Solutions Manual for Trigonometry 4th Edition by Dugopolski IBSN 9780321900340 Full clear download (no formatting errors) at: http://downloadlink.org/p/solutions-manual-for-trigonometry-4th-edition-bydugopolski-ibsn-9780321900340/ 80

For Thought 1. True 2. False, since the range of y = 4 sin(x) is [−4, 4], we get that the range of y = 4 sin(x) + 3 is [−4 + 3, 4 + 3] or [−1, 7]. 3. True, since the range of y = cos(x) is [−1, 1], we find that the range of y = cos(x) − 5 is [−1 − 5, 1 − 5] or [−6, −4]. 4. False, the phase shift is −π/6.

Chapter 2

Graphs of the Trigonometric Functions

√ √ 3 sin(π/3) /2 13. = = 3. cos(π/3) 1/2 √ cos(π/6) 3/2 √ = = 3. 14. sin(π/6) 1/2 15. 1/2 16. 17.

1/2

1 1 = = 2. cos(π/3) 1/2

√ 1 2 3 1 =√ = . 18. 3 3/2 sin(π/3)

5. False, the graph of y = sin(x + π/6) lies π/6 to the left of the graph of y = sin(x).

19. 0

20.

1

21. 0

22.

0

6. True, since cos(5π/6 − π/3) = cos(π/2) = 0 and cos(11π/6 − π/3) = cos(3π/2) = 0.

23. −2 24. −2 √ 2 25. − 26. 1 2 9. starting, maximum, inflection, minimum, ending

7. False, for if x = π/2 we find sin(π/2) = 1 = cos(π/2 + π/2) = cos(π) = −1. 8. False, the minimum value is −3. 9. True, since the maximum value of y = −2 cos(x) is 2, we get that the maximum value of y = −2 cos(x) + 4 is 2 + 4 or 6. 10. True, since (−π/6 + π/3, 0) = (π/6, 0).

2.1 Exercises 1. sin α, cos α 2. sine 3. sine wave 4. periodic 5. fundamental cycle 6. amplitude 7. phase shift 8. cosine

10. maximum, inflection, minimum 11. 0 12.

1

27. (0 + π/4, 0) = (π/4, 0)

36. (2π/3 − π/3, −1) = (π/3, −1)

28. (π/2, 1)

37. (π − π/3, 1) = (2π/3, 1)

29. (π/2 + π/4, 3) = (3π/4, 3)

38. (2π − π/3, 4) = (5π/3, 4)

30. (3π/4 + π/4, −1) = (π, −1)

39. (π/2 − π/3, −1) = (π/6, −1)

31. (−π/2 + π/4, −1) = (−π/4, −1)

40. (π/4 − π/3, 2) = (−π/12, 2)

32. (−π/4 + π/4, −2) = (0, −2)

41. (−π − π/3, 1) = (−4π/3, 1)

33. (π + π/4, 0) = (5π/4, 0)

42. (−π/2 − π/3, −1) = (−5π/6, −1)

34. (2π + π/4, 0) = (9π/4, 0)

43. (π + π/6, −1 + 2) = (7π/6, 1)

35. (π/3 − π/3, 0) = (0, 0)

44. (π/6 + π/6, −2 + 2) = (π/3, 0)

2.1 The Unit Circle and Graphing

45. (π/2 + π/6, 0 + 2) = (2π/3, 2) 46. (π/3 + π/6, −1 + 2) = (π/2, 1) 47. (−3π/2 + π/6, 1 + 2) = (−4π/3, 3) 48. (−π/2 + π/6, 2 + 2) = (−π/3, 4) 49. (2π + π/6, −4 + 2) = (13π/6, −2) 50. (−π + π/6, 5 + 2) = (−5π/6, 7) 51.

π + 2π ,0 = 2

52.

π , −2 6

3π ,0 2

0 + π/4 ,2 = 2

54.

3π ,1 4

55.

π/6 + π/2 ,1 = 2

56.

3π ,2 8

58.

68. Amplitude 1, period 2π, phase shift −π/2, range [−1, 1] 69. Amplitude 2, period 2π, phase shift −π/3, range [−2, 2] 70. Amplitude 3, period 2π, phase shift π/6, range [−3, 3] 71. Amplitude 1, phase shift 0, range [−1, 1], π , −1 , some points are (0, 0), 2 3π (π, 0), , 1 , (2π, 0) 2 y

53.

57.

81

π ,2 8

π/3 + , −4 = π/2 2

1

Π €€€€ 2

π ,1 3

5π , −4 12

64. P (−π/8, 0), Q(π/8, 2), R(3π/8, 0), S(5π/8, −2) 65. Amplitude 2, period 2π, phase shift 0, range [−2, 2]

1

Π €€€€ 2

Π

Π €3€€€2€€€€

2Πx

1 -

73. Amplitude 3, phase shift 0, range [−3, 3], some 3π π points are (0, 0), ,3 , , −3 , (π, 0), 2 2 (2π, 0) y 3

66. Amplitude 4, period 2π, phase shift 0, range [−4, 4] 67. Amplitude 1, period 2π, phase shift π/2, range [−1, 1]

x

y

60. P (0, 1), Q(π/6, 3), R(π/3, 1), S(π/2, −1)

63. P (0, 2), Q(π/12, 3), R(π/6, 2), S(π/4, 1)

2Π

72. Amplitude 1, phase shift 0,range [−1, 1], some points are (0, −1), (π/2, 0), (π, 1), (3π/2, 0), (2π, −1)

59. P (0, 0), Q(π/4, 2), R(π/2, 0), S(3π/4, −2)

62. P (π/3, 0), Q(5π/6, 3), R(4π/3, 0), S(11π/6, −3)

3Π €€€€€€€€ 2

-1

7π ,5 8

61. P (π/4, 0), Q(5π/8, 2), R(π, 0), S(11π/8, −2)

Π

Π €€€€ 2

-3

Π

3Π €€€€2€€€€

2Π

x

82

Chapter 2

74. Amplitude 4, phase shift 0, range [−4, 4], some π 3π points are (0, 0), , −4 , , 4 , (π, 0), 2 2 (2π, 0)

Graphs of the Trigonometric Functions

77. Amplitude 1, phase shift −π, range [−1, 1], some points are (−π, 0), (−π/2, 1), (0, 0), π , −1 , (π, 0) 2

y

y

4 1

x

Π - €€€€ 2

-Π

Π €€€€ 2

Π €3€€€2€€€€

Π

2Π

x

€€€€

Π 2

Π

-1 4

75. Amplitude 1/2, phase shift 0, range [−1/2, 1/2], some points are (0, 1/2), (π/2, 0), (π, −1/2) (3π/2, 0), (2π, 1/2)

78. Amplitude 1, phase shift π, range [−1, 1], some points are (0, −1), (π/2, 0), (π, 1), (3π/2, 0), (2π, −1)

y

y

0.5

1

Π €€€€ 2

Π

Π €3€€€€€€€ 2

2Π

x

Π €€€€ 2

3Π €€€€€€€€ 2

2Π

5Π €€€€ 2

3Π

x

-1

-0.5

76. Amplitude 1/3, phase shift 0, range [−1/3, 1/3], some points are π

Π

, 0 , (π, −1/3), (3π/2, 0),

2

1 2π,

0,

1 , 3

79. Amplitude 1, phase€€€€ shift π/3, range [−1, 1], 5π 4π π some points are ,1 , ,0 , , −1 , 3 6 3 11π 7π

3

y

,0 ,

6

3

,1 ,

y

1 €€€€ 3

1 - €€€€ 3

Π €€€€ 2

Π

Π €3€€€€€€€ 2

2Π

x

Π €€€€ 3

5Π €€€€€€€€ 6

4Π €€€€€€€€ 3

11 Π €€€€€€€€€€ 6

7Π €€€€€€€€ 3

x

2.1 The Unit Circle and Graphing

83

80. Amplitude 1, phase shift −π/4, range [−1, 1], π π 3π some points are − , 1 , ,0 , , −1 , 4 4 4 y

83. Amplitude 1, phase shift 0, range [−2, 0], some π 3π points are (0, −1), , 2 , (π, −1), ,0 , 2 − 2 (2π, −1) y

1

Π - €€€€ 4

€€€€

Π 4

€€€€ €3€€€ Π 4

€€€€ €5€€€ Π 4

Π

3Π €€€€2€€€€

2Π

x

-1

-1

5π ,0 , 4

Π €€€€ 2

x €€€€ €7€€€ Π 4

7π ,1 4

-2

81. Amplitude 1, phase shift 0, range [1, 3], π 3π some points are (0, 3), , 2 , (π, 1) ,2 , 2 2 y

84. Amplitude 1, phase shift 0, range [1, 3], some points are (0, 2),

2

π , 1 , (π, 2),

3π ,3 , 2

(2π, 2) y

3

2 3 1 2 Π €€€€ 2

3Π €€€€€€€€ 2

Π

2Π

x 1

(2π, 3) 82. Amplitude 1, phase shift 0, range [−4, −2], π some points are (0, −2), , −3 , (π, −4), 2 3π , −3 , (2π, −2) 2

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

Π €€€€ 2

Π

3Π €€€€€€€€ 2

2Π

x

85. Amplitude 1, phase shift −π/4, range [1, 3], π π 3π some points are − , 2 , ,3 , ,2 , 4 4 4 7π y ,2 5π , 1 , 4 4

x

y 3

-2

2

-3 1 -4 Π - €€€€ 4

€€€€

Π 4

€€€€ €3€€€ Π 4

€€€€ €5€€€ Π 4

x 7Π €€€€ €€€€ 4

84

Chapter 2

86. Amplitude 1, phase shift π/2, range π π [−3, −1], some points are , −2 , 2 3π 5π (π, −1), , −2 , (2π, −3), , −2 2 2

Graphs of the Trigonometric Functions

89. Amplitude 2, phase shift π/3, range [−1, 3], 5π 4π ,1 , , −1 , ,1 , some points are 3 6 3 11π 7π ,3 ,1 6 3

y

y Π €€€€ 2

3Π €€€€€€€€ 2

Π

2Π

5Π €€€€€€€€ 2

x 3

-1 1 -2 Π €€€€ 3

-3

5Π €€€€€€€€ 6

4Π €€€€€€€€ 3

11 Π €€€€€€€€€€€ 6

7Π €€€€€€ €€ 3

x

-1

87. Amplitude 2, phase shift −π/6, range [−1, 3], π π 5π some points are − , 3 , ,1 , , −1 , 6 3 6 y

3

90. Amplitude 3, phase shift −π/3, π range [−4, 2], some points are − , −4 , 3 2π 7π π 5π ,2 , , −1 , , −1 , , −4 3 6 6 3 2

y

1

Π - €€€€ 6

€€€€ Π 3

-1

€€€€€5€€€ Π 6

x Π€€ €€€€€€€€4 Π €€€€€€€€11 3 6

4π 11π ,3 ,1 , 3 6 88. Amplitude 3, phase shift −2π/3, 2π ,1 , 3 5π 4π , −2 , and ,1 6 3

range [−5, 1], some points are π π − , −2 , 6

3

, −5 ,

−

Π - €€€€ 3

€€€€ -1

Π 6

€€€€

2Π €€€€ €€€€€€€€ 3

7 Π€€€€€€€€ 6

x

5Π 3

-4

91. Note the amplitude is 2, phase shift is π/2, and the period is 2π. Then A = 2, C = π/2, and B = 1. An equation is y = 2 sin x −

y 1 2Π - €€€€€€€€ 3

Π - €€€€ 6 -2

-5

€€€€

€€€€

Π 3

x €€€€€5€€€ Π 6

4Π €€€€ 3

π . 2

92. Note the amplitude is 2, phase shift is π/3, period is 2π, and the vertical shift is 1 unit up. Then A = 2, C = π/3, B = 1, and D = 1. π An equation is y = 2 sin x − + 1. 3 93. Note the amplitude is 2, phase shift −π, the period is 2π, and the vertical shift is 1 unit up. Then A = 2, C = −π, B = 1, and D = 1. An equation is y = 2 sin (x + π) + 1. 94. Note the amplitude is 1, phase shift is −π/2, period is 2π, and the vertical shift is 1 unit up. Then A = 1, C = −π/2, B = 1, and D = 1.

2.1 The Unit Circle and Graphing

An equation is y = sin x +

85

π + 1. 2

111. π/6

95. Note the amplitude is 2, phase shift is π/2, and the period is 2π. Then A = 2, C = π/2, and π . B = 1. An equation is y = 2 cos x − 2 96. Note, A = −2, C = π/2, and B = 1. π An equation is y = −2 cos x − . 2 97. Note, the amplitude is 2, phase shift is π, the period is 2π, and the vertical shift is 1 unit up. Then A = 2, C = π, B = 1, and D = 1. An equation is y = 2 cos (x − π) + 1. 98. Note, the amplitude is 1, phase shift is −π/3, the period is 2π, and the vertical shift is 1 unit up. Then A = 1, C = −π/3, B = 1, and D = 1. An equation is y = cos (x + π/3) + 1. 99. y = sin x −

π 4

101. y = sin x +

100. π 2

y = cos x −

102.

π 6

y = cos x +

93 · 106 · 2π ≈ 67, 000 mph 113. 365(24) √ √ 2 114. a) − 1 b) c) 3 2 d) Undefined e) − 2 f ) Undefined √ √ 2 g) − 3 h) − 2 115. arcsin(0.36) ≈ 21.1◦ r √ 1 116. sin α = − 1 − cos2 α = − 1 − = 9 √ r 8 2 2 − =− 9 3

117. Let r be the radius of the small circle, and let x be the closest distance from the small circle to the point of tangency of any two circles with radius 1. By the Pythagorean theorem, we find 1 + (x + r)2 = (1 + r)2

π 3

π 103. y = − cos x − 5 104. y = − sin x +

112. 315◦

π 7

π +2 105. y = − cos x − 8

and 1 + (1 + 2r + x)2 = 22 . The second equation may be written as 1 + (r + 1)2 + 2(r + 1)(r + x) + (r + x)2 = 4. Using the first equation, the above equation simplifies to (r + 1)2 + 2(r + 1)(r + x) + (1 + r)2 = 4

π 106. y = − sin x + −3 9

or

π 107. y = −3 cos x + −5 4

Since (from first equation, again)

1 π 108. y = − sin x − +4 2 3 109. A determines the stretching, shrinking, or reflection about the x-axis, C is the phase shift, and D is the vertical translation. 110. f (x) = sin(x) is an odd function since sin(−x) = − sin(x), and f (x) = cos(x) is an even function since cos(−x) = cos(x)

(r + 1)2 + (r + 1)(r + x) = 2. q

(1 + r)2 − 1

x+r = we obtain

q

(1 + r)2 − 1 = 2.

(r + 1)2 + (r + 1) Solving for r, we find r=

2

√

3− 3 . 3

86

Chapter 2

118. Since sin x = 3 cos x, we find 1 = sin2 + cos2 x 1 = 9 cos2 x + cos2 x cos2 x =

Graphs of the Trigonometric Functions

x3 x5 + and 3! 5! y2 = sin(x), it follows that y and y2 differ by less than 0.1 if x lies in the interval (−2.46, 2.46).

b) From the graphs of y = x −

1 . 10

Then sin x cos x = 3 cos2 x =

y

1 10 . .5 x -2.45 -.5

2.1 Pop Quiz 1. Amplitude 5, period 2π, phase shift −2π/3, range [−5, 5] 2. Key points are (0, 0),

π , 3 , (π, 0), 2

3π , −3 , (2π, 0) 2 3. y = − cos x −

π 2

2.45

x3 x5 x7 x9 + − + 3! 5! 7! 9! and y2 = sin(x), one obtains that y and y2 differ by less than 0.1 if x lies in the interval (−4.01, 4.01).

From the graphs of y = x −

y

+3

4. [−2, 6] .5

5. Since the period is 5π/2 − π/2 = 2π, we get B = 1. The phase shift is π/2. The amplitude is 3. An equation is y = −3 sin x −

π . 2

2.1 Linking Concepts a) From the graphs of y1 = x and y2 = sin(x), we obtain that y1 and y2 differ by less than 0.1 if x lies in the interval (−0.85, 0.85).

x -4.01 -.5

4.01

From the graphs of y = x −

x11 x13 x15 x17 x19 + − + − and y2 = sin(x), 11! 13! 15! 17! 19! one finds that y and y2 differ by less than 0.1 if x lies in the interval (−7.82, 7.82). −

y

y .5

.5 x x

-.85 -.5

.85

x3 x5 x7 x9 + − + 3! 5! 7! 9!

-7.82 -.5

7.82

2.2 The General Sine Wave

87 y

c) From the graphs of y1 = 1 and y = cos(x), one obtains that y and y1 differ by less than 0.1 if x lies in the interval (−0.45, 0.45). y

.5 x -7.44 -.5

7.44

.5 x -.45 .45 -.5

From the graphs of y = 1 −

x2 x4 + and 2! 4!

y2 = cos(x), one finds y and y2 differ by less than 0.1 if x lies in the interval (−2.06, 2.06). y

d) To find sin(x), first let y be the reference angle π of x. Note, 0 ≤ y ≤ . 2 Next, we obtain that the difference between 3 y is less than f (y) = sin(y) and f (y) = y − 3! π 0.1 for 0 ≤ y ≤ . Thus, sin(x) = ± sin(y) = 2 ! y3 where the sign on the right side ± y− 3! depends on the sign of sin(x).

For Thought

.5 x -2.06 -.5

2.06

1. True, since B = 4 and the period is

2π π = . B 2

2π = 1. B 2π 3. True, since B = π and the period is = 2. B 2π 4. True, since B = 0.1π and the period is = 0.1π 20. π 5. False, the phase shift is − . 12 π 6. False, the phase shift is − . 8 2. False, since B = 2π and the period is

x2 x4 x6 x8 + − + 2! 4! 6! 8! and y2 = cos(x), one derives that y and y2 From the graphs of y = 1 −

differ by less than 0.1 if x lies in the interval (−3.63, 3.63). y

.5

7. True, since the period is P = 2π the frequency 1 1 is = . P 2π

x -3.63 -.5

3.63

From the graphs of y = 1 −

x2 x4 x6 x8 + − + 2! 4! 6! 8!

x10 x12 x14 x16 x18 + − + − and y2 = cos(x), 10! 12! 14! 16! 18! one finds that y and y2 differ by less than 0.1 if x lies in the interval (−7.44, 7.44). −

8. True, since the period is P = 2 the frequency 1 1 is = . P 2 9. False, rather the graphs of y = cos(x) and π y = sin x + are identical. 2 10. True

88

Chapter 2

2.2 Exercises

Graphs of the Trigonometric Functions

12. Period 6π, phase shift 0, range [−1, 1], labeled 3π 9π points are (0, 1), , 0 , (3π, −1), ,0 , 2 2 (6π, 1)

1. period 2. frequency

y

3. amplitude 1

4. phase shift π 2π or , and phase shift 0 2 4 6. Amplitude 1, period 4π, and phase shift 0 5. Amplitude 3, period

7. Since y = −2 cos 2 x + amplitude 2, period π phase shift − . 4

π 4

2π or π, and 2

2π , we get 3 2π amplitude 4, period , and 3 2π phase shift . 3

9. Since y = −2 sin (π (x − 1)), we get 2π amplitude 2, period or 2, and π phase shift 1.

period

-3Pi

− 1, we get

8. Since y = 4 cos 3 x −

10. Since y = sin

-6Pi

π (x + 2) , we get amplitude 1, 2

2π or 4, and phase shift −2. π/2

11. Period 2π/3, phase shift 0, range [−1, 1], π labeled points are (0, 0), ,1 , 6 π π 2π ,0 , ,0 , −1 , 3 3 2

3Pi

x 6Pi

-1

13. Period π, phase shift 0, range [−1, 1], labeled 3π π π points are (0, 0), ,0 , ,1 , , −1 , 4 4 2 (π, 0) y

1

-3Pi/4

-Pi/4

x Pi

Pi/4

-1

14. Period 2π/3, phase shift 0, range [−1, 1], π π labeled points are (0, −1), ,0 , ,1 , 6 3 π 2π ,0 , , −1 2 3 y

1

y

-2Pi/3 -Pi/3

1 x -2Pi/3

-Pi/6 -1

Pi/6

Pi/2

Pi/3 -1

x 2Pi/3

2.2 The General Sine Wave

89

15. Period π/2, phase shift 0, range [1, 3], labeled 3π π π ,2 , points are (0, 3), ,2 , ,1 , 8 4 8 π ,3 2

18. Period 10π, phase shift 0, range [2, 4], labeled 5π 15π , 3 , (5π, 4), ,3 , points are (0, 2), 2 2 (10π, 2) y

y 4 3 2 1 -Pi/2

-5Pi

-Pi/4

Pi/4

x 10Pi

5Pi

x Pi/2

19. Period 6, phase shift 0, range [−1, 1], labeled 16. Period 2π/3, phase shift 0, range [−2, 0], π π labeled points are (0, −1), ,0 , , −1 , 3 6 π 2π , −1 , −2 , 3 2

points are (0, 0), (1.5, 1), (3, 0), (4.5, −1), (6, 0) y 1

y x -2Pi/3

-Pi/6

Pi/6

x 2Pi/3

-6

-1.5

1.5

4.5

-1

-1

20. Period 8, phase shift 0, range [−1, 1], labeled

-2

points are (0, 0), (2, 1), (4, 0), (6, −1), (8, 0) y

17. Period 8π, phase shift 0, range [1, 3], labeled points are (0, 2), (2π, 1), (4π, 2), (6π, 3), (8π, 2)

1 x -8

y

-2

2 -1

3 1 -4Pi

4Pi

x 8Pi

6

90

Chapter 2

21. Period π, phase shift π/2, range [−1, 1], 3π π ,1 , labeled points are ,0 , 4 2 3π 5π ,0 (π, 0), , −1 , 2 4 y

Graphs of the Trigonometric Functions

24. Period 6, phase shift 1, range [−1, 1], labeled points are (1, 1), (2.5, 0), (4, −1), (5.5, 0), (7, 1) y 1

x 1

-Pi/4

x 2Pi

3Pi/4

-1

22. Period 2π/3, phase shift −π/3, range [−1, 1], π π π labeled points are − , 0 , − , 1 , 3 6 π π ,0 , −1 , (0, 0), 3 6

2.5

4

7

-1

25. Period π, phase shift −π/6, range [−1, 3], π π labeled points are − , 3 , ,1 , 6 12 7π 5π , −1 , ,1 , ,3 3 12 6 y

y

3

1 x -Pi/3

-Pi/6 -1

Pi/3

x Pi

-1

23. Period 4, phase shift −3, range [−1, 1], labeled points are (−3, 0), (−2, 1), (−1, 0), (0, −1), (1, 0) y

26. Period π/2, phase shift π/2, range [−4, 2], 5π π ,2 , labeled points are , −1 , 2 8 3π 7π , −4 , , −1 , (π, 2) 4 8 y 2

-4

-2

2 -1

-1 -4

Π 7Π Π Π 3 Π €€€€€€€€ €€€€ €5 €€€€ 2 €€€8 €€€€€€€€ 4 8

x

2.2 The General Sine Wave 2π π 3 1 , phase shift , range − , − , 3 6 2 2 π 3 π labeled points are ,− , , −1 , 3 2 6 π 1 5π 2π , −1 , ,− , , −1 2 3 2 6

27. Period

y

x Pi

Pi/2

91 π 32. Note, A = 2, B = 2, phase shift is C = − , 4 π and D = −1. Then y = 2 sin 2 x + −1. 4 π 33.

π −

=0 4 4 π π π 34. − = 2 4 4 35. f g

-0.5

36. f g -1.5

π

= f (0) = sin(0) = 0

4 π

=f

2

√ 2 = 2

π π = sin 4 4

π = h (f (0)) = h (sin (0)) = 4 h(0) = 3 · 0 = 0

37. h f g 28. Period π/2, phase shift −π/4, range [1/2, 3/2], π π labeled points are − , 1 , − , 0.5 , 4 8 π π (0, 1), , 1.5 , ,1 8 4 y

38. h f g h sin

π 2 π 4

π =h f = 4 √! √ 2 3 2 =h = 2 2

39. f (g (x)) = f x − 1.5

= sin x −

π 4

40. f (h (x)) = f (3x) = sin (3x)

1 0.5

−

x -Pi/4

π 4

Pi/4

29. Note, A = 2, period is π and so B = 2, phase π shift is C = , and D = 0 or no vertical shift. 4 π . Then y = 2 sin 2 x − 4 30. We can choose A = −1 with C = 0, i.e., no 1 phase shift. The period is 4π and so B = . 2 There is no vertical shift or D = 0. x Then y = − sin . 2 4π 3 31. Note, A = 3, period is and so B = , phase 3 2 π shift is C = − , and D = 3 since the vertical 3 shift is three units up. π 3 x+ Then y = 3 sin + 3. 3 2

π

= 41. h (f (g (x))) = h f x π π h sin x − = 3 sin x − 4 4 42. f (h (g (x))) = f h x − f 3 x−

π 4

4

π 4

= sin 3 x −

= π 4

43. 100 cycles/sec since the frequency is the reciprocal of the period 44. 1/2000 cycles per second 1 = 40 cycles per hour 0.025 1 46. Period is = 0.000025 second. 40, 000 45. Frequency is

92

Chapter 2

47. Substitute vo = 6, ω = 2, and xo = 0 vo · sin(ωt) + xo · cos(ωt). into x(t) = ω Then x(t) = 3 sin(2t). The amplitude is 3 and the period is π. y 3

Pi/4

Graphs of the Trigonometric Functions π π x+ + 25, 000 ≈ $17, 500. 6 2

15, 000 sin

54. Period is 12, amplitude is 150, phase-shift is −2, vertical translation is 350, a formula for the curve is πx π y = 150 sin + + 350; 6 3 for November (when x = 11), the utility bill is

x 5Pi/4

150 sin

-3

48. Substitute vo = −4, ω = π, and xo = 0 v into x(t) = o · sin(ωt) + xo · cos(ωt). ω 4 Then x(t) = − sin(πt). π The amplitude is 4/π and period is 2.

πx π + 6 3

+ 350 ≈ $425

55. a) period is 40, amplitude is 65, an equation for the sine wave is π y = 65 sin x 20 b) 40 days

y

π (36) ≈ −38.2 meters/second 20 d) The new planet is between Earth and Rho. c) 65 sin

4/Pi 0.5 x 2

4

-4/Pi

49. 11 years 50. Approximately 1.4 seconds 51. Note, the range of v = 400 sin(60πt) + 900 is [−400 + 900, 400 + 900] or [500, 1300]. (a) Maximum volume is 1300 cc and mininum volume is 500 cc (b) The runner takes a breath every 1/30 (which is the period) of a minute. So a runner makes 30 breaths in one minute. 52. (a) Maximum velocity is 8 cm/sec and mininum velocity is 0. (b) The rodent’s heart makes a beat every 1/3 (which is the period) of a second or it makes 180 beats in a minute.

56. a) Ganymede’s period is 7.155 days, or 7 days and 8 hours; Callisto’s period is 16.689 days, 16 days and 17 hours; Io’s period is 1.769 days, 1 day and 18 hours; Europa’s period is 3.551 days, or 3 days and 13 hours. To the nearest hour, it would be easiest to find Io’s period since it is the satellite with the smallest period. b) Ios’s amplitude is 262,000 miles, Europa’s amplitude is 417,000 miles, Ganymede’s amplitude is 666,000 miles, Callisto’s amplitude is 1,170,000 miles 2π π , we get B = . B 10 Also, the amplitude is 1 and the vertical translation is 1. An equation for the swell is

57. Since the period is 20 =

y = sin 53. Period is 12, amplitude is 15,000, phase-shift is −3, vertical translation is 25,000, a formula for the curve is y = 15, 000 sin

π π x+ 6 2

+ 25, 000;

for April (when x = 4), the revenue is

π x + 1. 10

58. Since the period is 200, the amplitude is 15, and the vertical translation is 15, an equation for the tsunami is y = 15 sin

π x + 15. 100

2.2 The General Sine Wave

93

59. The sine regression curve is y = 50 sin(0.214x − 0.615) + 48.8 or approximately y = 50 sin(0.21x − 0.62) + 48.8 The period is

63. Amplitude A =

1 , 2

2π 2π = = 2π, B 1 π phase shift , 2 1 1 period is [− + 3, + 3] or [2.5, 3.5] 2 2 period B =

64. y = − cos(x + π) + 2

2π 2π = ≈ 29.4 days. b 0.214 When x = 39, we find y = 50 sin(0.214(39) − 0.615) + 48.8 ≈ 98% On February 8, 2020, 98% of the moon is illuminated. Shown below is a graph of the regression equation and the data points.

y 90 80

65. If x is the height of the tree, then tan 30◦ = h/500 or h = 500 tan 30◦ ≈ 289 ft. p √ √ 66. Let r = (−3)2 + 62 = 45 = 3 5. √ 2 5 y 6 Then sin β = = √ = , r 5 3 5 √ x 5 −3 cos β = = √ =− , and 5 r 3 5 y 6 tan β = = = −2 x −3 67. Note, the sum of the two angles is 32◦ 370 + 48◦ 390 = 80◦ 760 = 81◦ 160 .

40

The third angle is 179◦ 600 − 81◦ 160 = 98◦ 440 . 10

20

x

68. Since s = rα, we find 5 = 60α

60. The sine regression curve is

α

=

1 radian 12

α

=

1 180◦ · 12 π

α

≈ 4.8◦

y = 95.4 sin(0.514x − 1.84) + 727.0 or approximately y = 95.4 sin(0.51x − 1.84) + 727.0

69. One possibility is The period is 2π 2π = ≈ 12.2 months. b 0.514 When x = 14, we obtain 95.4 sin(0.51(14) − 1.84) + 727.0 ≈ 648 min. between sunrise and sunset on Feb. 1, 2021. π 62. For instance, one can choose B = 4, C = , 4 and D = 5.

WRONG = 25938 and RIGHT = 51876. 70. Let x > y be the radii of the circles. By the Pythagorean theorem, y2 + 402 = x2 . Then the volume of water in the island is 2(πx2 − πy2 ) = 2π(40)2 ≈ 10, 053 ft3 .

94

Chapter 2

2.2 Pop Quiz 1. Since we have y = 4 sin 2 x −

π 3

we obtain amplitude 4, period 2π/B = 2π/2 or π, phase shift π/3. 2. Key points are (0, 0),

π , −3 , 4

π ,0 , 2

3π , 3 , (π, 0) 4 3. The period 2π/B = 2π/π or 2. Since the amplitude is 4 and there is a vertical upward shift of 2 units, the range is [−4 + 2, 4 + 2] or [−2, 6]. π 4. Note, A = 4 and C = − . Since the period is 6 π π 2π + = 2 6 3

π 6

5. True, since B = 2 and

2π 2π = = π. B 2

6. True, since B = π and

2π 2π = = 2. B π

7. False, rather the graphs of y = 2 csc x and 2 are identical. y= sin x 8. True, since the maximum and minimum of 0.5 csc x are ±0.5. π 9. True, since + kπ are the zeros of y = cos x 2 we get that the asymptotes of y = sec(2x) are π π kπ 2x = + kπ or x = + . If k = ±1, 2 4 2 π we get the asymptotes x = ± . 4 10. True, since if we substitute x = 0 in we get

1. domain .

5. Note, the period is

2. domain 3. asymptote

2π 2π 1 = = . B 500π 250 Since the frequency is the reciprocal of the

4. x-intercepts 5.

1 1 = =2 cos(π/3) 1/2

6.

√ 1 1 =√ = 2 cos(π/4) 2/2

period, the frequency is 250 cycles/minute. 7.

For Thought

2. True, since csc(x) =

1 1 = . cos(π/4) sin(π/4)

1 . sin(x)

3. True, since csc(π/2) = 4. False, since

1 sin(−π/4)

1. True, since sec(π/4) =

1 1 = = 1. sin(π/2) 1

1 1 or is undefined. cos(π/2) 0

8.

1 csc(4x)

1 1 or which is undefined. csc(0) 0

2.3 Exercises

we find B = 3. Thus, the curve is y = 4 sin 3 x +

Graphs of the Trigonometric Functions

=

√ 1 √ =− 2 −1/ 2

1 1 = =2 sin(π/6) 1/2

9. Undefined, since

1 1 = cos(π/2) 0

10. Undefined, since

1 1 = cos(3π/2) 0

11. Undefined, since

1 1 = sin(π) 0

Solutions Manual for Trigonometry 4th Edition by Dugopolski IBSN 9780321900340 Full clear download (no formatting errors) at: http://downloadlink.org/p/solutions-manual-for-trigonometry-4th-edition-bydugopolski-ibsn-9780321900340/

Solutions Manual for Trigonometry 4th Edition by Dugopolski IBSN 9780321900340

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Solutions Manual for Trigonometry 4th Edition by Dugopolski IBSN 9780321900340

Published on Oct 3, 2017

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