0 = (x + 1)(3x + 2)

x = −1, − 2 —

3

Check x = −1:

2(−1) − 3

— −1 + 1

=?

10 —

(−1)2 − 1 + 5

−5

— 0 =

? 10

— 0 + 5

Division by 0 is undefi ned.

Check x = − 2 —

3 :

2 ( −

2 —

3 ) − 3 —

− 2 —

3 + 1

=?

10 —

( − 2 —

3 ) 2 − 1

+ 5

−

4 —

3 − 3 —

1 —

3

=?

10 —

4 —

9 − 1

+ 5

−

13 —

3 —

1 —

3

=? 10

—

− 5 —

9

+ 5

−13 =? −18 + 5

−13 = −13 ✓

The solution is x = − 2 —

3 .

9.5–9.8 What Did You Learn? (p. 525)

1. As the angle at which you are looking down at the car

increases, the distance between you and the car increases.

2. Sample answer: The slope is the ratio of rise to run, and these

form the legs of a right triangle when drawn on the graph.

The tangent is the ratio of the lengths of the legs, so

m = rise

— run

= tan θ.

Chapter 9 Review (pp. 526–530)



has length 11.



opp. = √—

112 − 62 = √—

85 .

opp. = 85

6

11

θ


functions. Because cos θ = 6 —

11 , sec θ =

hyp. —

adj. =

11 —

6 .

The other values are:

sin θ = opp.

— hyp.

= √

— 85 —

11

tan θ = opp.

— adj.

= √

— 85 —

6

csc θ = hyp.

— opp.

= 11

— √

— 85 =

11 √—

85 —

85

cot θ = adj.

— opp.

= 6 —

√—

85 =

6 √—

85 —

85

2. tan 31° = h —

25

25(tan 31°) = h

15 = h


3. 382° − 360° = 22° 382° − 2 ⋅ 360° = −338°


180 degrees )

= π — 6


180 degrees )

= 5π — 4

Check x = − 16

— 3 :

12

—

− 16

— 3

+ 3 —

4 =

?

8 —

− 16

— 3

−9

— 4 +

3 —

4 =

? −

3 —

2

− 3 —

2 = −

3 —

2 ✓


Chapter 9

6. 3π — 4 =

3π — 4 radians ( 180° —

π radians )

= 135°

7. 5π — 3 =

5π — 3 radians ( 180° —

π radians )

= 300°

8.

140°

x

y

Convert 140° to radians.


180 degrees )

= 7π — 9

Find the area of the sector.

A = 1 —

2 r 2 θ

= 1 —

2 ( 35 ) 2 ( 7π —

9 )

= 8525

— 18

π

≈ 1497

The area is about 1497 square meters.


r = √—

x2 + y2

= √—

02 + 12

= √—

1

= 1

Using x = 0, y = 1, and r = 1, the values of the six


sin θ = y —

r =

1 —

1 = 1 csc θ =

r —

y =

1 —

1 = 1

cos θ = x —

r =

0 —

1 = 0 sec θ =

r —

x =

1 —

0 = undefi ned

tan θ = y —

x =

1 —

0 = undefi ned cot θ =

x —

y =

0 —

1 = 0


r = √—

x2 + y2

= √——

242 + (−7)2

= √—

625

= 25



sin θ = y —

r = −

7 — 25 csc θ =

r —

y = −

25 — 7

cos θ = x —

r = 24

— 25

sec θ = r —

x =

25 —

24

tan θ = y —

x = −

7 — 24 cot θ =

x —

y = −

24 — 7


r = √—

x2 + y2

= √—

( −4 ) 2 + 62

= √—

52

= 2 √—

13

Using x = −4, y = 6, and r = 2 √—



sin θ = y —

r =

3 √—

13 —

13 csc θ =

r —

y =

√—

13 —

3

cos θ = x —

r = −

2 √—

13 —

13 sec θ =

r —

x = −

√—

13 —

2

tan θ = y —

x = −

3 —

2 cot θ =

x —

y = −

2 —

3

12. The reference angle is 360° − 330° = 30°. The tangent

function is negative in Quadrant IV, so

tan 330° = −tan 30° = − √

— 3 —

3 .


angle is 360° − 315° = 45. The secant function is positive in

Quadrant IV, so

sec(−405°) = sec 45° = √—

2 .



π — 6 . The sine function is

positive in Quadrant I, so

sin 13π —

6 = sin

π — 6 =

1 —

2 .



5π — 3 . The reference angle is

2π − 5π — 3 =

π — 3 . The secant function is positive in Quadrant IV,

so sec 11π —

3 = sec

π — 3 = 2.


Chapter 9

16. The function is of the form g(x) = a cos bx, where a = 8 and

b = 1.

So, the amplitude is a = 8 and the period is 2π — b =

2π — 1 = 2π.

Intercepts: ( − π — 2 , 0 ) ; ( π —

2 , 0 ) ; ( 3π —

2 , 0 )

Maximum: (0, 8)

Minimum: (π, −8)

x

y8

2

−8

−6

−4

−π 2ππ


graph of f(x) = cos x.

17. The function is of the form g(x) = a sin bx, where a = 6 and

b = π.

So, the amplitude is a = 6 and the period is 2π — b =

2π — π

= 2.

Intercepts: (0, 0); (1, 0); (2, 0)

Maximum: ( 1 — 2 , 6 )

Minimum: ( 3 — 2 , −6 )

x

y

4

2

6

−6

−4

−21.50.5


π and a

vertical stretch by a factor of 6 of the graph of f(x) = sin x.


4 and

b = 4. So, the amplitude is a = 1 —

4 and the period is

2π — b =

2π — 4 =

π — 2 .

Intercepts: ( π — 8 , 0 ) ; ( 3π —

8 , 0 )


4 ) ; ( π —

2 ,

1 —

4 )

Minimum: ( π — 4 , −

1 —

4 )

x

y

0.25

0.5

−0.5

−0.25

8π

4π

2π


4 and a


4 of the graph of f(x) = cos x.


vertical shift.

Amplitude: a = 1 Horizontal shift: h = −π

Period: 2π — b =




On y = k: ( π — 2 , 2 ) ; ( 3π —

2 , 2 )

Maximum: (π, 3)

Minimums: (0, 2); (2π, 2)


x

y

2

3

4

−π 2ππ−2π−1


Chapter 9


vertical shift.


Period: 2π — b =




On y = k: (0, − 4); (π, − 4); (2π, −4)

Maximum: ( 3π — 2 , −3 )

Minimum: ( π — 2 , −5 )


x

y1

−π 2ππ−2π−1

−2

−4

−5

−6


vertical shift.

Amplitude: a = 2 Horizontal shift: h = − π — 2

Period: 2π — b =




On y = k: ( π — 2 , 0 ) ; ( 3π —

2 , 0 )

Maximums: (0, 2); (2π, 2)

Minimum: (π, –2)


x

y3

−π 2ππ−2π−1

−2

−3


b = 1 —


π — ∣ b ∣

= π — 1 —

2

= 2π.

Intercept: (0, 0)


2 ∣ b ∣ =

π —

2 ( 1 — 2 )

= π ;

x = − π —

2 ∣ b ∣ = −

π —

2 ( 1 — 2 ) = −π


, a ) = ( π —

4 ( 1 — 2 ) , 1 ) = ( π —

2 , 1 )

( − π — 4b

, −a ) = ( − π —

4 ( 1 — 2 ) , −1 ) = ( −

π — 2 , −1 )

2π x

y

1

3

2

4

−π π−1

−2

−3

−4

The graph of g is a horizontal stretch by a factor of 2 of the




∣ b ∣ =

π — 1 = π.

Intercepts: ( π — 2b

, 0 ) = ( π — 2 , 0 )


∣ b ∣ = π — 1 = π


, a ) = ( π — 4(1)

, 2 ) = ( π — 4 , 2 ) ;

( 3π — 4b

, −a ) = ( 3π — 4(1)

, −2 ) = ( 3π — 4 , −2 )

x

y

2

3

1

−4

−2

−1

−3

4π

2π 3

4π π




Chapter 9


b = 3π. So, the period is π —

∣ b ∣ =

π — 3π

= 1 —

3 .

Intercept: (0, 0)


2 ∣ b ∣ =

π — 6π

= 1 —

6 ;

x = − π —

2 ∣ b ∣ = −

π — 6π = −

1 —

6


, a ) = ( π — 4 ( 3π )

, 4 ) = ( 1 — 12

, 4 ) ; ( −

π — 4b

, −a ) = ( − π —

4(3π) , −4 ) = ( − 1 —

12 , −4 )

x

y

16

20

8

4

12

−12

−16

−20

−8

−416

16−


3π and a

vertical stretch by a factor of 4 of the graph of f (x) = tan x.

25. Step 1 Graph the function y = 5 sin x. The period is

2π — 1 = 2π.


of g occur when 5 sin x = 0, graph x = 0, x = π,

and x = 2π.


2 , −6 ) .


x

y

2

4

6

8

π−2

−4

−6

−8

−10

2π 2π3

2π

26. Step 1 Graph the function y = cos 1 —

2 x.

The period is 2π — 1 —

2

= 4π.


of g occur when cos 1 —

2 x = 0, graph x = –π, x = π,

and x = 3π.

Step 3 Plot points on g, such as (0, 1) and (2π, −1). Then


x

y8

4

2

6

−8

−6

−4

−23ππ−π

27. Step 1 Graph the function y = 5 cos π x.

The period is 2π — π

= 2.


of g occur when 5 cosπ x = 0, graph x = − 1 —

2 ,

x = 1 —

2 , and x =

3 —

2 .

Step 3 Plot points on g, such as (0, 5) and (1, −5).


x

y

16

20

8

4

12

−16

−20

−12

−8

−40.5 1.0 1.5−0.5


2 sin

π — 4 x.

The period is 2π — π — 4 = 8.


of g occur when 1 —

2 sin

π — 4 x = 0, graph x = 0, x = 4,

and x = 8.

Step 3 Plot points on g, such as (2, 1) and (6, –1). Then use

the asymptotes to sketch the curve.

x

y

2

1

3

−4

−3

−2

−1421 3 8


Chapter 9



value is −1.




———— 2

= 1 + (−1)

— 2 =

0 —

2 = 0



the midline y = 0 on the y−axis, the graph is a sine



The period is 4π = 2π — b . So, b =

1 —

2 . The amplitude is


———— 2

= 1 − (−1)

— 2 =

2 —

2 = 1.

The graph is a refl ection, so a < 0. So, a = −1. The function

is y = −sin 1 —

2 x.


graph, the maximum value is −1 and the minimum

value is –3.




———— 2

= −1 + (−3)

— 2 =

−4 —

2 = −2



maximum value on the y-axis, the graph is a cosine



The period is 2 = 2π — b . So, b = π. The amplitude is


———— 2

= −1 − (−3)

— 2 =

2 —

2 = 1

The graph is not a refl ection, so a > 0. So, a = 1.

The function is y = cosπ x −2.

31. The refl ector oscillates between 25 inches and 2 inches

above the ground. So, fi nd a sine or cosine function that may

be an appropriate model for the height over time.

Step 1 Identify the maximum and minimum values. The

maximum height is 25 inches. The minimum height

is 2 inches.



———— 2

= 25 + 2

— 2 =

27 —

2 = 13.5


sine or cosine function. When t = 0, the height is at

its minimum. So, use a cosine function whose graph

is a refl ection in the x-axis with no horizontal shift

(h = 0).

Step 4 Find the amplitude and period. The amplitude is


———— 2

= 25 − 2

— 2 = 11.5.

Because the graph is a refl ection in the x−axis,

a < 0. So, a = −11.5. Because the wheel is rotating

at a rate of 1 revolution per second, one revolution

is completed in 1 second. So, the period is 2π — b = 1,

and b = 2π.

A model for the height of the refl ector is

h = −11.5 cos 2π t + 13.5.




viewing window.

130

0

3


is P = 1.08 sin(0.585t − 2.33). The period represents the

amount of time it takes for the precipitation level to complete

one cycle, which is about 10.7 months.

33. cot2x − cot2x cos2x = cot2x (1 − cos2x)

= cot2x (sin2x)

= cos2x

— sin2x

∙ sin2x

= cos2x


Chapter 9

34. (sec x + 1)(sec x − 1) ——

tan x =

sec2x − 12

— tan x

= sec2x − 1

— tan x

= tan2x

— tan x

= tan x

35. sin ( π — 2 − x ) tan x = cos x ⋅ tan x

= cos x ⋅ sin x

— cos x

= sin x

36. cos x sec x — 1 + tan2x

= cos x ⋅ sec x

— sec2 x

= cos x

— sec x

= cos x cos x

= cos2 x

37. tan ( π — 2 − x ) cot x = cot x ⋅ cot x

= cot2 x

= csc2 x − 1

38. sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

= √

— 2 —

2 ( √

— 3 —

2 ) +

√—

2 —

2 ( 1 —

2 )

= √

— 2 + √

— 6 —

4

39. tan ( −15° ) = tan(30° − 45°)

= tan 30° − tan 45° ——

1 + tan 30° tan 45°

=

√

— 3 —

3 − 1

——

1 + ( √—

3 —

3 ) (1)

= √—

3 − 2

40. cos π — 12

= cos ( π — 3 −

π — 4 )

= cos π — 3 cos

π — 4 + sin

π — 3 sin

π — 4

= 1 —

2 ( √

— 2 —

2 ) +

√—

3 —

2 ( √

— 2 —

2 )

= √

— 2 + √

— 6 —

4

41. tan(a + b) = tan a + tan b

—— 1 − tan a tan b

= 1 — 4 +

3 —

7

—

1 − ( 1 — 4 ) ( 3 —

7 )

= 19

— 25

42. cos ( x + 3π — 4 ) + cos ( x −

3π — 4 ) = 1

cos x cos 3π — 4 − sin x sin

3π — 4 + cos x cos

3π — 4 + sin x sin

3π — 4 = 1

− √

— 2 —

2 cos x −

√—

2 —

2 sin x −

√—

2 —

2 cos x +

√—

2 —

2 sin x = 1

− √—

2 cos x = 1

cos x = − √

— 2 —

2

In the interval 0 ≤ x < 2π, the solutions are x = 3π — 4 and

x = 5π — 4 .

43. tan ( x + π ) + cos ( x + π — 2 ) = 0

tan x + tan π ——

1 − tan x tan π + cos x cos

π — 2 − sin x sin

π — 2 = 0

tan x − sin x = 0

tan x = sin x


Chapter 9 Test (p. 531)

1. cos2 x + sin2 x —— 1 + tan2 x

= 1 —

sec2 x

= cos2 x

2. 1 + sin x — cos x

+ cos x

— 1 + sin x

= (1 + sin x)(1 + sin x) + cos2 x

——— cos x(1 + sin x)

= 1 + 2 sin x + sin2 x + cos2 x

——— cos x (1 + sin x)

= 2 + 2 sin x

—— cos x(1 + sin x)

= 2 —

cos x

= 2 sec x

3. cos ( x + 3π — 2 ) = cos x cos

3π — 2 − sin x sin

3π — 2

= (cos x)(0) − (sin x)(−1)

= sin x

4. An angle coterminal to −300° is 60°. The secant function is

positive in Quadrant I, so

sec(−300°) = sec 60° = 2.


Chapter 9



value is −1.

Step 2 Identify the vertical shift k. The value of k is the

mean of the maximum value and minimum value.

k = (maximum value) − (mininum value)

———— 2

= 5 + ( −1 )

— 2 =

4 —

2 = 2



maximum value on the y-axis. The graph is a cosine



2 = 2π — b → b = π.

The amplitude is


———— 2

= 5 − ( −1 )

— 2 =

6 —

2 = 3.


The function is y = 3 cosπ x + 2.



value is −5.




———— 2

= 1 + ( −5 )

— 2 =

−4 —

2 = −2



the midline y = −2 on the y-axis, the graph is a sine



3π — 2 =

2π — b → b =

4 —

3 .

The amplitude is


———— 2

= 1 − ( −5 )

— 2 =

4 —

2 = 2.


The function is y = −3 sin 4 —

3 x −2.

7. The function is of the form g(x) = a tan bx, where a = −4

and b = 2.

So, the period is π —

∣ b ∣ =

π — 2 .

Intercept: (0, 0)


2 ∣ b ∣ =

π — 2 ( 2 )

= π — 4 ; x = −

π —

2 ∣ b ∣

= − π —

2 ( 2 ) = −

π — 4


, a ) = ( π — 4(2)

, −4 ) = ( π — 8 , −4 ) ;

( − π — 4b

, − a ) = ( − π —

4(2) , −(−4) ) = ( −

π — 8 , 4 )

x

y16

8

4

12

−16

−12

−8

−44−π8−π

4π


2 and

a vertical stretch by a factor of 4 followed by a refl ection

across the x-axis of the graph of f (x) = tan x.


vertical shift.

Amplitude: ∣ a ∣ = 2 Horizontal shift: h = 0

Period: 2π — b =

3π — 1 —

3




On y = k: ( 3π — 2 , 3 ) ; ( 9π —

2 , 3 )

Maximum: (3π, 5)

Minimums: (0, 1); (6π, 1)


x

y

1

2

3

4

5

6

π−1

3π 5π

The graph of g is a horizontal stretch by a factor of 3 and

a vertical stretch by a factor of 2 followed by a refl ection

across the x-axis and a translation 3 units up of the graph of

f (x) = cos x.


Chapter 9

9. Step 1 Graph the function y = 3 sinπ x. The period is

2π — π

= 2.


occur when 3 sin 2π = 0, graph x = 0, x = 1,

and x = 2.

Step 3 Plot points on g, such as ( 1 — 2 , 3 ) and ( 3 —

2 , −3 ) .


x

y

2

4

6

8

−2

−4

−6

−8

−10

0.5 1 1.5 2


vertical stretch by a factor of 3 of the graph of f (x) = csc x.

10. −50° = −50 degrees ( π radians —

180 degrees )

= − 5π — 18

−50° + 360° = 310° −50° − 360° = −410°

11. 4π — 5 =

4π — 5 radians ( 180 degrees

— π radians

) = 144°

4π — 5 + 2π =

14π — 5

4π — 5 − 2π = −

6π — 5

12. 8π — 3 =

8π — 3 radians ( 180 degrees

— π radians

) = 480°

8π — 3 − 2π =

2π — 3

8π — 3 − 2 ⋅ 2π = −

4π — 3

13. Convert 40° to radians: 40° = 2π — 9 .

Arc length: s = θr Area of sector: A = 1 —

2 θr2

= ( 2π — 9 ) (13) =

1 —

2 ( 2π —

9 ) (13)2

= 26π —

9 =

169π — 9

≈ 9.08 ≈ 59.00

The arc length is about 9.08 inches and the area is about

59.00 square inches.


r = √—

x2 + y2

= √—

22 + (−9)2

= √—

85

Using x = 2, y = −9, and r = √—



sin θ = y —

r = −

9 √—

85 —

85 csc θ =

r —

y = −

√—

85 —

9

cos θ = x —

r =

2 √—

85 —

85 sec θ =

r —

x =

√—

85 —

2

tan θ = y —

x = −

9 —

2 cot θ =

x —

y = −

2 —

9


r = √—

x2 + y2

= √—

(−1)2 + 02

= √—

1

= 1

Using x = −1, y = 0, and r = 1, the values of the six


sin θ = y —

r = 0 csc θ =

r —

y =

0 —

0 = undefi ned

cos θ = x —

r = −1 sec θ =

r —

x = −1

tan θ = y —

x = 0 cot θ =

x —

y = −

1 —

0 = undefi ned

16. The quadrant is Quadrant III because cos θ < 0, then θ must

be in Quadrant II or Quadrant III and because tan θ > 0, then

θ must be in Quadrant III.

17. sin 60° = d —

200 , where d is the distance from the base of the

crane to the top. So, d = 200 sin 60°. So, the height of the

building is h = 200 sin 60° + 5 ≈ 178 feet.


The scatter plot appears sinusoidal. So, perform a sinusional


viewing window.

120

0

100

The model appears to be a good fi t. So, a model for the

data is T = 23.14 sin (0.495 m − 1.95) + 63.7. The period

represents the amount of time it takes for the weather to

complete one cycle, which is about 12.7 months.


Chapter 9

Chapter 9 Standards Assessment (pp. 532–533)

1. tan x sec x cos x = sin x

— cos x

⋅ 1 —

cos x ⋅ cos x

= sin x

— cos x

= tan x

sin2 x + cos2 x = 1

cos2(−x) tan2 x ——

sin2(−x) =

cos2 x tan2 x —

sin2 x

=

cos2 x ⋅ sin2 x

— cos2 x

——

sin2 x

= 1

cos ( π — 2 − x ) csc x = sin x csc x

= sin x ⋅ 1 —

sin x

= 1

The expressions are sin2x + cos2x, cos2 (−x) tan2x

—— sin2(−x)

,

cos ( π — 2 − x ) csc x.

2. A; Perimeter of the playground

= 2x + 6x + (2x + x) + 2x + x + (6x − 2x)

= 18x

Area of the playground = (2x)(6x) + (x)(2x)

= 12x2 + 2x2

= 14x2

So, the ratio of the perimeter to the area is 18x

— 14x2

= 9 —

7x .

3. a. Using the sinusoidal regression feature, the functions that

model the data are

y1 = 28.53 sin(0.548t + 3.12) + 45.7 and

y2 = 7.89 sin(0.610t − 0.09) + 17.1

b.

130

0

80

Sample answer: The two graphs increase and decrease in

about the same intervals.

4. a. log 1000 = log 103

= 3 log 10

= 3

b. log2 15 = log2 (3 ⋅ 5)

= log2 3 + log2 5

≈ 1.585 + 2.322

= 3.907

c. ln e = 1

d. log2 9 = log2 (32)

= 2 log2 3

≈ 2(1.585)

= 3.17

e. log2 5 —

3 = log2 5 − log2 3

≈ 2.322 − 1.585

= 0.737

f. log2 1 = 0

The order is f, e, c, a, d, and b.

5. C; A. y = 5 sin x

B. y = 5 cos ( π — 2 − x )

= 5 sin x

C. y = 5 cos ( x + π — 2 )

= 5 [ cos x cos π — 2 − sin x sin

π — 2 ]

= −5 sin x

D. y = −5 sin ( π + x)

= −5 [sin π cos x + cos π sin x]

= −5 (−sin x)

= 5 sin x

6. a. s = rθ 4π = 6θ

4π — 6 = θ

2.09 ≈ θ θ < 3 radians

b. The tangent function is negative in Quadrant II.

tan θ < 0

c. θ′ = π − θ = π −

4π — 6

= π — 3

= π — 3 ⋅

180° — π

= 60°

θ′ > 45°

7. The factors of −6 are ±1, ±2, ±3, ±6.

The factors of 2 are ±1, ±2.

The possible rational roots are ±1, ± 1 — 2 , ±2, ±3,

± 3 — 2 , ±6.

From the graph, you can see that the real zeros of the

functions are −2, − 1 — 2 , and 3.


Chapter 9

8. −210° + 360° = 150°

150° = 150° ( π — 180°

) =

5π — 6

yes; Both angles lie in Quadrant II and have a reference

angle of 180° − 150° = 30° or π − 5π — 6 =

π — 6 and, therefore,

–210° is coterminal with 5π — 6 .

9. a. For Company A, the fi rst term is a1 = 20,000 and the

common difference is d = 1000.

an = a1 + (n − 1) d

= 20000 + (n − 1) 1000

= 1000n + 19,000

The sequence is represented by an = 1000n + 19,000, and

it is an arithmetic sequence.

For Company B, the fi rst term is a1 = 20,000 and the

common ratio is r = 0.04 + 1 = 1.04.

an = a1 r n−1

= 20,000 (1.04)n−1

The sequence is represented by an = 20,000 (1.04)n−1,

and it is a geometric sequence.

b.

0

20,000

30,000

40,000

0 5 10 15 20 n

an

anbn

Years

Sala

ry (

do

llars

)

c. Sample answer: You would choose to work for Company B

if you would be working for longer than 12 years.

d. ∑ n=1

20

an = ∑ n=1

20

(1000n + 19,000)

= 1000 [ 20 (21) —

2 ] + 19,000(20)

= 590,000

∑ n=1

20

bn = ∑ n=1

20

20,000 (1.04)n−1

= −20,000 [ 1 + 1.04 + 1.042 + 1.043 + . . . + 1.0419 ]

≈ 595, 562

The total earnings for Company A is $590,000 and for

Company B is about $595,562.

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Chapter 9 · Worked-Out Solutions Chapter 9 4. Step 1 Draw a right triangle with acute angle θ...

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