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Chapter 2 Trigonometric Functions 16.

Section 2.1 1. C = 2π r ; A = π r 2 1 2π (2) = 2π 2. A = 2

3. standard position

17.

18.

4. central angle 5. d 6. rθ ;

1 2 rθ 2

19.

7. b 8.

s θ ; t t

20.

9. True 10. False; υ = rω

21.

11.

12. 22.

13.

14.

15.

_

1 1 1 º  23. 40º10 ' 25" =  40 + 10 ⋅ + 25 ⋅ ⋅  60 60 60   ≈ (40 + 0.1667 + 0.00694)º ≈ 40.17º 1 1 1 º  24. 61º 42 ' 21" =  61 + 42 ⋅ + 21 ⋅ ⋅  60 60 60   ≈ (61 + 0.7000 + 0.00583)º ≈ 61.71º

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Section 2.1: Angles and Their Measure 32. 29.411º = 29º + 0.411º = 29º + 0.411(60 ') = 29º +24.66 ' = 29º +24 '+ 0.66 ' = 29º +0.66(60") = 29º +24 '+ 39.6" ≈ 29º 24 ' 40"

1 1 1 º + 20 ⋅ ⋅ 60 60 60 ≈ (50 + 0.2333 + 0.00556)º

25. 50º14 '20" = 50 + 14 ⋅ ≈ 50.24º

1 1 1 º  26. 73º 40 ' 40" =  73 + 40 ⋅ + 40 ⋅ ⋅  60 60 60   ≈ (73 + 0.6667 + 0.0111)º ≈ 73.68º

33. 19.99º = 19º + 0.99º = 19º + 0.99(60 ') = 19º +59.4 ' = 19º +59 '+ 0.4 ' = 19º +59 '+ 0.4(60") = 19º +59 '+ 24" = 19º 59 ' 24"

1 1 1 º  27. 9º 9 '9" =  9 + 9 ⋅ + 9 ⋅ ⋅  60 60 60   = (9 + 0.15 + 0.0025)º ≈ 9.15º

1 1 1 º  28. 98º 22 ' 45" =  98 + 22 ⋅ + 45 ⋅ ⋅  60 60 60   ≈ (98 + 0.3667 + 0.0125)º ≈ 98.38º

34. 44.01º = 44º + 0.01º = 44º + 0.01(60 ') = 44º +0.6 ' = 44º +0 '+ 0.6 ' = 44º +0 '+ 0.6(60") = 44º +0 '+ 36" = 44º 0 '36"

29. 40.32º = 40º + 0.32º = 40º + 0.32(60 ') = 40º +19.2 ' = 40º +19 '+ 0.2 ' = 40º +19 '+ 0.2(60") = 40º +19 '+ 12" = 40º19 '12"

35. 30° = 30 ⋅

π π radian = radian 180 6 π 2π radian = radians 180 3

30. 61.24º = 61º + 0.24º = 61º + 0.24(60 ') = 61º +14.4 ' = 61º +14 '+ 0.4 ' = 61º +14 '+ 0.4(60") = 61º +14 '+ 24" = 61º14 ' 24"

36. 120° = 120 ⋅

31. 18.255º = 18º + 0.255º = 18º + 0.255(60 ') = 18º +15.3' = 18º +15'+ 0.3' = 18º +15'+ 0.3(60") = 18º +15'+ 18" = 18º15'18"

39. − 60° = − 60 ⋅

37. 240° = 240 ⋅

4π π radian = radians 180 3

38. 330° = 330 ⋅

11π π radian = radians 180 6

40. −30° = −30 ⋅ 41. 180° = 180 ⋅

π π radian = − radian 180 3

π π radian = − radian 180 6

π radian = π radians 180

42. 270° = 270 ⋅

π 3π radian = radians 180 2

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Chapter 2: Trigonometric Functions

43. −135° = −135 ⋅

π 3π radian = − radians 180 4

44. − 225° = − 225 ⋅ 45. − 90° = − 90 ⋅

5π π radian = − radians 180 4

π π radian = − radians 180 2

46. −180° = −180 ⋅

π π 180 = ⋅ degrees = 60° 3 3 π

48.

5π 5π 180 degrees = 150° = ⋅ 6 6 π

50. − 51.

54.

2π radian 9 ≈ − 0.70 radian π radian 180 17π radian =− 60 ≈ − 0.89 radian

62. − 51° = − 51 ⋅

63. 125° = 125 ⋅

2π 2π 180 degrees = −120° =− ⋅ 3 3 π

25π radians 36 ≈ 2.18 radians

64. 350° = 350 ⋅

180 degrees = 720° π

π radian 180

35π radians 18 ≈ 6.11 radians =

π π 180 degrees = 15° = ⋅ 12 12 π

65. 3.14 radians = 3.14 ⋅

5π 5π 180 degrees = 75° = ⋅ 12 12 π

66. 0.75 radian = 0.75 ⋅

π π 180 =− ⋅ degrees = − 90° 2 2 π

56. −π = −π ⋅

π radian 180

=

π π 180 degrees = 90° = ⋅ 2 2 π

55. −

π radian 180

=−

5π 5π 180 =− ⋅ degrees = − 225° 4 4 π

52. 4π = 4π ⋅ 53.

61. − 40° = − 40 ⋅

π radian = −π radians 180

47.

49. −

π radian 180 73π = radians 180 ≈ 1.27 radians

60. 73° = 73 ⋅

67. 2 radians = 2 ⋅

180 degrees = −180° π

68. 3 radians = 3 ⋅

π π 180 degrees = − 30° 57. − = − ⋅ 6 6 π

180 degrees ≈ 42.97º π

180 degrees ≈ 114.59º π

180 degrees ≈ 171.89º π

69. 6.32 radians = 6.32 ⋅

3π 3π 180 =− ⋅ degrees = − 135° 58. − 4 4 π

70.

17π π 59. 17° = 17 ⋅ radian = radian ≈ 0.30 radian 180 180

180 degrees ≈ 179.91º π

2 radians = 2 ⋅

180 degrees ≈ 362.11º π

180 degrees ≈ 81.03º π

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Section 2.1: Angles and Their Measure

71. r = 10 meters; θ = s = rθ = 10 ⋅

1 radian; A = 2 ft 2 3 1 A = r 2θ 2 1 2 1 2= r   2 3 1 2 = r2 6 12 = r 2

1 radian; 2

81. θ =

1 = 5 meters 2

72. r = 6 feet; θ = 2 radian; s = rθ = 6 ⋅ 2 = 12 feet

1 radian; s = 2 feet; 3 s = rθ s 2 r= = = 6 feet θ (1/ 3)

73. θ =

r = 12 = 2 3 ≈ 3.464 feet 1 radian; A = 6 cm 2 4 1 A = r 2θ 2 1 1 6 = r2   2 4 1 6 = r2 8 48 = r 2

82. θ =

1 radian; s = 6 cm; 4 s = rθ s 6 r= = = 24 cm θ (1/ 4 )

74. θ =

75. r = 5 miles; s = 3 miles; s = rθ s 3 θ = = = 0.6 radian r 5

r = 48 = 4 3 ≈ 6.928 cm

83. r = 5 miles; A = 3 mi 2 1 A = r 2θ 2 1 2 3 = (5) θ 2 25 3= θ 2 6 θ= = 0.24 radian 25

76. r = 6 meters; s = 8 meters; s = rθ s 8 4 θ = = = ≈ 1.333 radians r 6 3 77. r = 2 inches; θ = 30º = 30 ⋅ s = rθ = 2 ⋅

π π = radian; 180 6

π π = ≈ 1.047 inches 6 3

78. r = 3 meters; θ = 120º = 120 ⋅ s = rθ = 3 ⋅

84. r = 6 meters; A = 8 m 2 1 A = r 2θ 2 1 2 8 = ( 6) θ 2 8 = 18θ 8 4 θ = = ≈ 0.444 radian 18 9

π 2π = radians 180 3

2π = 2π ≈ 6.283 meters 3

1 radian 2 1 1 100 21 A = r 2θ = (10 )   = =25 m 2 2 2 2 4  

79. r = 10 meters; θ =

85. r = 2 inches; θ = 30º = 30 ⋅

80. r = 6 feet; θ = 2 radians

A=

1 1 2 A = r 2θ = ( 6 ) ( 2 ) =36 ft 2 2 2

π π = radian 180 6

1 2 1 π 2 π  r θ = ( 2 )   = ≈ 1.047 in 2 2 2 6 3

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Chapter 2: Trigonometric Functions

86. r = 3 meters; θ = 120º = 120 ⋅ A=

1 2 1 2  2π r θ = ( 3)  2 2  3

87. r = 2 feet; θ =

π

π 2π = radians 180 3

92. r = 40 inches; θ = 20º =

 2  =3π ≈ 9.425 m 

s = rθ = 40 ⋅

π 6

π π = radian 180 4 1 2 1 2 π  A = r θ = ( 4 )   = 2π ≈ 6.28 m 2 2 2 4

π π = radians 180 3 1 1 2  π  3π A = r 2θ = ( 3)   = ≈ 4.71 cm 2 2 2 3 2  

94. r = 3 cm; θ = 60º = 60 ⋅

radian

π 2π = ≈ 2.094 meters 6 3 1 1 4π 2 π  A = r 2θ = ( 4 )   = ≈ 4.189 m 2 2 2 6 3

s = rθ = 4 ⋅

89. r = 12 yards; θ = 70º = 70 ⋅

π 3π = radians 180 4 1 1 675π 2  3π  A = r 2θ = ( 30 )   = ≈ 1060.29 ft 2 2 2 4 2  

95. r = 30 feet; θ = 135º = 135 ⋅

π 7π = radians 180 18

96. r = 15 yards; A = 100 yd 2 1 A = r 2θ 2 1 2 100 = (15 ) θ 2 100 = 112.5θ 100 8 = ≈ 0.89 radian θ= 112.5 9

7π ≈ 14.661 yards 18 1 1 2  7π  2 A = r 2θ = (12 )   = 28π ≈ 87.965 yd 2 2 18  

s = rθ = 12 ⋅

90. r = 9 cm; θ = 50º = 50 ⋅

π 40π = ≈ 13.96 inches 9 9

93. r = 4 m; θ = 45º = 45 ⋅

radians

3 π 2π s = rθ = 2 ⋅ = ≈ 2.094 feet 3 3 1 1 2π 2 π  A = r 2θ = ( 2 )   = = ≈ 2.094 ft 2 2 2 3 3

88. r = 4 meters; θ =

π radian 9

π 5π = radian 180 18

°

5π ≈ 7.854 cm 18 1 1 2  5π  45π A = r 2θ = ( 9 )  ≈ 35.343 cm 2 = 2 2 18 4  

s = rθ = 9 ⋅

or

8 180  160  ⋅ =  ≈ 50.93° 9 π  π  1 2 1 2π θ = 120° = r1 θ − r2 2θ 2 2 3 1 2π 1 2π = (34) 2 − (9) 2 2 3 2 3

97. A =

91. r = 6 inches In 15 minutes, 15 1 π rev = ⋅ 360º = 90º = radians θ= 60 4 2 π s = rθ = 6 ⋅ = 3π ≈ 9.42 inches 2

=

1 2π 1 2π (1156) − (81) 2 3 2 3

= (1156)

In 25 minutes, 25 5 5π rev = ⋅ 360º = 150º = radians θ= 60 12 6 5π s = rθ = 6 ⋅ = 5π ≈ 15.71 inches 6

π 3

− (81)

π 3

1156π 81π − 3 3 1075π = ≈ 1125.74 in 2 3 =

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Section 2.1: Angles and Their Measure 104. r = 5 m; ω = 5400 rev/min = 10800π rad/min v = rω = (5) ⋅10800π m/min = 54000π cm/min cm 1m 1km 60 min ⋅ ⋅ ⋅ v = 54000π min 100cm 1000m 1hr ≈ 101.8 km/hr

1 2 1 25π θ = 125° = r1 θ − r2 2θ 2 2 36 1 25π 1 25π = (30) 2 − (6) 2 2 36 2 36

98. A =

=

1 25π 1 25π (900) − (36) 2 36 2 36

= (450)

105. d = 26 inches; r = 13 inches; v = 35 mi/hr 35 mi 5280 ft 12 in. 1 hr v= ⋅ ⋅ ⋅ hr mi ft 60 min = 36,960 in./min v 36,960 in./min ω= = 13 in. r ≈ 2843.08 radians/min 2843.08 rad 1 rev ≈ ⋅ min 2π rad ≈ 452.5 rev/min

25π 25π − (18) 36 36

11250π 450π − 36 36 10800π = = 300π ≈ 942.48 in 2 36 =

99. r = 5 cm; t = 20 seconds; θ =

1 radian 3

(1 / 3) 1 1 1 radian/sec = = ⋅ = 20 3 20 60 t 1 s rθ 5 ⋅ (1 / 3) 5 1 cm/sec v= = = = ⋅ = 20 3 20 12 t t

ω=

θ

106. r = 15 inches; ω = 3 rev/sec = 6π rad/sec v = rω = 15 ⋅ 6π in./sec = 90π ≈ 282.7 in/sec in. 1ft 1mi 3600sec v = 90π ⋅ ⋅ ⋅ ≈ 16.1 mi/hr sec 12in. 5280ft 1hr

100. r = 2 meters; t = 20 seconds; s = 5 meters

ω=

θ

( s / r ) (5 / 2)

= = 20 t t s 5 1 = m/sec v= = t 20 4

107. r = 3960 miles θ = 35º 9 '− 29º 57 ' = 5º12 ' = 5.2º π = 5.2 ⋅ 180 ≈ 0.09076 radian s = rθ = 3960(0.09076) ≈ 359 miles

5 1 1 = ⋅ = radian/sec 2 20 8

101. r = 25 feet; ω = 13 rev/min = 26π rad/min v = rω = 25 ⋅ 26π ft./min = 650π ≈ 2042.0 ft/min ft. 1mi 60 min ⋅ ⋅ ≈ 23.2 mi/hr v = 650π min 5280ft 1hr

108. r = 3960 miles θ = 38º 21'− 30º 20 ' = 8º1' ≈ 8.017º π = 8.017 ⋅ 180 ≈ 0.1399 radian s = rθ = 3960(0.1399) ≈ 554 miles

102. r = 6.5 m; ω = 22 rev/min = 44π rad/min v = rω = (6.5) ⋅ 44π m/min = 286π ≈ 898.5 m/min m 1km 60 min ⋅ ⋅ ≈ 53.9 km/hr v = 286π min 1000m 1hr 103. r = 4 m; ω = 8000 rev/min = 16000π rad/min v = rω = (4) ⋅16000π m/min = 64000π cm/min cm 1m 1km 60 min v = 64000π ⋅ ⋅ ⋅ min 100cm 1000m 1hr ≈ 120.6 km/hr

109. r = 3429.5 miles

ω = 1 rev/day = 2π radians/day = v = rω = 3429.5 ⋅

π radians/hr 12

π ≈ 898 miles/hr 12

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Chapter 2: Trigonometric Functions 115. r = 4 feet; ω = 10 rev/min = 20π radians/min v = rω = 4 ⋅ 20π ft = 80π min 80π ft 1 mi 60 min = ⋅ ⋅ min 5280 ft hr ≈ 2.86 mi/hr

110. r = 3033.5 miles π ω = 1 rev/day = 2π radians/day = radians/hr 12 π v = rω = 3033.5 ⋅ ≈ 794 miles/hr 12

111. r = 2.39 × 105 miles ω = 1 rev/27.3 days = 2π radians/27.3 days π radians/hr = 12 ⋅ 27.3 π v = rω = ( 2.39 × 105 ) ⋅ ≈ 2292 miles/hr 327.6

116. d = 26 inches; r = 13 inches; ω = 480 rev/min = 960π radians/min v = rω = 13 ⋅ 960π in = 12480π min 12480π in 1 ft 1 mi 60 min = ⋅ ⋅ ⋅ min 12 in 5280 ft hr ≈ 37.13 mi/hr v ω= r 80 mi/hr 12 in 5280 ft 1 hr 1 rev = ⋅ ⋅ ⋅ ⋅ 13 in 1 ft 1 mi 60 min 2π rad ≈ 1034.26 rev/min

112. r = 9.29 × 107 miles ω = 1 rev/365 days = 2π radians/365 days π radians/hr = 12 ⋅ 365 v = rω = ( 9.29 × 107 ) ⋅

π ≈ 66, 633 miles/hr 4380

113. r1 = 2 inches; r2 = 8 inches; ω1 = 3 rev/min = 6π radians/min Find ω2 : v1 = v2

117. d = 8.5 feet;

r = 4.25 feet;

v = 9.55 mi/hr

v 9.55 mi/hr = 4.25 ft r 9.55 mi 1 5280 ft 1 hr 1 rev = ⋅ ⋅ ⋅ ⋅ hr 4.25 ft mi 60 min 2π ≈ 31.47 rev/min

ω=

r1ω1 = r2ω2 2(6π) = 8ω2 12π 8 = 1.5π radians/min 1.5π = rev/min 2π 3 = rev/min 4

ω2 =

118. Let t represent the time for the earth to rotate 90 miles. t 24 = 90 2π(3559) 90(24) t= ≈ 0.0966 hours ≈ 5.8 minutes 2π(3559)

114. r = 30 feet 1 rev 2π π = = ≈ 0.09 radian/sec ω= 70 sec 70 sec 35 π rad 6π ft v = rω = 30 feet ⋅ = ≈ 2.69 feet/sec 35 sec 7 sec

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Section 2.1: Angles and Their Measure 123. We know that the distance between Alexandria and Syene to be s = 500 miles. Since the measure of the Sun’s rays in Alexandria is 7.2° , the central angle formed at the center of Earth between Alexandria and Syene must also be 7.2° . Converting to radians, we have

119. A = π r 2 = π (9) 2 = 81π 3 243 π . Now 81π = 4 4 we calculate the small area. A = πr2

We need ¾ of this area.

= π (3) = 9π

7.2° = 7.2° ⋅ s = rθ

2

500 = r ⋅

1 9 We need ¼ of the small area. 9π = π 4 4 243 9 252 π+ π= π = 63π So the total area is: 4 4 4 square feet.

r=

25

π

π 180°

=

π 25

radian . Therefore,

π 25

⋅ 500 =

12,500

C = 2π r = 2π ⋅

π

≈ 3979 miles

12,500

π

= 25, 000 miles.

The radius of Earth is approximately 3979 miles, and the circumference is approximately 25,000 miles.

120. First we find the radius of the circle. C = 2π r 8π = 2π r 4=r

124. a.

The area of the circle is A = π r 2 = π (4) 2 = 16π . The area of the sector of the circle is 4π . Now we calculate the area of the rectangle. A = lw A = (4)(4 + 7) A = 44

The length of the outfield fence is the arc length subtended by a central angle θ = 96° with r = 200 feet. s = r ⋅ θ = 200 ⋅ 96° ⋅

π

≈ 335.10 feet 180° The outfield fence is approximately 335.1 feet long.

b. The area of the warning track is the difference between the areas of two sectors with central angle θ = 96° . One sector with r = 200 feet and the other with r = 190 feet. 1 1 θ A = R 2θ − r 2θ = ( R 2 − r 2 ) 2 2 2 96° π 2 = ⋅ ( 200 − 1902 ) 2 180° 4π = ( 3900 ) ≈ 3267.26 15 The area of the warning track is about 3267.26 square feet.

So the area of the rectangle that is outside of the circle is 44 − 4π u 2 . 121. The earth makes one full rotation in 24 hours. The distance traveled in 24 hours is the circumference of the earth. At the equator the circumference is 2π(3960) miles. Therefore, the linear velocity a person must travel to keep up with the sun is: s 2π(3960) v= = ≈ 1037 miles/hr t 24 122. Find s, when r = 3960 miles and θ = 1'. 1 degree π radians ⋅ ≈ 0.00029 radian θ = 1'⋅ 60 min 180 degrees s = rθ = 3960(0.00029) ≈ 1.15 miles Thus, 1 nautical mile is approximately 1.15 statute miles.

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Chapter 2: Trigonometric Functions 125. r1 rotates at ω1 rev/min , so v1 = r1ω1 . r2 rotates at ω2 rev/min , so v2 = r2ω2 . Since the linear speed of the belt connecting the pulleys is the same, we have: v1 = v2

135. x 2 − 9 cannot be zero so the domain is:

{ x | x ≠ ±3} 136. Shift to the left 3 units would give y = x + 3 .

r1ω1 = r2ω2

Reflecting about the x-axis would give y = − x + 3 . Shifting down 4 units would result

r1ω1 r2ω2 = r2ω1 r2ω1

in y = − x + 3 − 4 .

r1 ω2 = r2 ω1

x1 + x2 y1 + y2 , 2 2

137.

126. Answers will vary. 127. If the radius of a circle is r and the length of the arc subtended by the central angle is also r, then the measure of the angle is 1 radian. Also, 180 1 radian = degrees .

=

3 + ( −4) 6 + 2 , 2 2

=

1 −1 8 , = − ,4 2 2 2

π

1° =

1 revolution 360

Section 2.2

 π radians  128. Note that 1° = 1° ⋅   ≈ 0.017 radian  180°   180°  and 1 radian ⋅   ≈ 57.296° .  π radians  Therefore, an angle whose measure is 1 radian is larger than an angle whose measure is 1 degree.

1. c 2 = a 2 + b 2 2.

3. True 4. equal; proportional

129. Linear speed measures the distance traveled per unit time, and angular speed measures the change in a central angle per unit time. In other words, linear speed describes distance traveled by a point located on the edge of a circle, and angular speed describes the turning rate of the circle itself.

 1 3 5.  − ,   2 2 

6. −

can be rewritten as follows: s = rθ =

180

8.

rθ .

10. a 11.

f ( x) = 3x + 7 0 = 3x + 7 3 x = −7 → x = −

( 0,1)

 2 2 9.  ,   2 2 

131 – 133. Answers will vary. 134.

1 2

7. b

130. This is a true statement. That is, since an angle measured in degrees can be converted to radian measure by using the formula 180 degrees = π radians , the arc length formula

π

f ( 5 ) = 3 ( 5 ) − 7 = 15 − 7 = 8

y x ; r r

12. False 7 3

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Section 2.2: Trigonometric Functions: Unit Circle Approach

 3 1 3 13. P =  ,   x = , y= 2  2 2 1 sin t = y = 2 3 cos t = x = 2 1   y 1 2 2 = tan t = =   = ⋅ x  3 2 3    2  1 1 2 csc t = = = 1⋅ = 2 y 1 1   2

 2 21  2 21 15. P =  − ,  x=− , y = 5 5  5 5 

1 2

21 5 2 cos t = x = − 5  21  y  5  21  5  21 tan t = = = − = − x  2 5  2 2 −   5

sin t = y =

1 3

3 3

=

3 3

1 1 5 5 21 5 21 = = 1⋅ = ⋅ = 21 y  21  21 21 21    5  1 1 5  5 sec t = = = 1 −  = − 2  2 x  2 −   5  2 − 2 5 x  5  cot t = = =− ⋅ y  21  5 21    5  csc t =

1 1 2 2 3 2 3 = = 1⋅ = ⋅ = x  3 3 3 3 3    2   3   x  2  3 2 cot t = =  = ⋅ = 3 y 2 1 1   2

sec t =

=−

1 3 1 3 14. P =  , −  x= , y=− 2 2  2 2 sin t = y = −

3 2

2 21

21 21

=−

2 21 21

 1 2 6 1 2 6 16. P =  − ,  x=− , y = 5 5  5 5 

1 2  3 − y  2  3 2 =− ⋅ =− 3 tan t = = x 2 1 1   2

cos t = x =

2 6 5 1 cos t = x = − 5 2 6  2 6 y   5 tan t = =  5  = − = −2 6 x 5  1  1 −   5

sin t = y =

1 1 2 2 3 2 3 = =− =− ⋅ =− 3 y  3 3 3 3 −   2  1 1 2 sec t = = = 1⋅ = 2 1 x 1   2 1   x 1 2 1 3 3 cot t = =  2  = − ⋅ =− ⋅ =− y  2 3  3 3 3 3 −   2  csc t =

csc t =

1 1 5 5 6 5 6 = = 1⋅ = ⋅ = y 2 6 12 2 6 2 6 6    5 

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Chapter 2: Trigonometric Functions

1 1  5 = = 1 −  = −5 x  1  1 −   5  1 −  1 5  x cot t = =  5  = −  5  2 6  y 2 6    5 

sec t =

=−

1 2 6

6 6

=−

csc t =

1 = x      x  cot t = =  y   

sec t =

6 12

 2 2 2 2 17. P =  − , , y=  x=− 2 2  2 2  sin t =

2 2

cos t = x = −

1 2  2  2  2  2  2 

= 1⋅

2 2

=

2 2

2 2

= 2

=1

 2 2 1 2 2 1 19. P =  ,−  x= , y=− 3 3 3  3 1 sin t = y = − 3

2 2

 2   y tan t = =  2  = −1 x  2 −   2 

csc t =

1 1 2 2 2 = = 1⋅ = ⋅ = 2 y  2 2 2 2    2 

cos t = x =

2 2 3

 1 −  y 1 3 tan t = =  3  = − ⋅ x 2 2 3 2 2    3 

1 1 2 2 2 = = 1⋅ = ⋅ = 2 y  2 2 2 2    2 

=−

1 1 2 2  2  sec t = = = 1 − =− ⋅ =− 2  x  2 2 2 2  −   2 

csc t =

 2 − x  2  cot t = = = −1 y  2    2 

1 2 2

2 2

=−

2 4

1 1  3 = = 1 −  = −3 y  1  1 −   3

1 1 3 2 3 2  3  = = 1 ⋅ = = x 2 2 4 2 2 2 2 2    3  2 2  2 2 x   3 cot t = =  3  =  −  = −2 2 y 3  1  1 −   3 sec t =

 2 2 2 2 , , y= 18. P =    x = 2 2  2 2  2 2 2 cos t = x = 2  2   y  2  tan t = = =1 x  2    2 

sin t = y =

 5 2 5 2 20. P =  − , −   x = − , y=− 3 3 3  3 2 sin t = y = − 3 cos t = x = −

5 3

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Section 2.2: Trigonometric Functions: Unit Circle Approach

 11π   3π 8π  25. csc   = csc  +  2 2     2  3π  = csc  + 4π   2 

 2 −  y 2 3  3 tan t = =  = − −  x  3 5 5  −   3 

=

2

=

 3π  = csc  + 2 ⋅ 2π   2   3π  = csc    2  = −1

2 5 5

5 1 1 3  3 = 1 −  = − csc t = = y  2 2 2  −   3 sec t =

5

5

1 1  3  = = 1 −  x   5 5  −  3   

=−

3

5

=−

26. sec ( 8π ) = sec ( 0 + 8π )

= sec ( 0 + 4 ⋅ 2π ) = sec ( 0 ) = 1

 3π   3π  27. cos  −  = cos    2   2   π 4π  = cos  −  2 2  π  = cos  + (−1) ⋅ 2π  2 

3 5 5

5 5  5 −   3  x 5  3 5 cot t = =  =− −  = y 3  2 2  2 −   3

π = cos   2 =0

 11π   3π 8π  21. sin   = sin  +  2 2     2  3π  = sin  + 4π   2 

28. sin ( −3π ) = − sin ( 3π )

= − sin ( π + 2π ) = − sin ( π ) = 0

 3π  = sin  + 2 ⋅ 2π  2    3π  = sin    2  = −1

29. sec ( −π ) = sec ( π ) = −1 30. tan ( −3π ) = − tan(3π)

= − tan ( 0 + 3π ) = − tan ( 0 ) = 0

22. cos ( 7π ) = cos ( π + 6π )

= cos ( π + 3 ⋅ 2π ) = cos ( π ) = −1

23. tan ( 6π ) = tan(0 + 6π) = tan ( 0 ) = 0  7π   π 6π  24. cot   = cot  +   2  2 2  π  π = cot  + 3π  = cot   = 0 2   2

31. sin 45º + cos 60º =

2 1 1+ 2 + = 2 2 2

32. sin 30º − cos 45º =

1 2 1− 2 − = 2 2 2

33. sin 90º + tan 45º = 1 + 1 = 2 34. cos180º − sin180º = −1 − 0 = −1 35. sin 45º cos 45º =

2 2 2 1 ⋅ = = 2 2 4 2

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Chapter 2: Trigonometric Functions

36. tan 45º cos 30º = 1 ⋅

2π 1  2 = = 1 −  = −2 3  1  1 −   2  1 − 2π  2  1 2 1 3 3 cot = =− ⋅ =− ⋅ =− 3  3 2 3 3 3 3    2 

3 3 = 2 2

cos

37. csc 45º tan 60º = 2 ⋅ 3 = 6 38. sec 30º cot 45º =

2 3 2 3 ⋅1 = 3 3

39. 4sin 90º − 3 tan180º = 4 ⋅1 − 3 ⋅ 0 = 4

48. The point on the unit circle that corresponds to  3 1 5π = 150º is  − θ= , .  2 2 6 5π 1 sin = 6 2

40. 5cos 90º − 8sin 270º = 5 ⋅ 0 − 8(−1) = 8 41. 2sin

π π 3 3 − 3 tan = 2 ⋅ − 3⋅ = 3− 3 =0 3 6 2 3

42. 2sin

π π 2 + 3 tan = 2 ⋅ + 3 ⋅1 = 2 + 3 4 4 2

43. 2sec

π π 3 4 3 + 4 cot = 2 ⋅ 2 + 4 ⋅ =2 2+ 4 3 3 3

44. 3csc

π π 2 3 + cot = 3 ⋅ +1 = 2 3 +1 3 4 3

45. csc

5π 3 =− 6 2 1   5π 1  2  3 3 tan =  2  = ⋅ − =− ⋅ 6  3 3 3 3 2  −   2  5π 1 2 csc = = 1⋅ = 2 1 6   1   2 cos

π π + cot = 1 + 0 = 1 2 2

46. sec π − csc

sec

π = −1 − 1 = − 2 2

 3 −  5π  2  3 2 cot = =− ⋅ =− 3 6 2 1 1   2

47. The point on the unit circle that corresponds to  1 3 2π = 120º is  − , θ= . 3  2 2 

49. The point on the unit circle that corresponds to  7π 3 1 θ = 210º = is  − ,− . 6  2 2 1 sin 210º = − 2 3 cos 210º = − 2  1 −  1  2  3 3 2 = − ⋅ − = tan 210º =  ⋅ 2 3   3 3 3    − 2    1  2 csc 210º = = 1⋅  −  = − 2  1  1 −  2  

2π 3 = 3 2 2π 1 cos =− 3 2

sin

 3 2π  2  3  2 tan = = ⋅ −  = − 3 3  1 2  1 −   2 csc

 2  3 5π 1 2 3 = = 1⋅  − =− ⋅ 6  3 3 3 3  −   2 

2π 1 2 3 2 3 = = ⋅ = 3  3 3 3 3    2 

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Section 2.2: Trigonometric Functions: Unit Circle Approach

 2  3 2 3 = 1⋅  − =− ⋅ 3  3 3 3   −   2   3  −  2  3  2  cot 210º = =− ⋅ −  = 3 2  1  1 −    2

sec 210º =

1

csc

sec

3π 1 2 2  2  2 = = 1⋅  − =− =− 2 ⋅ 4  2  2 2   2 −   2 

 2 − 3π  2  2 2 = =− ⋅ = −1 cot 4 2  2 2    2 

50. The point on the unit circle that corresponds to  1 4π 3 θ = 240º = is  − , − . 3  2 2 

52. The point on the unit circle that corresponds to  2 2 11π θ= , = 495º is  − .  2 2  4

3 2 1 cos 240º = − 2  3 −   2  = − 3 ⋅ − 2  = 3 tan 240º =   1 2  1   −   2

sin 240º = −

11π 2 = 4 2 11π 2 cos =− 4 2  2   11π 2  2  =  2  = ⋅ − tan  = −1 4 2   2  2 −   2 

sin

 2  3 2 3 = 1⋅  − =− ⋅ 3  3 3 3  −   2  1  2 sec 240º = = 1⋅  −  = − 2 1    1 −   2  1 −  1  2  3 3 cot 240º =  2  = − ⋅  − = ⋅ 2  3  3 3 3 −   2  csc 240º =

3π 1 2 2 2 2 = = 1⋅ ⋅ = = 2 4  2 2 2 2    2 

1

csc

sec

11π 1 2 2 2 2 = = 1⋅ ⋅ = = 2 4 2  2 2 2    2  11π 1 2 2  2  2 = = 1⋅  − =− =− 2 ⋅ 4 2   2 2 2   −   2 

 2 − 11π  2  2 2 = =− ⋅ = −1 cot 4 2  2 2    2 

51. The point on the unit circle that corresponds to  2 2 3π θ= , = 135º is  − .  2 2  4 3π 2 = 4 2 3π 2 cos =− 4 2  2   3π 2  2  =  2  = ⋅ − tan  = −1 4  2  2 2 −   2 

sin

53. The point on the unit circle that corresponds to  1 3 8π θ= = 480º is  − , . 3  2 2  8π 3 = 3 2 8π 1 cos =− 3 2

sin

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Chapter 2: Trigonometric Functions 55. The point on the unit circle that corresponds to  2 2 9π , θ = 405º = is  .  2 2  4

 3 8π  2  3  2 = = ⋅ −  = − 3 tan 3  1 2  1 −    2

2 2 2 cos 405º = 2  2   2 2 tan 405º =  2  = ⋅ =1 2  2 2    2 

sin 405º =

8π 1 2 3 2 3 csc = = 1⋅ ⋅ = 3  3 3 3 3    2  8π 1  2 sec = = 1 ⋅  −  = −2 3  1  1 −   2  1 − 8π  2  1 2 3 3 = =− ⋅ ⋅ =− cot 3  3 2 3 3 3    2 

csc 405º =

sec 405º =

54. The point on the unit circle that corresponds to  3 1 13π = 390º is  θ= , .  2 2 6 13π 1 sin = 6 2

1  2    2 

  cot 405º =    

13π 3 = 6 2 1   13π 1 2 3 3 2 =   = ⋅ ⋅ = tan 6 3  3 2 3 3  2    13π 1 2 csc = = 1⋅ = 2 6 1 1   2  

cos

sec

1 2 2 = 1⋅ ⋅ = 2  2 2 2    2  = 1⋅

2 2

2 2

= 2

2  2  =1 2  2 

56. The point on the unit circle that corresponds to  3 1 13π , . θ = 390º = is  6  2 2 1 sin 390º = 2 3 cos 390º = 2 1   1 2 3 3 tan 390º =  2  = ⋅ ⋅ = 3  3 2 3 3    2  1 2 csc 390º = = 1⋅ = 2 1 1   2

13π 1 2 3 2 3 = = 1⋅ ⋅ = 6 3  3 3 3  2   

 3   13π  2  3 2 cot = = ⋅ = 3 1 6 2 1     2

sec 390º =

1  3    2 

= 1⋅

2 3

3 3

=

2 3 3

 3   3 2 cot 390º =  2  = ⋅ = 3 2 1 1   2

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Section 2.2: Trigonometric Functions: Unit Circle Approach 57. The point on the unit circle that corresponds to  3 1 π θ = − = − 30º is  , −  .  2 2 6 1  π sin  −  = − 2  6

59. The point on the unit circle that corresponds to  2 2 3π θ = −135º = − is  − ,− .  2 2  4 2 2 2 cos ( −135º ) = − 2  2 −  2  2  tan ( −135º ) =  2  = − ⋅ −  =1 2   2 2 −   2 

sin ( −135º ) = −

3  π cos  −  = 6 2    1 −  1 2 3 3 2  π tan  −  =  =− ⋅ ⋅ =− 2 3 3 3  6  3    2  1  π −2 csc  −  =  6  − 1     2 1 2 3 2 3  π = ⋅ = sec  −  = 6 3   3 3   3    2   3    3  π  2 cot  −  =  2  =  ⋅ −  = − 3  6   − 1   2   1     2

3

3 3

sec ( −135º ) =

 2  2 = 1⋅  − = 2 ⋅  2 2 2  −   2  1

2  2  =1 2  2 

60. The point on the unit circle that corresponds to  1 3 4π is  − , θ = − 240º = − . 3  2 2 

3  π sin  −  = − 2  3  π 1 cos  −  =  3 2  3 −   3 2  π ⋅ =− 3 tan  −  =  2  = − 1 3 2 1       2 3

1  2  2 = 1⋅  − =− 2 ⋅   2 2 2  −   2 

 − cot ( −135º ) =   − 

58. The point on the unit circle that corresponds to 1 π 3 θ = − = − 60º is  , − . 3 2 2 

1  2   π csc  −  = = 1⋅  − ⋅ 3  3  3  −   2  1 2  π = 1⋅ = 2 sec  −  = 1  3 1   2 1   π 1  2  2   cot  −  =   = ⋅  − ⋅ 3  3  3 2  −   2 

csc ( −135º ) =

3 2 1 cos ( − 240º ) = − 2  3   3  2 ⋅ −  = − 3 tan ( − 240º ) =  2  = 2  1  1 −   2

sin ( −240º ) =

=−

1  2  3 2 3 = 1⋅  = ⋅ 3  3  3 3    2  1  2 = 1 ⋅  −  = −2 sec ( − 240º ) =  1  1 −   2  1 −  1  2  3 3 =− cot ( −240º ) =  2  = − ⋅  ⋅ 2  3 3 3  3    2 

2 3 3

=−

csc ( −240º ) =

3 3

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Chapter 2: Trigonometric Functions 61. The point on the unit circle that corresponds to 5π θ= = 450º is ( 0, 1) . 2 5π 5π 1 =1 = =1 sin csc 2 2 1 5π 5π 1 sec =0 = = undefined cos 2 0 2 5π 1 5π 0 tan = = undefined cot = =0 2 0 2 1

64. The point on the unit circle that corresponds to  3 1 13π θ =− = −390° is  , −  . 6 2  2

1 3  13π   13π  sin  − cos  − =− = 2  6   6  2  1 −  1 2 3 3  13π   2  =− ⋅ ⋅ =− tan  − = 2 3 3 3  6   3    2  1  13π  csc  − = 1 − 2  6  −     2

62. The point on the unit circle that corresponds to θ = 5π = 900º is ( −1, 0 ) .

cos 5π = −1 tan 5π =

1 = undefined 0 1 = −1 sec 5π = −1 −1 cot 5π = = undefined 0 csc 5π =

sin 5π = 0

0 =0 −1

1 2 3 2 3  13π  sec  − = ⋅ = = 3 3 3  6   3    2   3    13π   2   3   2  cot  − = ⋅ −  = − 3 =  6   − 1   2   1     2

63. The point on the unit circle that corresponds to  1 3 14π θ =− = −840° is  − , − . 2  3  2

65. Set the calculator to degree mode: sin 28º ≈ 0.47 .

 14π sin  −  3

3 1   14π  cos  − =− =− 2 2   3   3 −   3  2  14π   2  tan  − = =− ⋅ −  = 3  2  1  3  − 1     2

 14π csc  −  3  14π sec  −  3

 14π cot  −  3

1  2   = 1⋅  − ⋅ =  3   3  −   2  1   2 = 1⋅  −  = − 2 =  − 1   1   2   1   −  1  2    2 = − ⋅ − ⋅  2  3  3 −   2 

3 3

3 3

=−

=

66. Set the calculator to degree mode: cos14º ≈ 0.97 .

2 3 3

67. Set the calculator to degree mode: 1 sec 21º = ≈ 1.07 . cos 21°

3 3

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Section 2.2: Trigonometric Functions: Unit Circle Approach 74. Set the calculator to radian mode: tan 1 ≈ 1.56 .

68. Set the calculator to degree mode: 1 cot 70º = ≈ 0.36 . tan 70º

75. Set the calculator to degree mode: sin1º ≈ 0.02 . 69. Set the calculator to radian mode: tan

π ≈ 0.32 . 10 76. Set the calculator to degree mode: tan1º ≈ 0.02 .

70. Set the calculator to radian mode: sin

π ≈ 0.38 . 8 77. For the point (−3, 4) , x = −3 , y = 4 ,

r = x 2 + y 2 = 9 + 16 = 25 = 5 sin θ =

4 5

csc θ =

3 5 4 tan θ = − 3

5 3 3 cot θ = − 4

cos θ = −

71. Set the calculator to radian mode: π 1 cot = ≈ 3.73 . 12 tan π 12

5 4

sec θ = −

78. For the point (5, −12) , x = 5 , y = −12 ,

r = x 2 + y 2 = 25 + 144 = 169 = 13 12 13 5 cos θ = 13 12 tan θ = − 5

13 12 13 sec θ = 5 5 cot θ = − 12

sin θ = −

72. Set the calculator to radian mode: 5π 1 csc = ≈ 1.07 . 13 sin 5π 13

csc θ = −

79. For the point (2, −3) , x = 2 , y = −3 ,

r = x 2 + y 2 = 4 + 9 = 13 73. Set the calculator to radian mode: sin 1 ≈ 0.84 .

sin θ = cos θ =

−3 13 2

13 3 tan θ = − 2

⋅ ⋅

13 13 13 13

=− =

3 13 13

2 13 13

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csc θ = −

13 3

13 2 2 cot θ = − 3

sec θ =


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Chapter 2: Trigonometric Functions 84. For the point (0.3, 0.4) , x = 0.3 , y = 0.4 ,

80. For the point (−1, −2) , x = −1 , y = − 2 , 2

2

r = x 2 + y 2 = 0.09 + 0.16 = 0.25 = 0.5

r = x + y = 1+ 4 = 5 −2

sin θ =

5

cos θ =

5

−1

5 5

5 5 −2 tan θ = =2 −1

=−

2 5 5

=−

5 5

csc θ =

0.4 4 = 0.5 5 0.3 3 cos θ = = 0.5 5 0.4 4 tan θ = = 0.3 3

5 5 =− −2 2

5 =− 5 −1 −1 1 cot θ = = −2 2 sec θ =

2 2  2  2 + +  −  +  −  2 2  2   2  =0 =

r = x2 + y 2 = 4 + 4 = 8 = 2 2 −2

2 2

cos θ = tan θ =

−2

2 2 −2 −2

2 2 2 2

=−

2 2

=−

2 2

csc θ = sec θ = cot θ =

=1

2 2 −2 2 2 −2 −2 −2

=− 2

 3 86. tan 60º + tan150º = 3 +  −   3 

=− 2

=

=1

= sin 40º + sin130º + sin ( 40º +180º ) +

r = x2 + y 2 = 1 + 1 = 2

cos θ =

1 2 −1 2

⋅ ⋅

2 2 2 2

1 tan θ = = −1 −1

=

2 2

=−

csc θ = 2 2

3 3− 3 2 3 = 3 3

87. sin 40º + sin130º + sin 220º + sin 310º

82. For the point (−1, 1) , x = −1 , y = 1 ,

sin θ =

csc θ =

85. sin 45º + sin135º + sin 225º + sin 315º

81. For the point (−2, −2) , x = − 2 , y = − 2 ,

sin θ =

0.5 5 = 0.4 4 0.5 5 sec θ = = 0.3 3 0.3 3 cot θ = = 0.4 4

sin θ =

sin (130º +180º )

2 = 2 1

= sin 40º + sin130º − sin 40º − sin130º =0

2 =− 2 −1 −1 cot θ = = −1 1

sec θ =

88. tan 40º + tan140º = tan 40º + tan (180º −40º ) = tan 40º − tan 40º =0

1 1 1 1 83. For the point  ,  , x = , y = , 3 4 3 4  

89. If f (θ ) = sin θ = 0.1 , then

f (θ + π ) = sin(θ + π) = − 0.1 .

1 1 25 5 + = = 9 16 144 12 1 5 1 12 3 5 4 5 4 12 sin θ = csc θ = = ⋅ = = ⋅ = 5 4 5 5 1 12 1 3 12 4 1 5 5 3 5 1 12 4 sec θ = 12 = ⋅ = cos θ = 3 = ⋅ = 1 12 1 4 5 3 5 5 3 12 1 1 1 3 3 1 4 4 tan θ = 4 = ⋅ = cot θ = 3 = ⋅ = 1 4 1 4 1 3 1 3 3 4 r = x2 + y 2 =

90. If f (θ ) = cos θ = 0.3 , then

f (θ + π ) = cos(θ + π) = − 0.3 .

91. If f (θ ) = tan θ = 3 , then f (θ + π ) = tan(θ + π) = 3 .

92. If f (θ ) = cot θ = − 2 , then f (θ + π ) = cot(θ + π) = −2 .

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Section 2.2: Trigonometric Functions: Unit Circle Approach

93. If sin θ =

94. If cos θ =

95.

1 1 5 , then csc θ = = 1⋅ = 5 . 1 5 1   5

f h

π 6

108. g ( p (60°) ) = cos

6

π 3

3 2

=

60° 2

= cos 30° =

3 2

f ( 60º ) = sin ( 60º ) =

π

= sin 2 = sin

2 1 3 3 = 1⋅ = . , then sec θ = 2 3 2 2     3

96. g ( 60º ) = cos ( 60º ) = 97.

107.

3 2

cos 315° 2 1 = cos 315° 2 1 2 2 = ⋅ = 2 2 4

109. p ( g (315°) ) =

1 2

1  60º   60º  f  = sin   = sin ( 30º ) = 2  2   2 

3  60º   60º  98. g   = cos   = cos ( 30º ) = 2  2   2 

110. h f

5π 6

= 2 sin

5π 6

= 2⋅

1 =1 2

2

2

99.  f ( 60º )  = ( sin 60º )

2

 3 3 =   = 4  2 

111. a.

2

2 1 2 1 100.  g ( 60º )  = ( cos 60º ) =   = 2 4  

101.

103. 2 f ( 60º ) = 2sin ( 60º ) = 2 ⋅ 104. 2 g ( 60º ) = 2 cos ( 60º ) = 2 ⋅ 105.

 2 π ,  is on the graph of f −1 . b. The point   2 4

3 f ( 2 ⋅ 60º ) = sin ( 2 ⋅ 60º ) = sin (120º ) = 2

102. g ( 2 ⋅ 60º ) = cos ( 2 ⋅ 60º ) = cos (120º ) = −

c.

1 2

= −2 π  The point  , − 2  is on the graph of 4  π  y = f x + −3 .  4

1 =1 2

106. g ( − 60º ) = cos ( − 60º ) = cos ( 300º ) =

π π  π  f  + −3 = f  −3 4 4 2 π  = sin   − 3 2 = 1− 3

3 = 3 2

f ( − 60º ) = sin ( − 60º ) = sin ( 300º ) = −

2 π  π  f   = sin   = 2 4 4 π 2 The point  ,  is on the graph of f. 4 2 

3 2

1 2

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Chapter 2: Trigonometric Functions

112. a.

3 π  π  g   = cos   = 6 6 2 π 3 The point  ,  is on the graph of g. 6 2 

116.

 3 π ,  is on the graph of g −1 . b. The point   2 6

c.

π π  2 g  −  = 2 g (0) 6 6 = 2 cos(0) = 2 ⋅1

0.5

−0.1224

0.4

−0.0789

−0.1973

0.2

−0.0199

−0.0997

0.1

−0.0050

−0.0050

0.01

−0.00005

−0.0050

θ

−0.2448

0.001

0.0000

−0.0005

0.0001

0.0000

−0.00005

0.00001

0.0000

−0.000005

cos θ − 1

θ

approaches 0 as θ

approaches 0. 117. Use the formula R (θ ) =

v0 2 sin ( 2θ ) g

with

g = 32.2ft/sec 2 ; θ = 45º ; v0 = 100 ft/sec :

113. Answers will vary. One set of possible answers 11π 5π π 7π 13π ,− , , , is − . 3 3 3 3 3

R ( 45º ) =

(100) 2 sin(2 ⋅ 45º ) ≈ 310.56 feet 32.2

Use the formula H (θ ) =

114. Answers will vary. One ser of possible answers 13π 5π 3π 11π 19π is − ,− , , , 4 4 4 4 4 115.

cos θ − 1

g (θ ) =

=2 π  Thus, the point  , 2  is on the graph of 6  π   y = 2g  x −  . 6 

cos θ − 1

θ

v0 2 ( sin θ )

2

with

2g

g = 32.2ft/sec 2 ; θ = 45º ; v0 = 100 ft/sec :

H ( 45º ) =

sin θ

1002 (sin 45º ) 2 ≈ 77.64 feet 2(32.2)

θ

sin θ

0.5

0.4794

0.9589

0.4

0.3894

0.9735

0.2

0.1987

0.9933

g = 9.8 m/sec 2 ; θ = 30º ; v0 = 150 m/sec :

0.1

0.0998

0.9983

R ( 30º ) =

θ

0.01

0.0100

1.0000

0.001

0.0010

1.0000

0.0001

0.0001

1.0000

118. Use the formula R (θ ) =

f (θ ) =

sin θ

θ

g

with

1502 sin(2 ⋅ 30º ) ≈ 1988.32 m 9.8

Use the formula H (θ ) =

0.00001 0.00001 1.0000

v0 2 sin ( 2θ )

v0 2 ( sin θ )

2

with

2g

g = 9.8 m/sec 2 ; θ = 30º ; v0 = 150 m/sec :

approaches 1 as θ approaches 0.

H ( 30º ) =

1502 (sin 30º ) 2 ≈ 286.99 m 2(9.8)

119. Use the formula R (θ ) =

v0 2 sin ( 2θ ) g

with

g = 9.8 m/sec 2 ; θ = 25º ; v0 = 500 m/sec : R ( 25º ) =

5002 sin(2 ⋅ 25º ) ≈ 19, 541.95 m 9.8

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Section 2.2: Trigonometric Functions: Unit Circle Approach

Use the formula H (θ ) =

v0 2 ( sin θ )

2

distance on road rate on road 8 − 2x = 8 x = 1− 4 1 = 1 − tan θ 4 1 = 1− 4 tan θ

123. Note: time on road =

with

2g

g = 9.8 m/sec ; θ = 25º ; v0 = 500 m/sec : 2

H ( 25º ) =

5002 (sin 25º ) 2 ≈ 2278.14 m 2(9.8)

120. Use the formula R (θ ) =

v0 2 sin ( 2θ ) g

with

g = 32.2ft/sec 2 ; θ = 50º ; v0 = 200 ft/sec : R ( 50º ) =

2002 sin(2 ⋅ 50º ) ≈ 1223.36 ft 32.2

Use the formula H (θ ) =

v0 2 ( sin θ )

a.

2

2g

2 1 − 3sin 30º 4 tan 30º 2 1 = 1+ − 1 1 3⋅ 4⋅ 2 3

T (30º ) = 1 +

with

g = 32.2ft/sec 2 ; θ = 50º ; v0 = 200 ft/sec : H ( 50º ) =

2002 (sin 50º ) 2 ≈ 364.49 ft 2(32.2)

121. Use the formula t (θ ) =

= 1+

Sally is on the paved road for 1 1− ≈ 0.57 hr . 4 tan 30º

2a with g sin θ cos θ

g = 32 ft/sec 2 and a = 10 feet : a.

t ( 30 ) =

b.

t ( 45 ) =

c.

t ( 60 ) =

2 (10 ) 32sin 30º ⋅ cos 30º 2 (10 ) 32sin 45º ⋅ cos 45º 2 (10 ) 32sin 60º ⋅ cos 60º

4 3 − ≈ 1.9 hr 3 4

b.

≈ 1.20 seconds

2 1 − 3sin 45º 4 tan 45º 2 1 = 1+ − 1 4 ⋅1 3⋅ 2

T (45º ) = 1 +

≈ 1.12 seconds

2 2 1 − ≈ 1.69 hr 3 4 Sally is on the paved road for 1 1− = 0.75 hr . 4 tan 45o = 1+

≈ 1.20 seconds

122. Use the formula x (θ ) = cos θ + 16 + 0.5cos(2θ ) . x ( 30 ) = cos ( 30º ) + 16 + 0.5cos(2 ⋅ 30º )

c.

2 1 − o 3sin 60 4 tan 60o 2 1 = 1+ − 3 4⋅ 3 3⋅ 2 4 1 = 1+ − 3 3 4 3 ≈ 1.63 hr

T (60º ) = 1 +

= cos ( 30º ) + 16 + 0.5cos ( 60º ) ≈ 4.90 cm x ( 45 ) = cos ( 45º ) + 16 + 0.5cos(2 ⋅ 45º ) = cos ( 45º ) + 16 + 0.5cos ( 90º ) ≈ 4.71 cm

Sally is on the paved road for 1 1− ≈ 0.86 hr . 4 tan 60o

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Chapter 2: Trigonometric Functions

d.

2 1 . − 3sin 90º 4 tan 90º But tan 90º is undefined, so we cannot use the function formula for this path. However, the distance would be 2 miles in the sand and 8 miles on the road. The total 2 5 time would be: + 1 = ≈ 1.67 hours. The 3 3 path would be to leave the first house walking 1 mile in the sand straight to the road. Then turn and walk 8 miles on the road. Finally, turn and walk 1 mile in the sand to the second house. T (90º ) = 1 +

127.

H ≈ 71

The tree is approximately 71 feet tall. 128.

124. When θ = 30º :

(1 + sec 30º ) 1 V ( 30º ) = π (2)3 ≈ 251.42 cm3 2 3 ( tan 30º ) 3

(1 + sec 60º ) 1 V ( 60º ) = π (2)3 ≈ 75.40 cm3 2 3 ( tan 60º ) 3

H 2 2D 22° 6 tan = 2 2D 2 D tan11° = 6

2 0.52° H tan = 2 2(384400) H = 768800 tan 0.26°

x + (1 − x 2 ) =

=

41 49

8 =0 49 Using the quadratic formula: 8 a = 1, b = −1, c = − 49 x2 − x −

3 = 15.4 tan11°

x=

Arletha is 15.4 feet from the car.

θ

H 126. tan = 2 2D 8° 555 tan = 2 2D 2 D tan 4° = 555 D=

H 2D

129. cos θ + sin 2 θ =

When θ = 60º :

D=

=

41 ; cos 2 θ + sin 2 θ = 1 49 Substitute x = cos θ ; y = sin θ and solve these simultaneous equations for y. 41 2 ; x + y2 = 1 x + y2 = 49 y 2 = 1 − x2

3

θ

θ

The moon has a radius of 1744 km.

(1 + sec 45º ) 1 V ( 45º ) = π (2)3 ≈ 117.88 cm3 2 3 ( tan 45º )

tan

tan

H = 3488

When θ = 45º :

125.

θ

H = 2 2D 20° H tan = 2 2(200) H = 400 tan10° tan

=

− ( −1) ± ( −1) 2 − 4(1)( − 498 ) 2

1± 1+ 2

32 49

=

81 49

2

=

1 ± 97 8 1 = or − 2 7 7

Since the point is in quadrant III then x = − and y 2 = 1 − −

555 = 3968 2 tan 4°

y=−

The tourist is 3968 feet from the monument.

1 7

2

= 1−

1 48 = 49 49

48 4 3 =− 49 7

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Section 2.2: Trigonometric Functions: Unit Circle Approach c.

1 130. cos 2 θ − sin θ = − ;cos 2 θ + sin 2 θ = 1 9 Substitute x = cos θ ; y = sin θ and solve these simultaneous equations for y. 1 x2 − y = − ; x2 + y 2 = 1 9 2 2 x = 1− y 1 (1 − y 2 ) − y = − 9 10 y2 + y − = 0 9 2 9 y + 9 y − 10 = 0 Using the quadratic formula: a = 9, b = 9, c = −10

y=

0

90°

R is largest when θ = 67.5º . sin θ − 0 sin θ = = tan θ . cos θ − 0 cos θ Since L is parallel to M, the slope of L is equal to the slope of M. Thus, the slope of L = tan θ .

132. Slope of M =

133. a.

When t = 1 , the coordinate on the unit circle is approximately (0.5, 0.8) . Thus, 1 ≈ 1.3 0.8 1 sec1 ≈ = 2.0 cos1 ≈ 0.5 0.5 0.8 0.5 tan1 ≈ = 1.6 cot1 ≈ ≈ 0.6 0.5 0.8 Set the calculator on RADIAN mode: sin1 ≈ 0.8

−9 ± 81 + 360 −9 ± 441 −9 ± 21 = = 18 18 18 Since the point is in quadrant II then −3 + 21 12 2 y= = = and 18 18 3 2 2 4 5 x2 = 1 − = 1− = 3 9 9 =

131. a.

20

45°

−(9) ± (9) 2 − 4(9)( −10) 18

x=−

Using the MAXIMUM feature, we find:

csc1 ≈

5 5 =− 9 3

R ( 60 ) = =

32

2

2

[sin ( 2 ⋅ 60º ) − cos ( 2 ⋅ 60º ) − 1]

2

[sin (120º ) − cos (120º ) − 1]

32 32

2

32

b. When t = 5.1 , the coordinate on the unit circle is approximately (0.4, −0.9) . Thus, 1 ≈ −1.1 −0.9 1 cos 5.1 ≈ 0.4 sec 5.1 ≈ = 2.5 0.4 −0.9 0.4 tan 5.1 ≈ ≈ −2.3 cot 5.1 ≈ ≈ −0.4 0.4 −0.9

sin 5.1 ≈ −0.9

 3  1  −  −  − 1  2  2 

≈ 32 2 

≈ 16.56 ft

b. Let Y1 = 20

45°

0

322 2 sin ( 2 x ) − cos ( 2 x ) − 1 32 

csc 5.1 ≈

Set the calculator on RADIAN mode:

90°

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Chapter 2: Trigonometric Functions 134. a.

When t = 2 , the coordinate on the unit circle is approximately (−0.4, 0.9) . Thus,

138.

1 ≈ 1.1 0.9 1 sec 2 ≈ = −2.5 cos 2 ≈ −0.4 −0.4 0.9 −0.4 tan 2 ≈ = −2.3 cot 2 ≈ ≈ −0.4 −0.4 0.9 Set the calculator on RADIAN mode:

sin 2 ≈ 0.9

csc 2 ≈

b. When t = 4 , the coordinate on the unit circle is approximately (−0.7, −0.8) . Thus, 1 ≈ −1.3 −0.8 1 cos 4 ≈ −0.7 sec 4 ≈ ≈ −1.4 −0.7 −0.8 −0.7 tan 4 ≈ cot 4 ≈ ≈ 1.1 ≈ 0.9 −0.7 −0.8 Set the calculator on RADIAN mode: sin 4 ≈ −0.8

csc 4 ≈

Answers will vary. 139.

135 – 137. Answers will vary.

2 x−7 2 x= y−7 x( y − 7) = 2 xy − 7 x = 2 y=

xy = 7 x + 2 7x + 2 2 = 7+ x x 2 f 1 ( x) = 7 + x y=

140. 180

141.

−13π = −780° 3

f ( x + 3) = = =

( x + 3) − 1 ( x + 3) 2 + 2 x+2 x2 + 6 x + 9 + 2 x+2 x 2 + 6 x + 11

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is


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Section 2.3: Properties of the Trigonometric Functions

142. d = (1 − 3) 2 + (0 − ( −6)) 2

19. cos

= ( −2) 2 + (6) 2 = 4 + 36 = 40 = 2 10

Section 2.3

20. sin

1  1 1. All real numbers except − ;  x | x ≠ −  2 2 

22. csc

3. False 4. True 5. 2π , π

π

23. sec

2

7. b. 8. a 9. 1 10. False; sec θ =

24. cot

1 cos θ

11. sin 405º = sin(360º + 45º ) = sin 45º =

2 2

1 12. cos 420º = cos(360º + 60º ) = cos 60º = 2

25. tan

13. tan 405º = tan(180º + 180º + 45º ) = tan 45º = 1

26. sec

14. sin 390º = sin(360º + 30º ) = sin 30º =

9π π 2 π  = sin  + 2π  = sin = 4 4 4 2  

21. tan ( 21π ) = tan(0 + 21π) = tan ( 0 ) = 0

2. even

6. All real number, except odd multiples of

33π π  π  = cos  + 8π  = cos  + 4 ⋅ 2π  4 4 4     π = cos 4 2 = 2

1 2

9π π  π  = csc  + 4π  = csc  + 2 ⋅ 2π  2 2  2  π = csc 2 =1 17 π π  π  = sec  + 4π  = sec  + 2 ⋅ 2π  4 4  4  π = sec 4 = 2 17 π π  π  = cot  + 4π  = cot  + 2 ⋅ 2π  4 4  4  π = cot 4 =1

19π π 3 π  = tan  + 3π  = tan = 6 6 6 3   25π π  π  = sec  + 4π  = sec  + 2 ⋅ 2π  6 6 6     π = sec 6 =

15. csc 450º = csc(360º + 90º ) = csc 90º = 1 16. sec 540º = sec(360º + 180º ) = sec180º = −1

2 3 3

27. Since sin θ > 0 for points in quadrants I and II, and cos θ < 0 for points in quadrants II and III, the angle θ lies in quadrant II.

17. cot 390º = cot(180º + 180º + 30º ) = cot 30º = 3

28. Since sin θ < 0 for points in quadrants III and IV, and cos θ > 0 for points in quadrants I and IV, the angle θ lies in quadrant IV.

18. sec 420º = sec(360º + 60º ) = sec 60º = 2

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Chapter 2: Trigonometric Functions 29. Since sin θ < 0 for points in quadrants III and IV, and tan θ < 0 for points in quadrants II and IV, the angle θ lies in quadrant IV.

1 1 5 = =− cos θ  3  3 −   5 1 3 cot θ = =− tan θ 4

sec θ =

30. Since cos θ > 0 for points in quadrants I and IV, and tan θ > 0 for points in quadrants I and III, the angle θ lies in quadrant I.

2 5 5 , cos θ = 5 5 2 5   sin θ  5  2 5 5 tan θ = = = ⋅ =2 cos θ 5  5 5    5 

37. sin θ =

31. Since cos θ > 0 for points in quadrants I and IV, and tan θ < 0 for points in quadrants II and IV, the angle θ lies in quadrant IV. 32. Since cos θ < 0 for points in quadrants II and III, and tan θ > 0 for points in quadrants I and III, the angle θ lies in quadrant III. 33. Since sec θ < 0 for points in quadrants II and III, and sin θ > 0 for points in quadrants I and II, the angle θ lies in quadrant II.

csc θ =

34. Since csc θ > 0 for points in quadrants I and II, and cos θ < 0 for points in quadrants II and III, the angle θ lies in quadrant II.

sec θ =

1 1 5 5 5 = = 1⋅ ⋅ = sin θ  2 5  2 2 5 5    5 

1 1 5 5 = = ⋅ = 5 cos θ  5  5 5    5  1 1 cot θ = = tan θ 2

3 4 35. sin θ = − , cos θ = 5 5  3 − sin θ  5  3 5 3 tan θ = = =− ⋅ =− 4 cos θ 5 4 4     5 1 1 5 csc θ = = =− sin θ  3  3 −   5 1 1 5 sec θ = = = cos θ  4  4   5 1 4 cot θ = =− tan θ 3

5 2 5 , cos θ = − 5 5   5 −  5 sin θ  =  − 5 ⋅ − 5  = 1 tan θ = =    cos θ  2 5   5   2 5  2 −   5 

38. sin θ = −

csc θ =

1 1  5  5 = = 1⋅  − =− 5 ⋅ sin θ  5 5 5  −   5 

1 1  5  5 5 = = − =− ⋅ cos θ  2 5   2 5  5 2 −  5   1 1 2 cot θ = = = 1⋅ = 2 tan θ  1  1   2 sec θ =

4 3 36. sin θ = , cos θ = − 5 5 4   sin θ 4  5 4 5 tan θ = =   = ⋅ −  = − cos θ  3  5  3  3 −   5 1 1 5 csc θ = = = 4 sin θ   4   5

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Section 2.3: Properties of the Trigonometric Functions

1 3 39. sin θ = , cos θ = 2 2 1   sin θ 1 2 3 3 2 =   = ⋅ ⋅ = tan θ = cos θ  3  2 3 3 3    2  1 1 2 csc θ = = = 1⋅ = 2 sin θ  1  1   2 sec θ =

cot θ =

sec θ =

cot θ =

1 1 4 2 = =− ⋅ = −2 2 tan θ  2 2 2  −   4 

2 2 1 , cos θ = − 3 3 2 2   sin θ  3  2 2 3 tan θ = = = ⋅− = −2 2 cos θ 3 1  1 −   3

42. sin θ =

1 1 2 3 2 3 = = ⋅ = cos θ  3  3 3 3    2  1 1 3 3 = = ⋅ = 3 tan θ  3  3 3  3   

1 1 3 2 3 2 = = 1⋅ ⋅ = sin θ  2 2  4 2 2 2    3  1 1  3 = = 1 ⋅  −  = −3 sec θ = cos θ  1   1 −   3 csc θ =

3 1 , cos θ = 2 2  3   sin θ  2  3 2 tan θ = = = ⋅ = 3 cos θ 2 1 1   2

40. sin θ =

cot θ =

1 1 2 3 2 3 = = 1⋅ ⋅ = sin θ  3  3 3 3    2  1 1 2 sec θ = = = 1⋅ = 2 cos θ  1  1   2

1 1 1 2 2 = =− ⋅ =− tan θ − 2 2 4 2 2 2

12 , θ in quadrant II 13 Solve for cos θ : sin 2 θ + cos 2 θ = 1

csc θ =

cot θ =

1 1 3 2 3 2 = = ⋅ = cos θ  2 2  2 2 2 4    3 

43. sin θ =

cos 2 θ = 1 − sin 2 θ cos θ = ± 1 − sin 2 θ Since θ is in quadrant II, cos θ < 0 .

1 1 1 3 3 = = ⋅ = tan θ 3 3 3 3

cos θ = − 1 − sin 2 θ 2

144 25 5  12  = − 1−   = − 1− =− =− 169 169 13  13   12    12 sin θ 12 13  13  =   = ⋅ −  = − tan θ = cos θ  5  13  5  5 −   13  1 1 13 = = csc θ = sin θ  12  12    13 

1 2 2 41. sin θ = − , cos θ = 3 3  1 −  sin θ 1 3 2 2 3 tan θ = =  =− ⋅ ⋅ =− cos θ  2 2  3 2 2 2 4    3  1 1  3 csc θ = = = 1 ⋅  −  = −3 sin θ  1   1 −  3  

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Chapter 2: Trigonometric Functions

1 1 13 = =− cos θ  5  5 −   13  1 1 5 = =− cot θ = tan θ  12  12 −  5  

1 1 5 = =− sin θ  3  3 −   5 1 1 5 = =− sec θ = cos θ  4  4 −   5 1 1 4 = = cot θ = tan θ  3  3   4 csc θ =

sec θ =

3 44. cos θ = , θ in quadrant IV 5 Solve for sin θ : sin 2 θ + cos 2 θ = 1

5 , θ in quadrant III 13 Solve for cos θ : sin 2 θ + cos 2 θ = 1

46. sin θ = −

sin 2 θ = 1 − cos 2 θ sin θ = ± 1 − cos 2 θ Since θ is in quadrant IV, sin θ < 0 .

cos 2 θ = 1 − sin 2 θ

2

16 4 3 sin θ = − 1 − cos 2 θ = − 1 −   = − =− 25 5 5

tan θ

csc θ

sec θ

cot θ

cos θ = ± 1 − sin 2 θ Since θ is in quadrant III, cos θ < 0 .  5 cos θ = − 1 − sin 2 θ = − 1 −  −   13 

 4 − sin θ  5  4 5 4 = = =− ⋅ =− cos θ 5 3 3 3   5   1 1 5 = = =− sin θ  4  4 −   5 1 1 5 = = = 3 cos θ   3   5 1 1 3 = = =− tan θ  4  4 −   3

2

144 12 =− 169 13  5 − sin θ  13  5  13  5 = = = − ⋅ −  = cos θ  12  13  12  12 −   13  1 1 13 = = =− sin θ  5  5 −   13  1 1 13 = = =− cos θ  12  12 −   13  1 1 12 = = = tan θ  5  5    12  =−

tan θ

csc θ

sec θ

4 45. cos θ = − , θ in quadrant III 5 Solve for sin θ : sin 2 θ + cos 2 θ = 1

cot θ

sin 2 θ = 1 − cos 2 θ

5 , 90º < θ < 180º , θ in quadrant II 13 Solve for cos θ : sin 2 θ + cos 2 θ = 1

47. sin θ =

sin θ = ± 1 − cos 2 θ Since θ is in quadrant III, sin θ < 0 .

cos 2 θ = 1 − sin 2 θ

sin θ = − 1 − cos 2 θ 2

16  4 = − 1−  −  = − 1− =− 25  5  3 − sin θ  5  3  5 = = − ⋅ −  = tan θ = cos θ  4  5  4 −  5  

cos θ = ± 1 − sin 2 θ Since θ is in quadrant II, cos θ < 0 .

9 3 =− 25 5

5 cos θ = − 1 − sin 2 θ = − 1 −    13 

3 4

= − 1−

2

25 144 12 =− =− 169 169 13

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Section 2.3: Properties of the Trigonometric Functions

tan θ

csc θ

sec θ

cot θ

5   sin θ 5  13  5 13 = =   = ⋅ −  = − cos θ  12  13  12  12 −   13  1 1 13 = = = sin θ  5  5    13  1 1 13 = = =− cos θ  12  12 −   13  1 1 12 = = =− tan θ  5  5 −  12  

sin θ = 1 − cos 2 θ 2

1 8 2 2  1 = 1−  −  = 1− = = 9 9 3  3

2 2   sin θ  3  2 2  3  = = ⋅−  = −2 2 tan θ = cos θ 3  1  1 −    3 1 1 3 2 3 2 = = ⋅ = sin θ  2 2  2 2 2 4    3  1 1 = = −3 sec θ = cos θ  1  −    3 csc θ =

4 , 270º < θ < 360º ; θ in quadrant IV 5 Solve for sin θ : sin 2 θ + cos 2 θ = 1

48. cos θ =

cot θ =

2 3π 50. sin θ = − , π < θ < , θ in quadrant III 3 2 Solve for cos θ : sin 2 θ + cos 2 θ = 1

sin θ = ± 1 − cos 2 θ Since θ is in quadrant IV, sin θ < 0 . 2

9 3 4 sin θ = − 1 − cos 2 θ = − 1 −   = − =− 25 5 5

sin θ tan θ = cos θ 1 csc θ = sin θ sec θ =

1 cos θ

cot θ =

1 tan θ

1 1 2 2 = ⋅ =− tan θ − 2 2 2 4

cos θ = ± 1 − sin 2 θ Since θ is in quadrant III, cos θ < 0 .

 3 −  3 5 3 5 = =− ⋅ =− 5 4 4 4   5 1 5 = =− 3  3 −  5   1 5 = = 4 4   5 1 4 = =− 3  3 −   4

 2 cos θ = − 1 − sin 2 θ = − 1 −  −   3

2

5 5 =− 9 3  2 −  sin θ 3 =  tan θ = cos θ  5  −   3  =−

2  3  5 2 5 = − ⋅ − = ⋅ 3  5 5 5 1 1 3 csc θ = = =− sin θ  2  2 −   3

1 π < θ < π , θ in quadrant II 49. cos θ = − , 3 2 Solve for sin θ : sin 2 θ + cos 2 θ = 1

sec θ =

sin 2 θ = 1 − cos 2 θ sin θ = ± 1 − cos 2 θ Since θ is in quadrant II, sin θ > 0 .

cot θ =

1 1 3 5 3 5 = =− ⋅ =− 5 cos θ  5 5 5  −   3  1 1 5 5 5 = ⋅ = = tan θ  2 5  2 5 5 2  5   

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Chapter 2: Trigonometric Functions

2 51. sin θ = , tan θ < 0, so θ is in quadrant II 3 Solve for cos θ : sin 2 θ + cos 2 θ = 1

sec θ =

cos 2 θ = 1 − sin 2 θ

cot θ =

cos θ = ± 1 − sin θ Since θ is in quadrant II, cos θ < 0 . 2

1 1 15 15 = ⋅ = tan θ 15 15 15

53. sec θ = 2, sin θ < 0, so θ is in quadrant IV

cos θ = − 1 − sin θ 2

1 1 = sec θ 2 Solve for sin θ : sin 2 θ + cos 2 θ = 1 Solve for cos θ : cos θ =

2

4 5 5 2 = − 1−   = − 1− = − =− 9 9 3 3 2   sin θ 2  3  2 5 3 =   = ⋅ − tan θ = =− cos θ  5 5 5 3   −   3  1 1 3 csc θ = = = 2 sin θ   2   3

sin 2 θ = 1 − cos 2 θ sin θ = ± 1 − cos 2 θ Since θ is in quadrant IV, sin θ < 0 . sin θ = − 1 − cos 2 θ 2

1 3 3 1 = − 1−   = − 1− = − =− 4 4 2 2  3 −  sin θ  2  = =− tan θ = cos θ 1   2 1 1 csc θ = = =− sin θ  3 −   2  1 1 3 cot θ = = =− tan θ − 3 3

1 1 3 3 5 sec θ = = =− =− cos θ  5 5 5 −  3   1 1 5 cot θ = = =− =− 5 tan θ  2 5  2 2 5 −  5   1 52. cos θ = − , tan θ > 0 4 sin θ Since tan θ = > 0 and cos θ < 0 , sin θ < 0 . cos θ Solve for sin θ : sin 2 θ + cos 2 θ = 1

3 2 ⋅ =− 3 2 1 2 3

=−

2 3 3

54. csc θ = 3, cot θ < 0, so θ is in quadrant II 1 1 = csc θ 3 Solve for cos θ : sin 2 θ + cos 2 θ = 1

Solve for sin θ : sin θ =

sin θ = ± 1 − cos 2 θ sin θ = − 1 − cos 2 θ

cos θ = ± 1 − sin 2 θ Since θ is in quadrant II, cos θ < 0 .

2

1  1 = − 1−  −  = − 1− 4 16  

cos θ = − 1 − sin 2 θ

15 15 =− 16 4  15   −  4  sin θ 15  4   tan θ = = =− ⋅  −  = 15 cos θ 4  1  1 −    4 =−

csc θ =

1 1 4 = − = −4 = cos θ  1  1 −   4

2

1 8 2 2 1 = − 1−   = − 1− = − =− 9 9 3 3 1   sin θ = 3 tan θ = cos θ  2 2  −  3  

1 1 4 15 4 15 = =− ⋅ =− sin θ  15  15 15 15 −   4 

1  3  2 2 = ⋅ − ⋅ =−  3  2 2 2 4

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Section 2.3: Properties of the Trigonometric Functions

sec θ =

cot θ =

sin θ = − 1 − cos 2 θ

1 1 3 3 ⋅ 2 =− 2 = =− cos θ  2 2  4 2 2 2  −  3  

2

16 9 3  4 = − 1−  −  = − 1− =− =− 25 25 5  5 1 1 5 = =− csc θ = sin θ  3  3 −   5

1 1 4 2 = =− ⋅ = −2 2 tan θ  2 2 2  −   4 

1 57. tan θ = − , sin θ > 0, so θ is in quadrant II 3 Solve for sec θ : sec2 θ = 1 + tan 2 θ

3 , sin θ < 0, so θ is in quadrant III 4 Solve for sec θ : sec2 θ = 1 + tan 2 θ

55. tan θ =

sec θ = ± 1 + tan 2 θ Since θ is in quadrant III, sec θ < 0 .

sec θ = ± 1 + tan 2 θ Since θ is in quadrant II, sec θ < 0 .

sec θ = − 1 + tan 2 θ

sec θ = − 1 + tan 2 θ 2

1 10 10  1 = − 1+  −  = − 1+ = − =− 9 9 3  3

2

9 25 5 3 = − 1+   = − 1+ =− =− 16 16 4 4 1 4 =− cos θ = sec θ 5

cos θ =

sin θ = − 1 − cos 2 θ

1 1 3 3 10 = =− =− sec θ  10  10 10  −   3 

sin θ = 1 − cos 2 θ

2

16 9 3  4 = − 1−  −  = − 1− =− =− 25 25 5  5 1 1 5 = =− csc θ = sin θ  3  3 −   5 1 1 4 = = cot θ = tan θ  3  3   4

2

 3 10  90 = 1 −  −  = 1 − 100  10  10 10 = 100 10 1 1 = = 10 csc θ = sin θ  10   10    1 1 = = −3 cot θ = tan θ  1  −    3 =

4 , cos θ < 0, so θ is in quadrant III 3 1 1 3 = = tan θ = cot θ  4  4   3 Solve for sec θ : sec2 θ = 1 + tan 2 θ

56. cot θ =

58. sec θ = − 2, tan θ > 0, so θ is in quadrant III

Solve for tan θ : sec2 θ = 1 + tan 2 θ

sec θ = ± 1 + tan 2 θ Since θ is in quadrant III, sec θ < 0 .

tan θ = ± sec 2 θ − 1

tan θ = sec 2 θ − 1 = (− 2) 2 − 1 = 4 − 1 = 3

sec θ = − 1 + tan 2 θ 2

9 25 5 3 = − 1+   = − 1+ =− =− 16 16 4 4 1 4 cos θ = =− sec θ 5

cos θ =

1 1 =− sec θ 2

cot θ =

1 1 3 3 = ⋅ = tan θ 3 3 3

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Chapter 2: Trigonometric Functions

π 2 3  π 75. sec  −  = sec = 6 3  6

sin θ = − 1 − cos 2 θ 2

1 3 3  1 = − 1−  −  = − 1− = − =− 4 4 2  2

π 2 3  π 76. csc  −  = − csc = − 3 3 3  

1 1 2 3 2 3 csc θ = = =− ⋅ =− sin θ  3 3 3 3 −   2 

77. sin 2 ( 40º ) + cos 2 ( 40º ) = 1 78. sec 2 (18º ) − tan 2 (18º ) = 1

3 59. sin(− 60º ) = − sin 60º = − 2

60. cos(−30º ) = cos 30º =

3 2

61. tan(−30º ) = − tan 30º = −

3 3

79. sin ( 80º ) csc ( 80º ) = sin ( 80º ) ⋅

1 =1 sin ( 80º )

80. tan (10º ) cot (10º ) = tan (10º ) ⋅

1 =1 tan (10º )

81. tan ( 40º ) −

2 62. sin(−135º ) = − sin135º = − 2

sin ( 40º )

cos ( 40º ) cos ( 20º )

= tan ( 40º ) − tan ( 40º ) = 0

63. sec(− 60º ) = sec 60º = 2

82. cot ( 20º ) −

64. csc(−30º ) = − csc 30º = − 2

83. cos ( 400º ) ⋅ sec ( 40º ) = cos ( 40º +360º ) ⋅ sec ( 40º )

sin ( 20º )

= cot ( 20º ) − cot ( 20º ) = 0

= cos ( 40º ) ⋅ sec ( 40º )

65. sin(−90º ) = − sin 90º = −1

= cos ( 40º ) ⋅

66. cos(− 270º ) = cos 270º = 0

1 =1 cos ( 40º )

84. tan ( 200º ) ⋅ cot ( 20º ) = tan ( 20º +180º ) ⋅ cot ( 20º )

π  π 67. tan  −  = − tan = −1  4 4

= tan ( 20º ) ⋅ cot ( 20º ) = tan ( 20º ) ⋅

68. sin(−π) = − sin π = 0 π 2  π 69. cos  −  = cos = 4 2  4

1 =1 tan ( 20º )

 π   25π   π   25π  85. sin  −  csc   = − sin   csc    12   12   12   12   π   π 24π  = − sin   csc  +   12   12 12  π  π  = − sin   csc  + 2π   12   12  π  π  = − sin   csc    12   12  1 π  = − sin   ⋅ = −1  12  π  sin    12 

π 3  π 70. sin  −  = − sin = − 3 2  3

71. tan(−π) = − tan π = 0 3π  3π  72. sin  −  = − sin = −(−1) = 1 2  2  π  π 73. csc  −  = − csc = − 2 4 4  

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Section 2.3: Properties of the Trigonometric Functions  π   37π 86. sec  −  cos  18    18

87.

sin ( −20º )

cos ( 380º ) = =

92. If cot θ = − 2 , then

 π   37π   = sec   cos   18     18  π   π 36π  = sec   cos  +  18    18 18  π  π  = sec   cos  + 2π   18   18  π  π  = sec   cos   18    18  1 π  = sec   ⋅ =1  18  sec  π     18 

cot θ + cot (θ − π ) + cot (θ − 2π ) = − 2 + ( −2 ) + ( −2 ) = −6

93. sin1º + sin 2º + sin 3º +... + sin 357º + sin 358º + sin 359º = sin1º + sin 2º + sin 3º + ⋅⋅⋅ + sin(360º −3º ) + sin(360º − 2º ) + sin(360º −1º ) = sin1º + sin 2º + sin 3º + ⋅⋅⋅ + sin(−3º ) + sin( − 2º ) + sin(−1º ) = sin1º + sin 2º + sin 3º + ⋅⋅⋅ − sin 3º − sin 2º − sin1º = sin (180º ) = 0

+ tan ( 200º )

− sin ( 20º )

cos ( 20º +360º ) − sin ( 20º ) cos ( 20º )

94. cos1º + cos 2º + cos 3º + ⋅⋅⋅ + cos 357º + cos 358º + cos 359º = cos1º + cos 2º + cos 3º +... + cos(360º −3º ) + cos(360º − 2º ) + cos(360º −1º ) = cos1º + cos 2º + cos 3º +... + cos(−3º ) + cos(− 2º ) + cos(−1º ) = cos1º + cos 2º + cos 3º +... + cos 3º + cos 2º + cos1º = 2 cos1º +2 cos 2º +2 cos 3º +... + 2 cos178º + 2 cos179º + cos180º = 2 cos1º +2 cos 2º +2 cos 3º +... + 2 cos(180º − 2º )

+ tan ( 20º +180º )

+ tan ( 20º )

= − tan ( 20º ) + tan ( 20º ) = 0

88.

sin ( 70º )

cos ( −430º ) = = =

sin ( 70º )

+ tan ( −70º )

cos ( 430º )

− tan ( 70º )

sin ( 70º )

cos ( 70º +360º ) sin ( 70º )

cos ( 70º )

+ 2 cos(180º −1º ) + cos (180º ) = 2 cos1º +2 cos 2º +2 cos 3º +... − 2 cos 2º − 2 cos1º + cos180º = cos180º = −1

− tan ( 70º )

− tan ( 70º )

95. The domain of the sine function is the set of all real numbers.

= tan ( 70º ) − tan ( 70º ) = 0

96. The domain of the cosine function is the set of all real numbers.

89. If sin θ = 0.3 , then sin θ + sin (θ + 2π ) + sin (θ + 4π ) = 0.3 + 0.3 + 0.3 = 0.9

97.

90. If cos θ = 0.2 , then cos θ + cos (θ + 2π ) + cos (θ + 4π )

f (θ ) = tan θ is not defined for numbers that are

odd multiples of

= −0.2 + 0.2 + 0.2 = 0.6

91. If tan θ = 3 , then tan θ + tan (θ + π ) + tan (θ + 2π ) = 3+3+3 = 9

π . 2

98.

f (θ ) = cot θ is not defined for numbers that are multiples of π .

99.

f (θ ) = sec θ is not defined for numbers that are

odd multiples of

π . 2

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Chapter 2: Trigonometric Functions

100.

f (θ ) = csc θ is not defined for numbers that are multiples of π .

114. a. b.

101. The range of the sine function is the set of all real numbers between −1 and 1, inclusive.

1 1 1 + + 4 4 4 3 = 4

115. a.

104. The range of the cotangent function is the set of all real numbers.

b.

105. The range of the secant function is the set of all real numbers greater than or equal to 1 and all real numbers less than or equal to −1 .

f (−a ) = − f (a ) = − 2 f ( a ) + f ( a + π) + f ( a + 2 π) = f (a) + f (a) + f (a) = 2+2+2 = 6

106. The range of the cosecant function is the set of all real number greater than or equal to 1 and all real numbers less than or equal to −1 . 107. The sine function is odd because sin(−θ ) = − sin θ . Its graph is symmetric with respect to the origin.

116. a.

f (−a ) = − f (a ) = − (−3) = 3

b.

f ( a ) + f ( a + π) + f ( a + 4 π) = f (a) + f (a) + f (a) = −3 + (−3) + (−3) = −9

117. a. b.

108. The cosine function is even because cos(−θ ) = cos θ . Its graph is symmetric with respect to the y-axis. 109. The tangent function is odd because tan(−θ ) = − tan θ . Its graph is symmetric with respect to the origin.

118. a. b.

110. The cotangent function is odd because cot(−θ ) = − cot θ . Its graph is symmetric with respect to the origin.

f (−a) = f (a) = − 4 f (a) + f (a + 2π) + f (a + 4π) = f (a ) + f (a ) + f (a) = − 4 + (− 4) + (− 4) = −12 f (−a ) = − f (a ) = − 2

f (a) + f (a + 2π) + f (a + 4π) = f (a ) + f (a ) + f (a) = 2+2+2 =6

111. The secant function is even because sec(−θ ) = sec θ . Its graph is symmetric with respect to the y-axis.

500 1 y = = , then 1500 3 x r 2 = x 2 + y 2 = 9 + 1 = 10

119. Since tan θ =

112. The cosecant function is odd because csc(−θ ) = − csc θ . Its graph is symmetric with respect to the origin.

b.

f (a) + f (a + 2π) + f (a − 2π) =

103. The range of the tangent function is the set of all real numbers.

f (−a) = − f (a) = −

1 4

= f (a ) + f (a ) + f (a)

102. The range of the cosine function is the set of all real numbers between −1 and 1, inclusive.

113. a.

f (−a) = f (a) =

r = 10 sin θ =

1

=

1

. 1+ 9 10 5 5 T = 5− +  1  1  3⋅     3   10 

1 3

f (a) + f (a + 2π) + f (a + 4π) = f (a ) + f (a ) + f (a)

= 5 − 5 + 5 10

1 1 1 = + + =1 3 3 3

= 5 10 ≈ 15.8 minutes

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Section 2.3: Properties of the Trigonometric Functions

120. a.

123. Suppose there is a number p, 0 < p < 2π for

1 y π = for 0 < θ < . 4 x 2 2 2 2 r = x + y = 16 + 1 = 17

tan θ =

which sin(θ + p ) = sin θ for all θ . If θ = 0 , then sin ( 0 + p ) = sin p = sin 0 = 0 ; so that

r = 17

Thus, sin θ =

1

. 17 2 1 T (θ ) = 1 + − 1   1   3⋅  4⋅  17   4   = 1+

2 17 2 17 −1 = ≈ 2.75 hours 3 3

124. Suppose there is a number p, 0 < p < 2π , for

1 b. Since tan θ = , x = 4 . Sally heads 4 directly across the sand to the bridge, crosses the bridge, and heads directly across the sand to the other house.

c.

which cos(θ + p) = cos θ for all θ . If θ =

p = π . Thus cos ( π ) = −1 = cos ( 0 ) = 1 , or

not be reached and she cannot get across the river.

−1 = 1 . This is impossible. The smallest positive number p for which cos(θ + p) = cos θ for all θ must then be p = 2π .

121. Let P = ( x, y ) be the point on the unit circle that corresponds to an angle t. Consider the equation y tan t = = a . Then y = ax . Now x 2 + y 2 = 1 , x 1 2 and so x + a 2 x 2 = 1 . Thus, x = ± 1 + a2 a . That is, for any real number a , y=± 1 + a2 there is a point P = ( x, y ) on the unit circle for which tan t = a . In other words, −∞ < tan t < ∞ , and the range of the tangent function is the set of all real numbers.

1 : Since cos θ has period 2π , so cos θ does sec θ .

125. sec θ =

1 : Since sin θ has period 2π , so sin θ does csc θ .

126. csc θ =

127. If P = (a, b) is the point on the unit circle corresponding to θ , then Q = (−a, −b) is the point on the unit circle corresponding to θ + π . −b b Thus, tan(θ + π) = = = tan θ . If there −a a exists a number p, 0 < p < π , for which

122. Let P = ( x, y ) be the point on the unit circle that corresponds to an angle t. Consider the equation x cot t = = a . Then x = ay . Now x 2 + y 2 = 1 , y

x=±

1 1 + a2

π , 2

π  π then cos  + p  = cos   = 0 ; so that p = π . 2   2 If θ = 0 , then cos ( 0 + p ) = cos ( 0 ) . But

θ must be larger than 14º , or the road will

so a 2 y 2 + y 2 = 1 . Thus, y = ±

π

π  π then sin  + p  = sin   . 2 2  2 π 3   π But p = π . Thus, sin   = −1 = sin   = 1 ,  2  2 or −1 = 1 . This is impossible. The smallest positive number p for which sin(θ + p ) = sin θ for all θ must then be p = 2π . p = π . If θ =

tan(θ + p ) = tan θ for all θ , then if θ = 0 , tan ( p ) = tan ( 0 ) = 0. But this means that p is a

and

multiple of π . Since no multiple of π exists in the interval ( 0, π ) , this is impossible. Therefore,

a

. That is, for any real number a , 1 + a2 there is a point P = ( x, y ) on the unit circle for which cot t = a . In other words, −∞ < cot t < ∞ , and the range of the tangent function is the set of all real numbers.

the fundamental period of f (θ ) = tan θ is π .

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Chapter 2: Trigonometric Functions

The graph would be shifted horizontally to the right 3 units, stretched by a factor of 2, reflected about the x-axis and then shifted vertically up by 5 units. So the graph would be:

1 : Since tan θ has period π , so tanθ does cot θ .

128. cot θ =

129. Let P = (a, b) be the point on the unit circle

corresponding to θ . Then csc θ =

1 1 = b sin θ

1 1 = a cos θ a 1 1 cot θ = = = b b   tan θ   a

sec θ =

130. Let P = (a, b) be the point on the unit circle corresponding to θ . Then b sin θ tan θ = = a cos θ a cos θ cot θ = = b sin θ

139.

f (25) = 3 25 − 9 + 6 = 3 16 + 6 = 12 + 6 = 18 So the point (25,18) is on the graph.

140. y = (0)3 − 9(0) 2 + 3(0) − 27 = −27 Thus the y-intercept is (0, −27) .

131. (sin θ cos φ ) 2 + (sin θ sin φ ) 2 + cos 2 θ = sin 2 θ cos 2 φ + sin 2 θ sin 2 φ + cos 2 θ = sin 2 θ (cos 2 φ + sin 2 φ ) + cos 2 θ = sin 2 θ + cos 2 θ =1

132 – 136. Answers will vary.

Section 2.4

( − x)4 + 3 x 4 + 3 = = f ( x) . Thus, ( − x) 2 − 5 x 2 − 5 f ( x) is even. 138. We need to use completing the square to put the function in the form f ( x ) = a ( x − h) 2 + k

137.

1. y = 3x 2

f ( − x) =

Using the graph of y = x 2 , vertically stretch the graph by a factor of 3.

f ( x) = −2 x 2 + 12 x − 13 = −2( x 2 − 6 x) − 13 = −2 x 2 − 6 x +

144 4( 2) 2

− 13 + 2

144 4( 2) 2

= −2( x 2 − 6 x + 9) − 13 + 18 = −2( x 2 − 6 x + 9) + 5 = −2( x − 3) 2 + 5

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Section 2.4: Graphs of the Sine and Cosine Functions

g.

2. y = 2 x Using the graph of y = x , compress horizontally by a factor of

1 . 2

The graph of y = cos x crosses the y-axis at the point (0, 1), so the y-intercept is 1. b. The graph of y = cos x is decreasing for 0< x<π. c. The smallest value of y = cos x is −1 .

12. a.

d. e.

π 3π , 2 2 cos x = 1 when x = − 2π, 0, 2π;

cos x = 0 when x =

cos x = −1 when x = −π, π.

π 3. 1; 2

3 11π π π 11π when x = − ,− , , 2 6 6 6 6 g. The x-intercepts of cos x are   ( 2k + 1) π , k an integer  x | x = 2   13. y = 2sin x cos x =

f.

4. 3; π 5. 3;

The x-intercepts of sin x are { x | x = kπ , k an integer}

2π π = 6 3

This is in the form y = A sin(ω x) where A = 2

6. True 7. False; The period is

π

and ω = 1 . Thus, the amplitude is A = 2 = 2 = 2.

and the period is T =

8. True

ω

=

2π = 2π . 1

14. y = 3cos x

9. d

This is in the form y = A cos(ω x) where A = 3

10. d

and ω = 1 . Thus, the amplitude is A = 3 = 3

11. a.

and the period is T =

The graph of y = sin x crosses the y-axis at the point (0, 0), so the y-intercept is 0.

π π <x< . 2 2

The largest value of y = sin x is 1.

d.

sin x = 0 when x = 0, π, 2π .

e.

sin x = 1 when x = −

T=

2π = 2π . 1

ω

=

2π = π. 2

1  16. y = − sin  x  2  This is in the form y = A sin(ω x) where A = −1

3π π , ; 2 2 π 3π sin x = −1 when x = − , . 2 2 sin x = −

=

A = − 4 = 4 and the period is

c.

f.

ω

15. y = − 4 cos(2 x) This is in the form y = A cos(ω x) where A = − 4 and ω = 2 . Thus, the amplitude is

b. The graph of y = sin x is increasing for −

1 . Thus, the amplitude is A = − 1 = 1 2 2π 2π = 1 = 4π . and the period is T =

and ω =

1 5π π 7π 11π when x = − ,− , , 2 6 6 6 6

ω

2

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Chapter 2: Trigonometric Functions 17. y = 6sin(π x) This is in the form y = A sin(ω x) where A = 6

9  3π 22. y = cos  − 5  2

and ω = π . Thus, the amplitude is A = 6 = 6 and the period is T =

ω

=

2π =2. π

ω

=

3π . Thus, the amplitude is 2 9 9 A = = and the period is 5 5 2π 2π 4 = = . T= ω 3π 3

2π . 3

2

23. F

1 3  19. y = − cos  x  2 2  This is in the form y = A cos(ω x) where

24. E 25. A

1 3 and ω = . Thus, the amplitude is 2 2 1 1 A = − = and the period is 2 2 2π 2π 4π = 3 = . T= ω 3 2

A=−

26. I 27. H 28. B 29. C

4 2  20. y = sin  x  3 3 

30. G

This is in the form y = A sin(ω x) where A =

4 3

31. J 32. D

2 4 4 . Thus, the amplitude is A = = 3 3 3 2π 2π = 2 = 3π . and the period is T = and ω =

ω

5  2π 21. y = sin  − 3  3

33. Comparing y = 4 cos x to y = A cos (ω x ) , we

find A = 4 and ω = 1 . Therefore, the amplitude 2π is 4 = 4 and the period is = 2π . Because 1 the amplitude is 4, the graph of y = 4 cos x will lie between −4 and 4 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [ 0, 2π ]

3

5  2π  x  = − sin  3  3 

 x 

This is in the form y = A sin(ω x) where A = −

5 3

2π . Thus, the amplitude is 3 5 5 A = − = and the period is 3 3 2π 2π = = 3. T=

and ω =

ω

9 5

and ω =

and ω = 3 . Thus, the amplitude is A = − 3 = 3 2π

 x 

This is in the form y = A cos(ω x) where A =

18. y = − 3cos(3x) This is in the form y = A cos(ω x) where A = − 3

and the period is T =

 9  3π x  = cos  5   2

into four subintervals, each of length

2π π = by 4 2

finding the following values: π 3π , and 2π 0, , π , 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = 4 cos x , we multiply the y-coordinates of the five key points for y = cos x by A = 4 . The five key points are

2π 3

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Section 2.4: Graphs of the Sine and Cosine Functions

direction to obtain the graph shown below.

π   3π   , 0  , (π , −4 ) ,  , 0  , ( 2π , 4 ) 2    2  We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

( 0, 4 ) ,

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −3,3] .

35. Comparing y = −4sin x to y = A sin (ω x ) , we

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

find A = −4 and ω = 1 . Therefore, the amplitude 2π is −4 = 4 and the period is = 2π . Because 1 the amplitude is 4, the graph of y = −4sin x will lie between −4 and 4 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [ 0, 2π ]

range is [ −4, 4] .

34. Comparing y = 3sin x to y = A sin (ω x ) , we

find A = 3 and ω = 1 . Therefore, the amplitude 2π is 3 = 3 and the period is = 2π . Because 1 the amplitude is 3, the graph of y = 3sin x will lie between −3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [ 0, 2π ] into four subintervals, each of length

2π π = by 4 2 π 3π finding the following values: 0, , π , , 2π 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = −4sin x , we multiply the y-coordinates of

into four subintervals, each of length

2π π = by 4 2

finding the following values: π 3π 0, , π , , and 2π 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = 3sin x , we multiply the y-coordinates of the five key points for y = sin x by A = 3 . The five key

the five key points for y = sin x by A = −4 . The five key points are π 3π ( 0, 0 ) ,  , −4  , (π , 0 ) ,  , 4  , ( 2π , 0 ) 2    2  We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

π   3π  points are ( 0, 0 ) ,  ,3  , (π , 0 ) ,  , −3  , 2 2     ( 2π , 0 )

We plot these five points and fill in the graph of the curve. We then extend the graph in either

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Chapter 2: Trigonometric Functions 37. Comparing y = cos ( 4 x ) to y = A cos (ω x ) , we

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

find A = 1 and ω = 4 . Therefore, the amplitude 2π π = . Because the is 1 = 1 and the period is 4 2 amplitude is 1, the graph of y = cos ( 4 x ) will lie

range is [ −4, 4] .

36. Comparing y = −3cos x to y = A cos (ω x ) , we

find A = −3 and ω = 1 . Therefore, the amplitude 2π = 2π . Because is −3 = 3 and the period is 1 the amplitude is 3, the graph of y = −3cos x will lie between −3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [ 0, 2π ]

into four subintervals, each of length

between −1 and 1 on the y-axis. Because the period is

π

2

, one cycle will begin at x = 0 and

π

 π . We divide the interval  0,   2 π /2 π = into four subintervals, each of length 4 8 by finding the following values: π π 3π π , and 0, , , 8 4 8 2 These values of x determine the x-coordinates of the five key points on the graph. The five key points are π π 3π π ( 0,1) ,  , 0  ,  , −1 ,  , 0  ,  ,1   8  2  8  4 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

end at x =

2π π = by 4 2

finding the following values: π 3π 0, , π , , and 2π 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = −3cos x , we multiply the y-coordinates of the five key points for y = cos x by A = −3 . The five key points are π 3π ( 0, −3) ,  , 0  , (π ,3) ,  , 0  , ( 2π , −3) 2    2  We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. y 5 (, 3) (, 3) 3 ( ––– , 0) 2 x 2   2  , 0) ( –– 2 (0, 3) (2, 3) 5

2

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −1,1] .

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −3,3] .

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Section 2.4: Graphs of the Sine and Cosine Functions

y-axis. Because the period is π , one cycle will begin at x = 0 and end at x = π . We divide the interval [ 0, π ] into four subintervals, each of

38. Comparing y = sin ( 3x ) to y = A sin (ω x ) , we

find A = 1 and ω = 3 . Therefore, the amplitude 2π . Because the is 1 = 1 and the period is 3 amplitude is 1, the graph of y = sin ( 3x ) will lie

length

π 4

by finding the following values:

π 3π , , , and π 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = − sin ( 2 x ) , we multiply the y-coordinates of

between −1 and 1 on the y-axis. Because the 2π period is , one cycle will begin at x = 0 and 3 2π  2π  . We divide the interval  0,  end at x = 3  3  2π / 3 π = into four subintervals, each of length 4 6 by finding the following values: π π π 2π 0, , , , and 6 3 2 3 These values of x determine the x-coordinates of the five key points on the graph. The five key points are π π π 2π ( 0, 0 ) ,  ,1 ,  , 0  ,  , −1 ,  , 0  6 3 2   3       We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

0,

π

the five key points for y = sin x by A = −1 .The five key points are π π 3π ( 0, 0 ) ,  , −1 ,  , 0  ,  ,1 , (π , 0 ) 4  2   4  We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. y 2 3  , 1) (, 0) ––– , 1) ( ––– ( 4 4 2

2



(0, 0)

x

 , 0) (–– 2

 , 1) 2 (–– 4

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −1,1] .

40. Since cosine is an even function, we can plot the equivalent form y = cos ( 2 x ) .

Comparing y = cos ( 2 x ) to y = A cos (ω x ) , we

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

find A = 1 and ω = 2 . Therefore, the amplitude 2π is 1 = 1 and the period is = π . Because the 2 amplitude is 1, the graph of y = cos ( 2 x ) will lie

range is [ −1,1] .

39. Since sine is an odd function, we can plot the equivalent form y = − sin ( 2 x ) .

between −1 and 1 on the y-axis. Because the period is π , one cycle will begin at x = 0 and end at x = π . We divide the interval [ 0, π ] into

Comparing y = − sin ( 2 x ) to y = A sin (ω x ) , we find A = −1 and ω = 2 . Therefore, the 2π amplitude is −1 = 1 and the period is =π . 2 Because the amplitude is 1, the graph of y = − sin ( 2 x ) will lie between −1 and 1 on the

four subintervals, each of length the following values: π π 3π , and π 0, , , 4 2 4 167

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π 4

by finding


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Chapter 2: Trigonometric Functions

These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = cos ( 2 x ) , we multiply the y-coordinates of

the five key points for y = sin x by A = 2 . The five key points are ( 0, 0 ) , (π , 2 ) , ( 2π , 0 ) , ( 3π , −2 ) , ( 4π , 0 ) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

the five key points for y = cos x by A = 1 .The five key points are π π 3π ( 0,1) ,  , 0  ,  , −1 ,  , 0  , (π ,1)   4  4  2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −2, 2] .

1  42. Comparing y = 2 cos  x  to y = A cos (ω x ) , 4  1 we find A = 2 and ω = . Therefore, the 4 2π amplitude is 2 = 2 and the period is = 8π . 1/ 4 Because the amplitude is 2, the graph of 1  y = 2 cos  x  will lie between −2 and 2 on 4  the y-axis. Because the period is 8π , one cycle will begin at x = 0 and end at x = 8π . We divide the interval [ 0,8π ] into four subintervals,

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −1,1] .

1  41. Comparing y = 2sin  x  to y = A sin (ω x ) , 2  1 we find A = 2 and ω = . Therefore, the 2 2π = 4π . amplitude is 2 = 2 and the period is 1/ 2 Because the amplitude is 2, the graph of 1  y = 2sin  x  will lie between −2 and 2 on the 2  y-axis. Because the period is 4π , one cycle will begin at x = 0 and end at x = 4π . We divide the interval [ 0, 4π ] into four subintervals, each

of length

each of length

8π = 2π by finding the following 4

values: 0, 2π , 4π , 6π , and 8π These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1  y = 2 cos  x  , we multiply the y-coordinates 4  of the five key points for y = cos x by A = 2 .The five key points are ( 0, 2 ) , ( 2π , 0 ) , ( 4π , −2 ) , ( 6π , 0 ) , (8π , 2 )

4π = π by finding the following 4

values: 0, π , 2π , 3π , and 4π These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1  y = 2sin  x  , we multiply the y-coordinates of 2 

We plot these five points and fill in the graph of the curve. We then extend the graph in either 168

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Section 2.4: Graphs of the Sine and Cosine Functions

direction to obtain the graph shown below.

direction to obtain the graph shown below. y (2, 0) (0, 2) (8, 2) 2 (8, 2) (2, 0) (6, 0) x 8 4 4 8 (4, 2)

2

(4, 2)

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −2, 2] .

 1 1 range is  − ,  .  2 2

1 43. Comparing y = − cos ( 2 x ) to y = A cos (ω x ) , 2 1 we find A = − and ω = 2 . Therefore, the 2 1 1 2π amplitude is − = and the period is =π . 2 2 2 1 Because the amplitude is , the graph of 2 1 1 1 y = − cos ( 2 x ) will lie between − and on 2 2 2 the y-axis. Because the period is π , one cycle will begin at x = 0 and end at x = π . We divide the interval [ 0, π ] into four subintervals, each of

length

π 4

1  44. Comparing y = −4sin  x  to y = A sin (ω x ) , 8  1 we find A = −4 and ω = . Therefore, the 8 amplitude is −4 = 4 and the period is

2π = 16π . Because the amplitude is 4, the 1/ 8 1  graph of y = −4sin  x  will lie between −4 8  and 4 on the y-axis. Because the period is 16π , one cycle will begin at x = 0 and end at x = 16π . We divide the interval [ 0,16π ] into

by finding the following values:

π 3π , , , and π 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 y = − cos ( 2 x ) , we multiply the y-coordinates 2 of the five key points for y = cos x by 0,

π

four subintervals, each of length

16π = 4π by 4

finding the following values: 0, 4π , 8π , 12π , and 16π These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1  y = −4sin  x  , we multiply the y-coordinates 8  of the five key points for y = sin x by A = −4 . The five key points are ( 0, 0 ) , ( 4π , −4 ) , (8π , 0 ) , (12π , 4 ) , (16π , 0 )

1 A = − .The five key points are 2 1   π   π 1   3π   1   0, −  ,  , 0  ,  ,  ,  , 0  ,  π , −  2  4   2 2  4   2  We plot these five points and fill in the graph of the curve. We then extend the graph in either

We plot these five points and fill in the graph of the curve. We then extend the graph in either

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Chapter 2: Trigonometric Functions

direction to obtain the graph shown below.

direction to obtain the graph shown below. y 5

(12, 4) (0, 0)

16

x (16, 0)

8

(8, 0) 5

(4, 4)

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −4, 4] .

range is [1,5] .

45. We begin by considering y = 2sin x . Comparing y = 2sin x to y = A sin (ω x ) , we find A = 2

46. We begin by considering y = 3cos x . Comparing

and ω = 1 . Therefore, the amplitude is 2 = 2

y = 3cos x to y = A cos (ω x ) , we find A = 3

2π = 2π . Because the 1 amplitude is 2, the graph of y = 2sin x will lie between −2 and 2 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [ 0, 2π ]

and the period is

into four subintervals, each of length

and ω = 1 . Therefore, the amplitude is 3 = 3 2π = 2π . Because the 1 amplitude is 3, the graph of y = 3cos x will lie between −3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [ 0, 2π ]

and the period is

2π π = by 4 2

finding the following values: π 3π , and 2π 0, , π , 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = 2sin x + 3 , we multiply the y-coordinates of the five key points for y = sin x by A = 2 and then add 3 units. Thus, the graph of y = 2sin x + 3 will lie between 1 and 5 on the yaxis. The five key points are π 3π ( 0,3) ,  ,5  , (π ,3) ,  ,1 , ( 2π ,3) 2   2  We plot these five points and fill in the graph of the curve. We then extend the graph in either

into four subintervals, each of length

2π π = by 4 2

finding the following values: π 3π 0, , π , , and 2π 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = 3cos x + 2 , we multiply the y-coordinates of the five key points for y = cos x by A = 3 and then add 2 units. Thus, the graph of y = 3cos x + 2 will lie between −1 and 5 on the y-axis. The five key points are π 3π ( 0,5) ,  , 2  , (π , −1) ,  , 2  , ( 2π ,5) 2   2  We plot these five points and fill in the graph of the curve. We then extend the graph in either

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Section 2.4: Graphs of the Sine and Cosine Functions

direction to obtain the graph shown below. y

direction to obtain the graph shown below.

(2, 2)

3 (0, 2) 2 



3 ( –– , 3) 2

2 x 3 (–– , 3) 2 1 (–– , 3) 2

(1, 8)

9

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ −8, 2] .

range is [ −1,5] .

π  48. We begin by considering y = 4sin  x  . 2  π  Comparing y = 4sin  x  to y = A sin (ω x ) , 2 

47. We begin by considering y = 5cos (π x ) .

Comparing y = 5cos (π x ) to y = A cos (ω x ) , we

find A = 5 and ω = π . Therefore, the amplitude 2π is 5 = 5 and the period is = 2 . Because the

we find A = 4 and ω =

π

amplitude is 5, the graph of y = 5cos (π x ) will

2

. Therefore, the

2π = 4. π /2 Because the amplitude is 4, the graph of π  y = 4sin  x  will lie between −4 and 4 on 2  the y-axis. Because the period is 4 , one cycle will begin at x = 0 and end at x = 4 . We divide the interval [ 0, 4] into four subintervals, each of

amplitude is 4 = 4 and the period is

lie between −5 and 5 on the y-axis. Because the period is 2 , one cycle will begin at x = 0 and end at x = 2 . We divide the interval [ 0, 2] into four subintervals, each of length

π

2 1 = by 4 2

finding the following values: 1 3 0, , 1, , and 2 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = 5cos (π x ) − 3 , we multiply the y-coordinates

4 = 1 by finding the following values: 4 0, 1, 2, 3, and 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for π  y = 4sin  x  − 2 , we multiply the y2  coordinates of the five key points for y = sin x by A = 4 and then subtract 2 units. Thus, the π  graph of y = 4sin  x  − 2 will lie between −6 2  and 2 on the y-axis. The five key points are ( 0, −2 ) , (1, 2 ) , ( 2, −2 ) , ( 3, −6 ) , ( 4, −2 )

length

of the five key points for y = cos x by A = 5 and then subtract 3 units. Thus, the graph of y = 5cos (π x ) − 3 will lie between −8 and 2 on

the y-axis. The five key points are 1 3 ( 0, 2 ) ,  , −3  , (1, −8 ) ,  , −3  , ( 2, 2 ) 2   2  We plot these five points and fill in the graph of the curve. We then extend the graph in either

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Chapter 2: Trigonometric Functions

direction to obtain the graph shown below.

3  9   , −2  , ( 3, 4 ) ,  ,10  , ( 6, 4 ) 2 2     We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

( 0, 4 ) ,

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the range is [ −6, 2] .

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

π  49. We begin by considering y = −6sin  x  . 3  π  Comparing y = −6sin  x  to y = A sin (ω x ) , 3 

we find A = −6 and ω =

π 3

range is [ −2,10] .

π  50. We begin by considering y = −3cos  x  . 4  π  Comparing y = −3cos  x  to y = A cos (ω x ) , 4 

. Therefore, the

2π = 6. π /3 Because the amplitude is 6, the graph of π  y = 6sin  x  will lie between −6 and 6 on the 3  y-axis. Because the period is 6, one cycle will begin at x = 0 and end at x = 6 . We divide the interval [ 0, 6] into four subintervals, each of

amplitude is −6 = 6 and the period is

we find A = −3 and ω =

π 4

. Therefore, the

2π =8. π /4 Because the amplitude is 3, the graph of π  y = −3cos  x  will lie between −3 and 3 on 4  the y-axis. Because the period is 8, one cycle will begin at x = 0 and end at x = 8 . We divide the interval [ 0,8] into four subintervals, each of amplitude is −3 = 3 and the period is

6 3 = by finding the following values: 4 2 3 9 0, , 3, , and 6 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for π  y = −6sin  x  + 4 , we multiply the y3  coordinates of the five key points for y = sin x by A = −6 and then add 4 units. Thus, the graph π  of y = −6sin  x  + 4 will lie between −2 and 3  10 on the y-axis. The five key points are

length

8 = 2 by finding the following values: 4 0, 2, 4, 6, and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for π  y = −3cos  x  + 2 , we multiply the y4  coordinates of the five key points for y = cos x by A = −3 and then add 2 units. Thus, the graph

length

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Section 2.4: Graphs of the Sine and Cosine Functions

on the y-axis. The five key points are π π 3π ( 0,5) ,  , 2  ,  ,5  ,  ,8  , (π ,5) 4  2   4  We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

π  of y = −3cos  x  + 2 will lie between −1 and 4  5 on the y-axis. The five key points are ( 0, −1) , ( 2, 2 ) , ( 4,5) , ( 6, 2 ) , (8, −1)

We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

y 10

(0,5)

 3π   4 ,8    π   ,5  2 

(π ,5 ) π   ,2 4 

−π

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

range is [ 2,8] .

range is [ −1,5] .

52. y = 2 − 4 cos ( 3x ) = −4 cos ( 3x ) + 2

51. y = 5 − 3sin ( 2 x ) = −3sin ( 2 x ) + 5

We begin by considering y = −4 cos ( 3 x ) .

We begin by considering y = −3sin ( 2 x ) .

Comparing y = −4 cos ( 3 x ) to y = A cos (ω x ) ,

Comparing y = −3sin ( 2 x ) to y = A sin (ω x ) ,

we find A = −4 and ω = 3 . Therefore, the 2π amplitude is −4 = 4 and the period is . 3 Because the amplitude is 4, the graph of y = −4 cos ( 3 x ) will lie between −4 and 4 on

we find A = −3 and ω = 2 . Therefore, the 2π amplitude is −3 = 3 and the period is =π . 2 Because the amplitude is 3, the graph of y = −3sin ( 2 x ) will lie between −3 and 3 on the

2π , one cycle 3 2π will begin at x = 0 and end at x = . We 3  2π  divide the interval 0,  into four  3  2π / 3 π = subintervals, each of length by 4 6 finding the following values: π π π 2π 0, , , , and 6 3 2 3 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = −4 cos ( 3x ) + 2 , we multiply the y-

the y-axis. Because the period is

y-axis. Because the period is π , one cycle will begin at x = 0 and end at x = π . We divide the interval [ 0, π ] into four subintervals, each of length

π 4

by finding the following values:

π 3π , , , and π 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y = −3sin ( 2 x ) + 5 , we multiply the y-

0,

π

π

coordinates of the five key points for y = sin x by A = −3 and then add 5 units. Thus, the graph of y = −3sin ( 2 x ) + 5 will lie between 2 and 8 173

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Chapter 2: Trigonometric Functions

coordinates of the five key points for y = cos x by A = −4 and then adding 2 units. Thus, the graph of y = −4 cos ( 3x ) + 2 will lie between −2

3 3 9 , , , and 3 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 5  2π  y = − sin  x  , we multiply the y3  3 

0,

and 6 on the y-axis. The five key points are π π π 2π ( 0, −2 ) ,  , 2  ,  , 6  ,  , 2  ,  , −2  6 3 2         3 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

3

coordinates of the five key points for y = sin x 5 by A = − .The five key points are 3 3 5 3 9 5 ( 0, 0 ) ,  , −  ,  , 0  ,  ,  , ( 3, 0 )  4 3  2   4 3 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

3

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the range is [ −2, 6] .

53. Since sine is an odd function, we can plot the 5  2π  equivalent form y = − sin  x . 3  3 

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

5  2π Comparing y = − sin  3  3

 x  to  5 2π y = A sin (ω x ) , we find A = − and ω = . 3 3 5 5 Therefore, the amplitude is − = and the 3 3 2π = 3 . Because the amplitude is period is 2π / 3 5 5  2π  , the graph of y = − sin  x  will lie 3 3  3 

 5 5 range is  − ,  .  3 3

54. Since cosine is an even function, we consider the 9  3π  equivalent form y = cos  x  . Comparing 5  2  9  3π  y = cos  x  to y = A cos (ω x ) , we find 5  2  9 3π A = and ω = . Therefore, the amplitude is 5 2 2π 4 9 9 = . Because = and the period is 5 5 3π / 2 3 9 the amplitude is , the graph of 5 9 9 9  3π  y = cos  x  will lie between − and 5 5 5  2 

5 5 and on the y-axis. Because the 3 3 period is 3 , one cycle will begin at x = 0 and end at x = 3 . We divide the interval [ 0,3] into

between −

four subintervals, each of length

3 by finding 4

the following values:

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Section 2.4: Graphs of the Sine and Cosine Functions

3 π 55. We begin by considering y = − cos  2 4 3 π  Comparing y = − cos  x  to 2 4 

4 , one 3 4 cycle will begin at x = 0 and end at x = . We 3  4 divide the interval 0,  into four subintervals,  3 4/3 1 = by finding the following each of length 4 3 values: 1 2 4 0, , , 1 , and 3 3 3 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 9  3π  y = cos  x  , we multiply the y-coordinates 5  2 

on the y-axis. Because the period is

of the five key points for y = cos x by A = 9  3π Thus, the graph of y = cos  − 5  2

between −

 x . 

3 π and ω = . 4 2 3 3 Therefore, the amplitude is − = and the 2 2 2π 3 = 8 . Because the amplitude is , period is π /4 2 3 π  the graph of y = − cos  x  will lie between 2 4  y = A cos (ω x ) , we find A = −

3 3 and on the y-axis. Because the period is 2 2 8, one cycle will begin at x = 0 and end at x = 8 . We divide the interval [ 0,8] into four −

9 . 5

8 = 2 by finding the 4 following values: 0, 2, 4, 6, and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 3 π  1 y = − cos  x  + , we multiply the y2 4  2 coordinates of the five key points for y = cos x

subintervals, each of length

 x  will lie 

9 9 and on the y-axis. The five key 5 5

points are  9 1  2 9 4 9  0,  ,  , 0  ,  , −  , (1, 0 ) ,  ,  5 3 3 5        3 5 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

3 1 and then add unit. Thus, the 2 2 3 π  1 graph of y = − cos  x  + will lie between 2 4  2 −1 and 2 on the y-axis. The five key points are 1 1 ( 0, −1) ,  2,  , ( 4, 2 ) ,  6,  , (8, −1) 2    2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

by A = −

y

(4, 2)

2

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

8 4

 9 9 range is  − ,  .  5 5

(8, 1)

2

1 (2, –– ) 2 (4, 2) 1 (6, –– ) 2 x 4 8

(0, 1) (8, 1)

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Chapter 2: Trigonometric Functions

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

y 2.5

range is [ −1, 2] .

(0,

3 ––) 2 3 (8, –– ) 2

1 π  56. We begin by considering y = − sin  x  . 2 8  1 π  Comparing y = − sin  x  to y = A sin (ω x ) , 2 8 

(4, 1)

3 (16, –– ) 2

x 1612 8 4 0.5

π 1 and ω = . Therefore, the 8 2 1 1 amplitude is − = and the period is 2 2 2π 1 = 16 . Because the amplitude is , the π /8 2 1 π  1 graph of y = − sin  x  will lie between − 2 8  2

4

8 12 16

From the graph we can determine that the domain is all real numbers, ( −∞, ∞ ) and the

we find A = −

range is [1, 2] .

57.

A = 3; T = π; ω =

2π 2π = =2 T π

y = ±3sin(2 x)

1 on the y-axis. Because the period is 16, 2 one cycle will begin at x = 0 and end at x = 16 . We divide the interval [ 0,16] into four

58.

and

subintervals, each of length

(12, 2)

(4, 2)

A = 2; T = 4π; ω =

2π 2π 1 = = T 4π 2

1  y = ±2sin  x  2 

16 = 4 by finding 4

59.

the following values: 0, 4, 8, 12, and 16 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 π  3 y = − sin  x  + , we multiply the y2 8  2

A = 3; T = 2; ω =

2π 2π = =π T 2

y = ±3sin(πx)

60.

A = 4; T = 1; ω =

2π 2π = = 2π T 1

y = ±4sin(2π x)

61. The graph is a cosine graph with amplitude 5 and period 8. 2π Find ω : 8 =

coordinates of the five key points for y = sin x 1 3 and then add units. Thus, the 2 2 1 π  3 graph of y = − sin  x  + will lie between 2 8  2 1 and 2 on the y-axis. The five key points are  3  3  3  0,  , ( 4,1) ,  8,  , (12, 2 ) , 16,   2  2  2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

by A = −

ω

8ω = 2π 2π π = ω= 8 4

π  The equation is: y = 5cos  x  . 4 

62. The graph is a sine graph with amplitude 4 and period 8π.

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Section 2.4: Graphs of the Sine and Cosine Functions

Find ω :

8π =

67. The graph is a reflected sine graph with 4π . amplitude 1 and period 3 4π 2π = Find ω : ω 3 4πω = 6π 6π 3 ω= = 4π 2 3  The equation is: y = − sin  x  . 2 

ω

8πω = 2π 2π 1 = ω= 8π 4

1  The equation is: y = 4sin  x  . 4 

63. The graph is a reflected cosine graph with amplitude 3 and period 4π. 2π Find ω : 4π =

ω

68. The graph is a reflected cosine graph with amplitude π and period 2π. 2π Find ω : 2π =

4πω = 2π 2π 1 ω= = 4π 2

ω

2πω = 2π 2π ω= =1 2π The equation is: y = −π cos x .

1  The equation is: y = −3cos  x  . 2 

64. The graph is a reflected sine graph with amplitude 2 and period 4. 2π Find ω : 4 =

69. The graph is a reflected cosine graph, shifted up 3 1 unit, with amplitude 1 and period . 2 3 2π Find ω : = 2 ω 3ω = 4π 4π ω= 3  4π  The equation is: y = − cos  x  +1 .  3 

ω

4ω = 2π 2π π ω= = 4 2 π  The equation is: y = − 2sin  x  . 2 

65. The graph is a sine graph with amplitude

3 and 4

period 1. Find ω : 1 =

70. The graph is a reflected sine graph, shifted down 4π 1 . 1 unit, with amplitude and period 3 2 4π 2π Find ω : = ω 3 4πω = 6π 6π 3 ω= = 4π 2 1 3  The equation is: y = − sin  x  − 1 . 2 2 

ω

ω = 2π 3 The equation is: y = sin ( 2π x ) . 4

66. The graph is a reflected cosine graph with 5 amplitude and period 2. 2 2π Find ω : 2 =

ω

2ω = 2π 2π ω= =π 2 5 The equation is: y = − cos ( πx ) . 2

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Chapter 2: Trigonometric Functions 71. The graph is a sine graph with amplitude 3 and period 4. 2π Find ω : 4 =

76.

f (π / 2 ) − f ( 0 )

π /2−0

ω

4ω = 2π 2π π ω= = 4 2

=

cos (π / 2 ) − cos ( 0 )

π /2 0 −1 2 = =− π /2 π

The average rate of change is −

π  The equation is: y = 3sin  x  . 2 

77.

72. The graph is a reflected cosine graph with amplitude 2 and period 2. 2π Find ω : 2 =

f (π / 2 ) − f ( 0 )

π /2−0

ω

2ω = 2π 2π ω= =π 2 The equation is: y = − 2 cos(π x) .

78.

2

π

π

The average rate of change is − 79.

.

 π cos  2 ⋅  − cos(2 ⋅ 0) f (π / 2 ) − f (0)  2 = π /2−0 π /2 cos(π ) − cos(0) −1 − 1 = = π /2 π /2 2 4 = −2 ⋅ = −

π

74. The graph is a sine graph with amplitude 4 and period π. 2π Find ω : π =

.

π

1 π  1  sin  ⋅  − sin  ⋅ 0  2 2 2  = π /2 sin (π / 4 ) − sin ( 0 ) = π /2 2 2 2 2 = 2 = ⋅ = π /2 2 π π

The average rate of change is

73. The graph is a reflected cosine graph with 2π . amplitude 4 and period 3 2π 2π Find ω : = ω 3 2πω = 6π 6π ω= =3 2π The equation is: y = − 4 cos ( 3 x ) .

2

4

π

f ( g ( x )) = sin ( 4 x )

ω

πω = 2π 2π ω= =2 π The equation is: y = 4sin ( 2 x ) .

75.

f (π / 2 ) − f ( 0 )

π /2−0

=

g ( f ( x)) = 4 (sin x ) = 4sin x

sin (π / 2 ) − sin ( 0 )

π /2

1− 0 2 = = π /2 π

The average rate of change is

2

π

.

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.


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Section 2.4: Graphs of the Sine and Cosine Functions

80.

f ( g ( x)) = cos

g ( f ( x)) = sin ( −3 x )

1 x 2

83. 1 1 g ( f ( x)) = ( cos x ) = cos x 2 2

81.

f ( g ( x)) = −2 ( cos x ) = −2 cos x

84.

g ( f ( x)) = cos ( −2 x )

85. I ( t ) = 220sin(60π t ), t ≥ 0 2π 1 = second ω 60π 30 Amplitude: A = 220 = 220 amperes Period: T =

82.

=

f ( g ( x)) = −3 (sin x ) = −3sin x

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Chapter 2: Trigonometric Functions c.

86. I (t ) = 120sin(30π t ), t ≥ 0 2π 2π 1 Period: T = = = second ω 30π 15 Amplitude: A = 120 = 120 amperes

V = IR 120sin(120π t ) = 20 I 6sin(120π t ) = I I (t ) = 6sin(120π t )

d.

A = 6 = 6 amperes

Amplitude: Period: T =

ω

=

2π 1 = second 120π 60

[V (t )]

2

89. a.

Amplitude: Period: T =

A = 220 = 220 volts 2π

ω

=

R

V0 sin ( 2πft )  = R 2 2 V sin ( 2πft ) = 0 R 2 V = 0 sin 2 ( 2πft ) R

87. V (t ) = 220sin(120π t ) a.

P (t ) =

2π 1 = second 120π 60

b, e.

b. The graph is the reflected cosine graph translated up a distance equivalent to the 1 , so ω = 4π f . amplitude. The period is 2f

The amplitude is c.

V = IR 220sin(120π t ) = 10 I I (t ) = 22sin(120π t ) Amplitude: Period: T =

c.

A = 22 = 22 amperes 2π

ω

=

2π 1 = second 120π 60

90. a.

88. V (t ) = 120sin(120π t ) a.

Amplitude: A = 120 = 120 volts Period: T =

ω

=

1 V0 2 V0 2 . ⋅ = 2 R 2R

The equation is: V2 V2 P ( t ) = − 0 cos ( 4πft ) + 0 2R 2R 2 V = 0 1 − cos ( 4πft )  2R

22sin(120π t ) = I

d.

2

2π 1 = second 120π 60

b, e.

Comparing the formulas: 1 sin 2 ( 2πft ) = (1 − cos ( 4πft ) ) 2 Since the tunnel is in the shape of one-half a sine cycle, the width of the tunnel at its base is one-half the period. Thus, 2π π T= = 2(28) = 56 or ω = . 28 ω The tunnel has a maximum height of 15 feet so we have A = 15 . Using the form y = A sin(ω x) , the equation for the sine curve that fits the opening is πx  y = 15sin  .  28 

b. Since the shoulders are 7 feet wide and the road is 14 feet wide, the edges of the road correspond to x = 7 and x = 21 . 180 Copyright © 2016 Pearson Education, Inc. Full file at https://testbankuniv.eu/Trigonometry-A-Unit-Circle-Approach-10th-Edition-Sullivan-Solutions-Manual


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Section 2.4: Graphs of the Sine and Cosine Functions

92. y = cos x , − 2π ≤ x ≤ 2π

 7π   π  15 2 15sin  ≈ 10.6  = 15sin   = 2  28  4

 21π   3π  15 2 ≈ 10.6 15sin   = 15sin   = 28 2    4  The tunnel is approximately 10.6 feet high at the edge of the road. 91. a.

2π ; 23 2π π = ; Emotional potential: ω = 28 14 2π Intellectual potential: ω = 33

Physical potential: ω =

93. y = sin x , − 2π ≤ x ≤ 2π

110

b.

0

0

#1, #2, #3

33

 2π  #1: P ( t ) = 50sin  t  + 50  23  π  # 2 : P ( t ) = 50sin  t  + 50  14 

94. Answers may vary.  7π 1   π 1   5π 1   13π 1  , , , , , , ,  −  6 2  6 2  6 2  6 2 95. Answers may vary.

 2π  #3 : P ( t ) = 50sin  t  + 50  33 

c.

 5π 1   π 1   π 1   5π 1   − , , − , , , , ,   3 2  3 2  3 2  3 2

No. 110

d.

96. 2sin x = −2

#2

7305

sin x = −1 Answers may vary.  π   3π   7π   11π  , −2  ,  , −2   − , −2  ,  , −2  ,   2   2   2   2 

#1

#3

7335 0 Physical potential peaks at 15 days after the 20th birthday, with minimums at the 3rd and 26th days. Emotional potential is 50% at the 17th day, with a maximum at the 10th day and a minimum at the 24th day. Intellectual potential starts fairly high, drops to a minimum at the 13th day, and rises to a maximum at the 29th day.

97. Answers may vary.  3π   π   5π   9π  ,1  − ,1 ,  ,1 ,  ,1 ,   4  4   4   4  98 – 102. Answers will vary.

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Chapter 2: Trigonometric Functions

103.

f ( x + h) − f ( x ) h ( x + h) 2 − 5( x + h) + 1 − ( x 2 − 5 x + 1) = h 2 2 ( x + 2 xh + h ) − (5 x + 5h) + 1 − x 2 + 5 x − 1 = h 2 2 x + 2 xh + h − 5 x − 5h + 1 − x 2 + 5 x − 1 = h 2 xh + h 2 − 5h h(2 x + h − 5) = = = 2x + h − 5 h h

8. y = cot x has no y-intercept. 9. The y-intercept of y = sec x is 1. 10. y = csc x has no y-intercept. 11. sec x = 1 when x = − 2π, 0, 2π; sec x = −1 when x = −π, π

y = 6 −1 = 5

3π π , ; 2 2 π 3π csc x = −1 when x = − , 2 2

12. csc x = 1 when x = −

(0,5)

The x-interecpts are: 0 = 3 x + 2 −1 1 = x+2 3 1 1 = x + 2 or − = x + 2 3 3 5 7 x=− or x = − 3 3 5 7 − ,0 , − ,0 3 3

13. y = sec x has vertical asymptotes when

1= 3 x+2 →

=

A=

= =

π 2

7. The y-intercept of y = tan x is 0.

y = 3 0 + 2 −1

180

4. y-axis; x = odd multiples of

6. True

11π 5π ≅ 4 4 5π 2 =− sin 4 2 105. The y-intercept is:

106. 40

π 2

5. b

104. −

π

3. origin; x = odd multiples of

x=−

3π π π 3π . ,− , , 2 2 2 2

14. y = csc x has vertical asymptotes when x = − 2π, − π, 0, π, 2π . 15. y = tan x has vertical asymptotes when x=−

2π 9 1 2 rθ 2 1 2 2π ( 5) 2 9 25π sq. units 9

3π π π 3π . ,− , , 2 2 2 2

16. y = cot x has vertical asymptotes when x = − 2π, − π, 0, π, 2π . 17. y = 3 tan x ; The graph of y = tan x is stretched vertically by a factor of 3.

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Section 2.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions

The domain is { x x ≠ kπ , k is an integer} . The

 kπ  The domain is  x x ≠ , k is an odd integer  . 2   The range is the set of all real number or (−∞, ∞) .

range is the set of all real number or (−∞, ∞) .

π 21. y = tan  2

 x  ; The graph of y = tan x is  2 horizontally compressed by a factor of .

18. y = −2 tan x ; The graph of y = tan x is stretched vertically by a factor of 2 and reflected about the x-axis.

π

 kπ  The domain is  x x ≠ , k is an odd integer  . 2   The range is the set of all real number or (−∞, ∞) .

The domain is { x x does not equal an odd integer} . The range is the set of all real number or (−∞, ∞) .

19. y = 4 cot x ; The graph of y = cot x is stretched vertically by a factor of 4.

1  22. y = tan  x  ; The graph of y = tan x is 2  horizontally stretched by a factor of 2.

The domain is { x x ≠ kπ , k is an integer} . The

The domain is { x x ≠ kπ , k is an integer} . The

range is the set of all real number or (−∞, ∞) .

range is the set of all real number or (−∞, ∞) . 1  23. y = cot  x  ; The graph of y = cot x is 4  horizontally stretched by a factor of 4.

20. y = −3cot x ; The graph of y = cot x is stretched vertically by a factor of 3 and reflected about the x-axis.

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Chapter 2: Trigonometric Functions

The domain is { x x ≠ 4kπ , k is an integer} . The

The domain is { x x ≠ kπ , k is an integer} . The

range is the set of all real number or (−∞, ∞) .

 1 1 range is  y y ≤ − or y ≥  . 2 2 

π 24. y = cot  4

 x  ; The graph of y = cot x is  4 horizontally stretched by a factor of .

27. y = −3csc x ; The graph of y = csc x is vertically stretched by a factor of 3 and reflected about the x-axis.

π

The domain is { x x ≠ 4k , k is an integer} . The

The domain is { x x ≠ kπ , k is an integer} . The

range is the set of all real number or (−∞, ∞) .

range is { y y ≤ −3 or y ≥ 3} .

25. y = 2sec x ; The graph of y = sec x is stretched vertically by a factor of 2.

28. y = −4sec x ; The graph of y = sec x is vertically stretched by a factor of 4 and reflected about the x-axis.

 kπ  The domain is  x x ≠ , k is an odd integer  . 2  

 kπ  The domain is  x x ≠ , k is an odd integer  . 2  

The range is { y y ≤ −2 or y ≥ 2} .

The range is { y y ≤ −4 or y ≥ 4} .

1 csc x ; The graph of y = csc x is vertically 2 1 compressed by a factor of . 2

26. y =

1  29. y = 4sec  x  ; The graph of y = sec x is 2  horizontally stretched by a factor of 2 and

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Section 2.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions 31. y = −2 csc (π x ) ; The graph of y = csc x is

vertically stretched by a factor of 4.

horizontally compressed by a factor of

1

π

,

vertically stretched by a factor of 2, and reflected about the x-axis.

The domain is { x x ≠ kπ , k is an odd integer} . The range is { y y ≤ −4 or y ≥ 4} .

1 csc ( 2 x ) ; The graph of y = csc x is 2 1 horizontally compressed by a factor of and 2 1 vertically compressed by a factor of . 2

The domain is { x x does not equal an integer} .

30. y =

The range is { y y ≤ −2 or y ≥ 2} .

π 32. y = −3sec  2

 x  ; The graph of y = sec x is  2 horizontally compressed by a factor of ,

π

vertically stretched by a factor of 3, and reflected about the x-axis.

 kπ  The domain is  x x ≠ , k is an integer  . The 2   1 1  range is  y y ≤ − or y ≥  . 2 2 

The domain is { x x does not equal an odd integer} . The range is { y y ≤ −3 or y ≥ 3} .

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Chapter 2: Trigonometric Functions

 2π 35. y = sec   3

 x  + 2 ; The graph of y = sec x is  3 horizontally compressed by a factor of and 2π shifted up 2 units.

1  33. y = tan  x  + 1 ; The graph of y = tan x is 4  horizontally stretched by a factor of 4 and shifted up 1 unit.

The domain is { x x ≠ 2kπ , k is an odd integer} .

 3  The domain is  x x ≠ k , k is an odd integer  . 4  

The range is the set of all real number or (−∞, ∞) .

The range is { y y ≤ 1 or y ≥ 3} .

34. y = 2 cot x − 1 ; The graph of y = cot x is vertically stretched by a factor of 2 and shifted down 1 unit.

 3π 36. y = csc   2

 x  ; The graph of y = csc x is  2 horizontally compressed by a factor of . 3π

The domain is { x x ≠ kπ , k is an integer} . The range is the set of all real number or (−∞, ∞) .  2  The domain is  x x ≠ k , k is an integer  . The 3  

range is { y y ≤ −1 or y ≥ 1} .

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Section 2.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions

1 1  tan  x  − 2 ; The graph of y = tan x is 2 4  horizontally stretched by a factor of 4, vertically 1 compressed by a factor of , and shifted down 2 2 units.

37. y =

1  39. y = 2 csc  x  − 1 ; The graph of y = csc x is 3  horizontally stretched by a factor of 3, vertically stretched by a factor of 2, and shifted down 1 unit.

The domain is { x x ≠ 3π k , k is an integer} .

The domain is { x x ≠ 2π k , k is an odd integer} .

The range is { y y ≤ −3 or y ≥ 1} .

The range is the set of all real number or (−∞, ∞) .

1  40. y = 3sec  x  + 1 ; The graph of y = sec x is 4  horizontally stretched by a factor of 4, vertically stretched by a factor of 3, and shifted up 1 unit.

1  38. y = 3cot  x  − 2 ; The graph of y = cot x is 2  horizontally stretched by a factor of 2, vertically stretched by a factor of 3, and shifted down 2 units.

The domain is { x x ≠ 2π k , k is an odd integer} . The range is { y y ≤ −2 or y ≥ 4} .

The domain is { x x ≠ 2π k , k is an integer} . The

π  3 f   − f ( 0) −0 tan (π / 6 ) − tan ( 0 ) 6  41. = = 3 π π /6 π /6 −0 6 3 6 2 3 = ⋅ = π 3 π 2 3 The average rate of change is .

range is the set of all real number or (−∞, ∞) .

π

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Chapter 2: Trigonometric Functions

42.

g ( f ( x)) = 4 ( tan x ) = 4 tan x

π  2 3 f   − f ( 0) −1 π sec / 6 sec 0 − ( ) ( ) 6   = = 3 π π /6 π /6 −0 6 2 3 −3 6 2 3 2− 3 = ⋅ = π π 3

The average rate of change is

(

)

2 3 2− 3

).

(

π

π  f   − f ( 0) tan ( 2 ⋅ π / 6 ) − tan ( 2 ⋅ 0 ) 6 = 43. π π /6 −0 6 3 −0 6 3 = = π /6 π 6 3 The average rate of change is .

46.

f ( g ( x)) = 2sec

1 x 2

π

π  f   − f ( 0) sec ( 2 ⋅ π / 6 ) − sec ( 2 ⋅ 0 ) 6 44. = π π /6 −0 6 2 −1 6 = = π /6 π 6 The average rate of change is .

π

45.

g ( f ( x)) =

f ( g ( x)) = tan ( 4 x )

47.

1 ( 2sec x ) = sec x 2

f ( g ( x)) = −2 ( cot x ) = −2 cot x

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Section 2.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions g ( f ( x)) = cot ( −2 x )

48.

f ( g ( x)) =

50.

1 ( 2 csc x ) = csc x 2

51. a.

g ( f ( x)) = 2 csc

1 x 2

Consider the length of the line segment in two sections, x, the portion across the hall that is 3 feet wide and y, the portion across that hall that is 4 feet wide. Then, 3 4 cos θ = and sin θ = x y 3 4 x= y= cos θ sin θ Thus, 3 4 + = 3sec θ + 4 csc θ . L = x+ y = cos θ sin θ

b. Let Y1 = 25

0

c.

3 4 + . cos x sin x

2

0

Use MINIMUM to find the least value: 25

49.

0

2

0

L is least when θ ≈ 0.83 .

d.

L≈

3 4 + ≈ 9.86 feet . cos ( 0.83) sin ( 0.83)

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Chapter 2: Trigonometric Functions

52. a.

d ( t ) = 10 tan(π t )

53.

b.

d ( t ) = 10 tan(π t ) is undefined at t =

1 and 2

3 , or in general at 2  k  k is an odd integer  . At these t = 2   instances, the length of the beam of light approaches infinity. It is at these instances in the rotation of the beacon when the beam of light being cast on the wall changes from one side of the beacon to the other. t=

c.

d.

e.

t

d ( t ) = 10 tan(π t )

0 0.1

0 3.2492

0.2

7.2654

0.3

13.764

0.4

30.777

Yes, the two functions are equivalent. 54. y 2 = x − 4 Test x-axis symmetry: Let y = − y

( − y )2 = x − 4

y 2 = x − 4 same

Test y-axis symmetry: Let x = − x y 2 = − x − 4 different

d (0.1) − d (0) 3.2492 − 0 = ≈ 32.492 0.1 − 0 0.1 − 0 d (0.2) − d (0.1) 7.2654 − 3.2492 = ≈ 40.162 0.2 − 0.1 0.2 − 0.1 d (0.3) − d (0.2) 13.764 − 7.2654 = ≈ 64.986 0.3 − 0.2 0.3 − 0.2 d (0.4) − d (0.3) 30.777 − 13.764 = ≈ 170.13 0.4 − 0.3 0.4 − 0.3

Test origin symmetry: Let x = − x and y = − y .

( − y )2 = − x − 4

y 2 = − x − 4 different

Therefore, the graph will have x-axis symmetry. 55.

( x − h) 2 + ( y − k ) 2 = r 2 ( x − ( −3)) 2 + ( y − 1) 2 = (3) 2

The first differences represent the average rate of change of the beam of light against the wall, measured in feet per second. For example, between t = 0 seconds and t = 0.1 seconds, the average rate of change of the beam of light against the wall is 32.492 feet per second.

( x + 3) 2 + ( y − 1) 2 = 9

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Section 2.6: Phase Shift; Sinusoidal Curve Fitting 4. y = 3sin(3 x − π)

4 −6 (4) 2 + 10 2−6 = 16 + 10 −4 2 = =− 26 13

56. h(4) =

Amplitude: Period:

A = 3 =3 T=

ω φ π Phase Shift: = ω 3

=

2π 3

Interval defining one cycle: φ φ  π  ω , ω + T  =  3 ,π      Subinterval width: T 2π / 3 π = = 4 4 6

57. The relation is a circle with center (0, 4) and radius 4 so it is not a function.

Key points:  π   π   2π   5π  , 0  ,  , −3  , (π , 0 )  , 0  ,  ,3  ,  3  2   3   6 

Section 2.6 1. phase shift 2. False 3. y = 4sin(2 x − π) Amplitude: Period:

A = 4 =4 T=

ω φ π Phase Shift: = ω 2

=

2π =π 2

Interval defining one cycle: φ φ   π 3π  ω , ω + T  =  2 , 2      Subinterval width: T π = 4 4 Key points:  π   3π   5π   3π   , 0  ,  , 4  , (π , 0 ) ,  , −4  ,  , 0  2 4 4        2 

π  5. y = 2 cos  3 x +  2  Amplitude: A = 2 = 2

2π 3 ω  π − π φ  2  Phase Shift: = =− 3 6 ω Interval defining one cycle: φ φ   π π ω , ω + T  = − 6 , 2      Subinterval width: T 2π / 3 π = = 4 4 6 Key points:  π  π  π  π   − , 2  , ( 0, 0 ) ,  , −2  ,  , 0  ,  , 2  6 6     3  2  Period:

T=

=

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Chapter 2: Trigonometric Functions

Subinterval width: T π = 4 4 Key points:  π   π   π   3π   − , 0  , ( 0, −3) ,  , 0  ,  ,3  ,  , 0  4  2   4   4 

6. y = 3cos ( 2 x + π ) Amplitude:

A = 3 =3

2π =π 2 π φ −π Phase Shift: = =− 2 ω 2 Interval defining one cycle: φ φ   π π ω , ω + T  = − 2 , 2      Subinterval width: T π = 4 4 Key points:  π   π  π  π   − ,3  ,  − , 0  , ( 0, −3) ,  , 0  ,  ,3  2 4     4  2  Period:

T=

ω

=

π  8. y = − 2 cos  2 x −  2  Amplitude: A = − 2 = 2 Period:

T=

ω

=

2π =π 2

π

φ 2 π Phase Shift: = = ω 2 4

Interval defining one cycle: φ φ   π 5π  ω , ω + T  =  4 , 4      Subinterval width: T π = 4 4 Key points: π   π   3π   5π   , −2  ,  , 0  ,  , 2  , (π , 0 ) ,  , −2  4  2   4   4 

π  7. y = −3sin  2 x +  2  Amplitude: A = − 3 = 3 Period:

T=

ω −

=

2π =π 2

π

π φ Phase Shift: = 2 =− 2 4 ω

Interval defining one cycle: φ φ   π 3π   ω , ω + T  = − 4 , 4     

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Section 2.6: Phase Shift; Sinusoidal Curve Fitting 9. y = 4sin(πx + 2) − 5 A = 4 =4

Amplitude:

2π =2 π ω φ −2 2 = =− Phase Shift: π π ω Interval defining one cycle: 2 φ φ   2  ω , ω + T  = − π , 2 − π      Subinterval width: T 2 1 = = 4 4 2 Key points:  2  1 2   2   − , −5  ,  − , −1 , 1 − , −5  , π π π 2       Period:

T=

3 2   − , −9  , 2 π 

=

11. y = 3cos(πx − 2) + 5 Amplitude: Period:

=

2π =2 π

Interval defining one cycle: 2 φ φ  2 ω , ω + T  = π , 2 + π      Subinterval width: T 2 1 = = 4 4 2 Key points:  2  1 2   2  3 2   ,8  ,  + ,5  , 1 + , 2  ,  + ,5  , π   2 π   π   2 π  2    2 + ,8  π  

A = 2 =2

2π =1 2π φ −4 2 Phase Shift: = =− ω 2π π Interval defining one cycle: 2 φ φ   2  ω , ω + T  =  − π ,1 − π      T 1 Subinterval width: = 4 4 Key points:  2  1 2  1 2   − ,6 ,  − , 4 ,  − , 2 ,  π  4 π  2 π  Period:

T=

ω φ 2 Phase Shift: = ω π

2    2 − , −5  π  

10. y = 2 cos(2πx + 4) + 4 Amplitude:

A = 3 =3

T=

ω

=

3 2   − ,4 , 4 π 

 2  1 − , 6   π 

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Chapter 2: Trigonometric Functions

Key points:  π   π   3π   5π   , 0  ,  ,3  ,  , 0  , (π , −3) ,  , 0  4 2 4        4 

12. y = 2 cos(2πx − 4) − 1 A = 2 =2

Amplitude:

2π =1 ω 2π φ 4 2 = = Phase Shift: ω 2π π Interval defining one cycle: 2 φ φ  2  ω , ω + T  =  π ,1 + π      Subinterval width: T 1 = 4 4 Key points:  2  1 2  1 2   ,1 ,  + , −1 ,  + , −3  , π π π 4 2       Period:

T=

3 2   + , −1  , 4 π 

=

π   π   14. y = −3cos  − 2 x +  = −3cos  −  2 x −   2 2     π  = −3cos  2 x −  2 

 2  1 + ,1  π 

Amplitude: Period:

A = −3 = 3 T=

ω

=

2π =π 2

π

φ 2 π Phase Shift: = = ω 2 4

Interval defining one cycle: φ φ   π 5π  ω , ω + T  =  4 , 4      Subinterval width: T π = 4 4 Key points: π   π   3π   5π   , −3  ,  , 0  ,  ,3  , (π , 0 ) ,  , −3  4  2   4   4 

π   π   13. y = −3sin  − 2 x +  = −3sin  −  2 x −   2 2     π  = 3sin  2 x −  2  Amplitude: Period:

A = 3 =3 T=

ω

=

2π =π 2

π

φ 2 π Phase Shift: = = ω 2 4

Interval defining one cycle: φ φ   π 5π  ω , ω + T  =  4 , 4      Subinterval width: T π = 4 4 194 Copyright © 2016 Pearson Education, Inc. Full file at https://testbankuniv.eu/Trigonometry-A-Unit-Circle-Approach-10th-Edition-Sullivan-Solutions-Manual


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Section 2.6: Phase Shift; Sinusoidal Curve Fitting

15.

19. y = 2 tan ( 4 x − π )

φ 1 = ω 2 2π 2π φ φ 1 ω= = = = =2 ω 2 2 T π φ =1 A = 2; T = π;

Begin with the graph of y = tan x and apply the following transformations: 1) Shift right π units  y = tan ( x − π ) 

Assuming A is positive, we have that y = A sin(ω x − φ ) = 2sin(2 x − 1)

2) Horizontally compress by a factor of  y = tan ( 4 x − π )    3) Vertically stretch by a factor of 2  y = 2 tan ( 4 x − π )   

  1  = 2sin  2  x −   2   

16.

A = 3; T =

π φ =2 ; 2 ω

1 4

2π 2π φ φ = =2 = =4 π T ω 4 2 φ =8 Assuming A is positive, we have that y = A sin(ω x − φ ) = 3sin(4 x − 8)

ω=

= 3sin  4 ( x − 2 ) 

17.

φ 1 =− ω 3 2π 2π 2 φ φ 1 = =− = = ω= 2 ω 3 T 3π 3 A = 3; T = 3π;

1 cot ( 2 x − π ) 2 Begin with the graph of y = cot x and apply the following transformations:

20. y =

3 1 2 2 3 3 9 Assuming A is positive, we have that 2 2 y = A sin(ω x − φ ) = 3sin  x +  3 9   2  1  = 3sin   x +   3  3 

φ =− ⋅ =−

18.

1) Shift right π units  y = cot ( x − π )  2) Horizontally compress by a factor of  y = cot ( 2 x − π )   

3) Vertically compress by a factor of

φ = −2 ω 2π 2π φ φ ω= = =2 = = −2 ω 2 T π φ = −4

1    y = 2 cot ( 2 x − π )   

A = 2; T = π;

Assuming A is positive, we have that y = A sin(ω x − φ ) = 2sin(2 x + 4) = 2sin  2 ( x + 2 ) 

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1 2

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Chapter 2: Trigonometric Functions

π  21. y = 3csc  2 x −  4  Begin with the graph of y = csc x and apply the following transformations:  π   units  y = csc  x −   4   4   1 2) Horizontally compress by a factor of 2   π   y = csc  2 x −   4     3) Vertically stretch by a factor of 3  π    y = 3csc  2 x −   4    

1) Shift right

π

π  23. y = − cot  2 x +  2  Begin with the graph of y = cot x and apply the following transformations: 1) Shift left

π 2

 π   units  y = cot  x +   2    

2) Horizontally compress by a factor of  π    y = cot  2 x +   2    

3) Reflect about the x-axis  π    y = − cot  2 x +   2     1 22. y = sec ( 3x − π ) 2 Begin with the graph of y = sec x and apply the following transformations: 1) Shift right π units  y = sec ( x − π )  1 2) Horizontally compress by a factor of 3  y = sec ( 3 x − π )    1 3) Vertically compress by a factor of 2 1    y = 2 sec ( 3 x − π )   

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Section 2.6: Phase Shift; Sinusoidal Curve Fitting

π  1 26. y = − csc  − π x +  2 4  Begin with the graph of y = csc x and apply the following transformations:  π π   1) Shift left units  y = csc  x +   4   4  

π  24. y = − tan  3 x +  2  Begin with the graph of y = tan x and apply the following transformations:  π π   1) Shift left units  y = tan  x +   2  2   1 2) Horizontally compress by a factor of 3

 π   2) Reflect about the y-axis  y = csc  − x +   4     2 3) Horizontally compress by a factor of

 π    y = tan  3x +   2    3) Reflect about the x-axis  π    y = − tan  x +   2   

π

 π   1  y = csc  − π x +   4     2 3) Reflect about the x-axis  π   1  y = − csc  − π x +   4    2 

π   ,1  12 

 π  , −1   4

25. y = − sec ( 2π x + π )

Begin with the graph of y = sec x and apply the following transformations: 1) Shift left π units  y = sec ( x + π )  1 2) Horizontally compress by a factor of 2π  y = sec ( 2π x + π )    3) Reflect about the x-axis  y = − sec ( 2π x + π )   

π  27. I ( t ) = 120sin  30π t −  , t ≥ 0 3  2π 2π 1 Period: T= = = second ω 30π 15 Amplitude: A = 120 = 120 amperes

π

φ 1 Phase Shift: second = 3 = ω 30π 90

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Chapter 2: Trigonometric Functions c.

π  28. I ( t ) = 220sin  60π t −  , t ≥ 0 6  2π 2π 1 Period: T= = = second ω 60π 30 Amplitude: A = 220 = 220 amperes π

d.

1 φ Phase Shift: second = 6 = ω 60π 360

y = 9.46sin (1.247 x + 2.906 ) + 24.088

e.

29. a. 30. a.

b.

33 − 16 17 = = 8.5 2 2 33+16 49 Vertical Shift: = = 24.5 2 2 2π 2π = ω= 5 5 Phase Shift (use y = 16, x = 6): Amplitude: A =

b.

 2π  16 = 8.5sin  ⋅ 6 − φ  + 24.5  5  12 π   −8.5 = 8.5sin  −φ   5 

79.8 − 36.0 43.8 = = 21.9 2 2 79.8+36.0 115.8 = = 57.9 Vertical Shift: 2 2 2π π = ω= 12 6 Phase Shift (use y = 36.0, x = 1): Amplitude: A =

36.0 = 21.9sin

 12π  −1 = sin  −φ  5   π 12π − = −φ 2 5 29π φ= 10 11π   2π x− Thus, y = 8.5sin   + 24.5 or 10   5

− 21.9 = 21.9sin −1 = sin −

 2π  11   y = 8.5sin   x −   + 24.5 . 4  5 

π ⋅1 − φ + 57.9 6 π −φ 6

π −φ 6

π π = −φ 2 6 2π φ= 3

Thus, y = 21.9sin y = 21.9sin

π 2π x− + 57.9 or 6 3

π ( x − 4) + 57.9 . 6

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Section 2.6: Phase Shift; Sinusoidal Curve Fitting c.

c.

d. d.

y = 24.25sin(0.493 x − 1.927) + 51.61

y = 21.68sin ( 0.516 x − 2.124) + 57.81

/

/

e.

90

e.

0

13

20

32. a.

31. a.

b.

b.

75.4 − 28.1 47.3 Amplitude: A = = = 23.65 2 2 75.4+28.1 103.5 = = 51.75 Vertical Shift: 2 2 2π π = ω= 12 6 Phase Shift (use y = 28.1, x = 1):

32.9 = 22.05sin

π ⋅1 − φ + 51.75 28.1 = 23.65sin 6 − 23.65 = 23.65sin −1 = sin

− 22.05 = 22.05sin

π −φ 6

−1 = sin

π −φ 6

π π − = −φ 2 6 2π φ= 3

Thus, y = 23.65sin y = 23.65sin

77.0 − 32.9 44.1 = = 22.05 2 2 77.0+32.9 109.9 = = 54.95 Vertical Shift: 2 2 2π π = ω= 12 6 Phase Shift (use y = 32.9, x = 1): Amplitude: A =

π π = −φ 2 6 2π φ= 3

y = 22.05sin

π ( x − 4) + 51.75 . 6

π −φ 6

π −φ 6

Thus, y = 22.05sin

π 2π x− + 51.75 or 6 3

π ⋅1 − φ + 54.95 6

π 2π x− + 54.95 or 6 3

π ( x − 4) + 54.95 . 6

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Chapter 2: Trigonometric Functions 34. a.

c.

b.

d.

y = 21.73sin(0.518 x − 2.139) + 54.82

80

e.

0.10 + 12.4167 = 12.5167 hours which is at 12:31 PM. 9.97 − (0.59) 9.38 = = 4.69 2 2 9.97 + (0.59) 10.56 Vertical Shift: = = 5.28 2 2 2π 24π π ω= = = 12.4167 6.20835 149 Phase Shift (use y = 9.97, x = 0.10): Ampl: A =

9.97 = 4.69sin

24π ⋅ 0.10 − φ + 5.28 149

4.69 = 4.69sin

24π ⋅ 0.10 − φ 149

1 = sin 0

33. a.

b.

π 2.4π = −φ 2 149 φ ≈ −1.5202

13

20

6.5 + 12.4167 = 18.9167 hours which is at 6:55 PM. 5.86 − ( −0.38)

Thus, y = 4.69sin

6.24 = 3.12 2 2 5.86 + ( −0.38) 5.48 Vertical Shift: = = 2.74 2 2 2π 24π π ω= = = 12.4167 6.20835 149 Phase Shift (use y = 5.86, x = 6.5): Ampl: A =

24π ⋅ 6.5 − φ + 2.74 149

3.12 = 3.12sin

24π ⋅ 6.5 − φ 149

π

or y = 4.69sin

=

5.86 = 3.12sin

1 = sin

c.

35. a.

156π −φ 2 149 φ ≈ 1.7184 =

or y = 3.12sin c.

y = 3.12sin

y = 4.69sin

24π x + 1.5202 + 5.28 149

24π ( x + 3.0042) + 5.28 . 149

24π (18) + 1.5202 + 5.28 149

≈ 0.90 feet

156π −φ 149

Thus, y = 3.12sin

2.4π −φ 149

24π x − 1.7184 + 2.74 149

13.75 − 10.52 = 1.615 2 13.75 + 10.52 Vertical Shift: = 12.135 2 2π ω= 365 Phase Shift (use y = 13.75, x = 172): Amplitude: A =

13.75 = 1.615sin

2π ⋅172 − φ + 12.135 365

1.615 = 1.615sin

2π ⋅172 − φ 365

344π −φ 365 π 344π = −φ 2 365 φ ≈ 1.3900 1 = sin

24π ( x − 3.3959) + 2.74 . 149

24π (15) − 1.7184 + 2.74 149

≈ 1.49 feet

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Section 2.6: Phase Shift; Sinusoidal Curve Fitting

Thus, y = 1.615sin y = 1.615sin

b.

c.

2π x − 1.39 + 12.135 or 365

2π ( x − 80.75) + 12.135 . 365

2π (91 − 80.75) + 12.135 365 ≈ 12.42 hours

y = 1.615sin

c. d. The actual hours of sunlight on April 1, 2014 were 12.75 hours. This is very close to the predicted amount of 12.71 hours. 37. a.

d. The actual hours of sunlight on April 1, 2014 were 12.43 hours. This is very close to the predicted amount of 12.42 hours. 36. a.

19.37 = 6.96sin

15.27 − 9.07 = 3.1 2 15.27 + 9.07 Vertical Shift: = 12.17 2 2π ω= 365 Phase Shift (use y = 15.27, x = 172): Amplitude: A =

15.27 = 3.1sin 3.1 = 3.1sin

6.96 = 6.96sin

2π ⋅172 − φ 365

Thus, y = 6.96sin

344π −φ 365 π 344π = −φ 2 365 φ ≈ 1.39

b.

2π ⋅172 − φ 365

344π −φ 365 π 344π = −φ 2 365 φ ≈ 1.39

y = 6.96sin

1 = sin

y = 3.1sin

2π ⋅172 − φ + 12.41 365

1 = sin

2π ⋅172 − φ + 12.17 365

Thus, y = 3.1sin

19.37 − 5.45 = 6.96 2 19.37 + 5.45 Vertical Shift: = 12.41 2 2π ω= 365 Phase Shift (use y = 19.37, x = 172): Amplitude: A =

b.

2π x − 1.39 + 12.41 or 365

2π ( x − 80.75) + 12.41 . 365

2π (91) − 1.39 + 12.41 365 ≈ 13.63 hours

y = 6.96sin

2π x − 1.39 + 12.17 or 365

2π ( x − 80.75) + 12.17 . 365

2π (91) − 1.39 + 12.17 365 ≈ 12.71 hours

y = 3.1sin

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Chapter 2: Trigonometric Functions c. d. The actual hours of sunlight on April 1, 2014 were 12.38 hours. This is very close to the predicted amount of 12.35 hours. 39 – 40. Answers will vary. 41.

13.42 − 10.83 = 1.295 2 13.42 + 10.83 Vertical Shift: = 12.125 2 2π ω= 365 Phase Shift (use y = 13.42, x = 172): Amplitude: A =

13.42 = 1.295sin

2π ⋅172 − φ + 12.125 365

1.295 = 1.295sin

2π ⋅172 − φ 365

344π −φ 365 π 344π = −φ 2 365 φ ≈ 1.39

Thus, y = 1.295sin

the

Use MAXIMUM and MINIMUM on the graph of y1 = 2 x3 − 3x 2 − 8 x + 14 .

1 = sin

b.

on

interval ( −4,5 )

d. The actual hours of sunlight on April 1, 2014 was 13.37 hours. This is close to the predicted amount of 13.63 hours. 38. a.

f ( x ) = 2 x3 − 3x 2 − 8 x + 14

local maximum: f ( −0.76) ≈ 17.47 local minimum:

2π x − 1.39 + 12.125 . 365

42.

2π y = 1.295sin (91) − 1.39 + 12.125 365 ≈ 12.35 hours

f (1.76) = 1.53

−16(2) 2 + 5(2) − −16( −1)2 + 5( −1) 2 − ( −1)

=

[ −16(4) + 10] − [ −16 − 5] 3

[ −54] − [ −21] = −54 + 21 = 3 −33 = = −11 3

c.

43.

3

7π 180 = 105° 12 π

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Chapter 2 Review Exercises

44. x 2 − 9 = −5

11. sin 2 20º +

2

x =4 x = ±2

1 = sin 2 20º + cos 2 20º = 1 sec2 20º

12. sec50º ⋅ cos 50º =

But only the −2 is in the domain of the part of the piecewise function. −3 x + 7 = −5 −3 x = −12 x=4

Thus the values are { −2, 4} .

1 ⋅ cos 50º = 1 cos 50º

13.

cos(−40º ) cos 40º = =1 cos 40º cos 40º

14.

sin(−40º ) − sin 40º = = −1 sin 40º sin 40º

15. sin 400º ⋅ sec ( −50º ) = sin 400º ⋅ sec 50º 1 cos 50º sin 40º sin 40º = = cos 50º sin(90º −50º ) sin 40º = =1 sin 40º = sin ( 40º +360º ) ⋅

Chapter 2 Review Exercises 1. 135° = 135 ⋅ 2. 18° = 18 ⋅

3.

3π π radian = radians 180 4

4 π and 0 < θ < , so θ lies in quadrant I. 5 2 Using the Pythagorean Identities: cos 2 θ = 1 − sin 2 θ

16. sin θ =

π π radian = radian 180 10

3π 3π 180 degrees = 135° = ⋅ 4 4 π

4. −

2

16 9 4 = cos 2 θ = 1 −   = 1 − 25 25 5 9 3 =± 25 5 Note that cos θ must be positive since θ lies in 3 quadrant I. Thus, cos θ = . 5 4 sin θ 5 4 5 4 = = ⋅ = tan θ = cos θ 53 5 3 3

5π 5π 180 =− ⋅ degrees = − 450° 2 2 π

5. tan

cos θ = ±

π π 1 1 − sin = 1 − = 4 6 2 2

6. 3sin 45º − 4 tan

7. 6 cos

π 2 3 3 2 4 3 = 3⋅ − 4⋅ = − 6 2 3 2 3

 3π 2  π + 2 tan  −  = 6  −  + 2 − 3 4  3  2 

(

)

= −3 2 − 2 3 π 5π  π  5π  8. sec  −  − cot  −  = sec + cot = 2 +1 = 3 3 4  3  4 

csc θ =

1 1 5 5 = 4 = 1⋅ = sin θ 5 4 4

sec θ =

1 1 5 5 = = 1⋅ = cos θ 53 3 3

cot θ =

1 1 3 3 = 4 = 1⋅ = tan θ 3 4 4

12 and sin θ < 0 , so θ lies in quadrant III. 5 Using the Pythagorean Identities:

17. tan θ =

9. tan π + sin π = 0 + 0 = 0 10. cos 540º − tan( − 405º ) = −1 − ( −1)

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Chapter 2: Trigonometric Functions

12 and θ lies in quadrant II. 13 Using the Pythagorean Identities: cos 2 θ = 1 − sin 2 θ

sec 2 θ = tan 2 θ + 1

19. sin θ =

2

144 169  12  sec 2 θ =   + 1 = +1 = 25 25  5 169 13 =± 25 5 Note that sec θ must be negative since θ lies in 13 quadrant III. Thus, sec θ = − . 5 1 1 5 cos θ = = =− sec θ − 135 13 sec θ = ±

tan θ =

2

144 25  12  cos 2 θ = 1 −   = 1 − = 169 169  13  25 5 =± 169 13 Note that cos θ must be negative because θ lies 5 in quadrant II. Thus, cos θ = − . 13 12 sin θ 12  13  12 = 13 =  −  = − tan θ = cos θ − 135 13  5  5 cos θ = ±

sin θ , so cos θ

12  5  12 −  = − 5  13  13 1 1 13 csc θ = = =− sin θ − 12 12 13

sin θ = ( tan θ )( cos θ ) =

cot θ =

1 1 5 = = tan θ 125 12

5 and tan θ < 0 , so θ lies in quadrant II. 4 Using the Pythagorean Identities: tan 2 θ = sec2 θ − 1

18. sec θ = −

2

1 1 13 = =− cos θ − 135 5

cot θ =

1 1 5 = 12 = − tan θ − 5 12

2

25 144  5 cos 2 θ = 1 −  −  = 1 − = 13 169 169  

9 3 =± 16 4

144 12 =± 169 13 Note that cos θ must be positive because θ lies 12 . in quadrant IV. Thus, cos θ = 13 sin θ − 135 5  13  5 = 12 = −   = − tan θ = cos θ 13  12  12 13 cos θ = ±

3 Note that tan θ < 0 , so tan θ = − . 4 1 1 4 cos θ = = =− sec θ − 54 5 sin θ , so cos θ

3 4 3 sin θ = ( tan θ )( cos θ ) = −  −  = . 4 5 5 1 1 5 csc θ = = = sin θ 35 3 cot θ =

sec θ =

5 3π and < θ < 2π (quadrant IV) 13 2 Using the Pythagorean Identities: cos 2 θ = 1 − sin 2 θ

25 9  5 tan θ =  −  − 1 = −1 = 4 16 16  

tan θ =

1 1 13 = 12 = sin θ 13 12

20. sin θ = −

2

tan θ = ±

csc θ =

1 1 4 = =− tan θ − 34 3

csc θ =

1 1 13 = =− sin θ − 135 5

sec θ =

1 1 13 = 12 = cos θ 13 12

cot θ =

1 1 12 = =− tan θ − 125 5

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Chapter 2 Review Exercises

1 and 180º < θ < 270º (quadrant III) 3 Using the Pythagorean Identities: sec 2 θ = tan 2 θ + 1

21. tan θ =

csc2 θ = 1 + ( −2 ) = 1 + 4 = 5 2

2

1 10 1 sec 2 θ =   + 1 = + 1 = 9 9 3 10 10 sec θ = ± =± 9 3 Note that sec θ must be negative since θ lies in

quadrant III. Thus, sec θ = − cos θ =

tan θ =

1 = sec θ

1 −

10 3

=−

3 10

π

< θ < π (quadrant II) 2 Using the Pythagorean Identities: csc2 θ = 1 + cot 2 θ

23. cot θ = − 2 and

10 . 3

10

10

=−

3 10 10

sin θ , so cos θ

1  3 10  10 sin θ = ( tan θ )( cos θ ) =  −  = − 3  10  10 1 1 10 = =− = − 10 csc θ = sin θ 10 10 − 10 1 1 = =3 cot θ = tan θ 13 3π < θ < 2π (quadrant IV) 2 Using the Pythagorean Identities: tan 2 θ = sec2 θ − 1

csc θ = ± 5 Note that csc θ must be positive because θ lies in quadrant II. Thus, csc θ = 5 .

1 1 5 5 = ⋅ = csc θ 5 5 5 cos θ , so cot θ = sin θ  5 2 5 . cos θ = ( cot θ )( sin θ ) = −2   = − 5  5  1 1 1 = =− tan θ = cot θ −2 2

sin θ =

sec θ =

1 1 5 5 = 2 5 =− =− cos θ − 5 2 2 5

24. y = 2sin(4 x) The graph of y = sin x is stretched vertically by a factor of 2 and compressed horizontally by a 1 factor of . 4

22. sec θ = 3 and

tan 2 θ = 32 − 1 = 9 − 1 = 8 tan θ = ± 8 = ±2 2 Note that tan θ must be negative since θ lies in

quadrant IV. Thus,. tan θ = −2 2 . 1 1 = cos θ = sec θ 3 sin θ tan θ = , so cos θ 2 2 1 . sin θ = ( tan θ )( cos θ ) = −2 2   = − 3 3 csc θ =

1 1 3 2 3 2 = 2 2 =− ⋅ =− sin θ − 3 4 2 2 2

cot θ =

1 1 2 2 = ⋅ =− tan θ −2 2 2 4

Domain: ( −∞, ∞ )

Range: [ −2, 2]

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Chapter 2: Trigonometric Functions 27. y = − 2 tan(3x ) The graph of y = tan x is stretched vertically by a factor of 2, reflected across the x-axis, and 1 compressed horizontally by a factor of . 3

25. y = −3cos(2 x) The graph of y = cos x is stretched vertically by a factor of 3, reflected across the x-axis, and 1 compressed horizontally by a factor of . 2

π π   Domain:  x | x ≠ + k ⋅ , k is an integer  6 3   Range: ( −∞, ∞ )

Domain: ( −∞, ∞ )

Range: [ −3,3]

26. y = tan( x + π) The graph of y = tan x is shifted π units to the left.

π  28. y = cot  x +  4 

The graph of y = cot x is shifted

π 4

units to the

left.

kπ   Domain:  x | x ≠ , k is an odd integer  2   Range: ( −∞, ∞ )

π   Domain:  x | x ≠ − + kπ , k is an integer  4   Range: ( −∞, ∞ )

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Chapter 2 Review Exercises 29. y = 4sec ( 2 x )

31. y = 4sin ( 2 x + 4 ) − 2

The graph of y = sec x is stretched vertically by a factor of 4 and compressed horizontally by a 1 factor of . 2 y 4 3 2 1

(⫺␲, 4)

3␲

⫺––– 4 ␲ , ⫺4 ⫺ –– 2

(

)

⫺ –– 4

⫺2 ⫺3 ⫺4 ⫺5

1 , 2 stretched vertically by a factor of 4, and shifted down 2 units.

compressed horizontally by a factor of

(␲, 4)

(0, 4)

–– 4

The graph of y = sin x is shifted left 4 units,

3␲ ––– 4

x

(––␲2 , ⫺4)

kπ   , k is an odd integer  Domain:  x | x ≠ 4   Range: { y | y ≤ −4 or y ≥ 4} π  30. y = csc  x +  4 

The graph of y = csc x is shifted

π 4

units to the

left.

Domain: ( −∞, ∞ )

Range: [ −6, 2]

x π 32. y = 5cot  −  3 4 The graph of y = cot x is shifted right π4 units,

stretched horizontally by a factor of 3, and stretched vertically by a factor of 5.

π   Domain:  x | x ≠ − + kπ , k is an integer  4   Range: { y | y ≤ −1 or y ≥ 1}

3π   Domain:  x | x ≠ + k ⋅ 3π , k is an integer  4   Range: ( −∞, ∞ )

33. y = sin(2 x)

Amplitude = 1 = 1 ; Period =

2π =π 2

34. y = − 2 cos(3π x)

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2π 2 = 3π 3


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Chapter 2: Trigonometric Functions 35. y = 4sin(3 x) Amplitude: Period:

A = 4 =4 T=

=

2π 3

ω φ 0 Phase Shift: = =0 ω 3

2 38. y = − cos ( πx − 6 ) 3 2 2 = 3 3 2π 2π Period: = =2 T= ω π φ 6 Phase Shift: = ω π Amplitude:

π 1 36. y = − cos  x +  2 2  Amplitude: A = −1 = 1 Period:

T=

ω

=

A = −

2π = 4π 1 2

π − φ Phase Shift: = 2 = −π 1 ω 2

39. The graph is a cosine graph with amplitude 5 and period 8π. 2π Find ω : 8π =

ω

8πω = 2π 2π 1 ω= = 8π 4 1 3  37. y = sin  x − π  2 2 

40. The graph is a reflected sine graph with amplitude 7 and period 8. 2π Find ω : 8 =

1 1 = 2 2 2π 2π 4π Period: = = T= 3 3 ω 2 φ π 2π Phase Shift: = = ω 3 3 2 Amplitude:

A =

ω

8ω = 2π 2π π ω= = 8 4 π  The equation is: y = −7 sin  x  . 4 

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Chapter 2 Review Exercises

41. Set the calculator to radian mode: sin

π 8

≈ 0.38 .

47. The domain of y = sec x is

π   x x ≠ odd multiple of  . 2  The range of y = sec x is { y y ≤ −1 or y ≥ 1} . The period is 2π .

42. Set the calculator to degree mode: 1 sec10o = ≈ 1.02 . cos10o

32o 20 '35" = 32 +

48. a.

20 35 + ≈ 32.34o 60 3600

63.18o 0.18o = (0.18)(60 ') = 10.8' 0.8' = (0.8)(60") = 48"

b.

Thus, 63.18o = 63o10 '48" 43. Terminal side of sin θ < 0 cos θ < 0 tan θ > 0

θ in quadrant III implies csc θ < 0 sec θ < 0 cot θ > 0

44. cos θ > 0, tan θ < 0 ; θ lies in quadrant IV.  1 2 2 45. P =  − ,   3 3 

2 2 1 3 2 3 2 ; csc t = = ⋅ = 3 4 2 2 2 2 2  3    1 1 cos t = − ; sec t = = −3 3 −1  3   2 2   2 2  3 tan t =  3  = ⋅ − = −2 2 ; 3  1  −1  3   1 2 2 ⋅ =− 4 −2 2 2

46. The point P = (−2, 5) is on a circle of radius

r = (−2) 2 + 52 = 4 + 25 = 29 with the center at the origin. So, we have x = −2 , y = 5 , and r = 29 . Thus, sin t = cos t =

π 6

π π = ≈ 1.047 feet 6 3 1 1 π 2 π A = ⋅ r 2θ = ⋅ ( 2 ) ⋅ = ≈ 1.047 square feet 2 2 6 3

s = rθ = 2 ⋅

50. In 30 minutes: r = 8 inches, θ = 180º or θ = π s = rθ = 8 ⋅ π = 8π ≈ 25.13 inches

sin t =

cot t =

49. r = 2 feet, θ = 30º or θ =

y 5 5 29 = = ; 29 r 29

y 5 x 2 29 −2 ; tan t = = − . = =− 2 r 29 x 29

In 20 minutes: r = 8 inches, θ = 120º or θ = s = rθ = 8 ⋅

2π 3

2π 16π = ≈ 16.76 inches 3 3

51. v = 180 mi/hr ;

1 mile 2 1 r = = 0.25 mile 4

d=

v 180 mi/hr = 0.25 mi r = 720 rad/hr 720 rad 1 rev = ⋅ hr 2π rad 360 rev = π hr ≈ 114.6 rev/hr

ω=

52. Since there are two lights on opposite sides and the light is seen every 5 seconds, the beacon makes 1 revolution every 10 seconds: 1 rev 2π radians π ⋅ = radians/second ω= 10 sec 1 rev 5

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Chapter 2: Trigonometric Functions

π  53. I (t ) = 220sin  30πt +  , 6 

a.

Period =

t≥0

Thus, y = 20sin

2π 1 = 30π 15

y = 20sin

π ( x − 4) + 75 . 6

c.

b. The amplitude is 220. c.

π 2π x− + 75 , or 6 3

The phase shift is: π

φ −6 1 π 1 = =− ⋅ =− ω 30π 6 30π 180 d. d.

y = 19.81sin ( 0.543 x − 2.296) + 75.66

e. 54. a.

55.

b.

95 − 55 40 = = 20 2 2 95+55 150 = = 75 Vertical Shift: 2 2 2π π = ω= 12 6 Phase shift (use y = 55, x = 1): Amplitude: A =

55 = 20sin −20 = 20sin −1 = sin −

π ⋅1 − φ + 75 6 π −φ 6

π −φ 6

π π = −φ 2 6 2π φ= 3

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Chapter 2 Test

Chapter 2 Test

2

 3  2 12. 2sin 2 60° − 3cos 45° = 2   − 3   2   2   3  3 2 3 3 2 3 1− 2 = 2  − = − = 2 2 2 2 4

1. 260° = 260 ⋅1 degree = 260 ⋅ =

π

(

radian

180

260π 13π radian = radian 180 9

)

13. Set the calculator to degree mode: sin17° ≈ 0.292

2. −400° = −400 ⋅1 degree = −400 ⋅ =−

π

180

radian

400π 20π radian = − radian 180 9

3. 13° = 13 ⋅1 degree = 13 ⋅ 4. −

π 8

radian = − =−

5.

6.

π 8

π 180

radian =

13π radian 180

π 180 ⋅ degrees = −22.5° 8 π

15. Set the calculator to degree mode: 1 sec 229° = ≈ −1.524 cos 229°

3π 3π radian = ⋅1 radian 4 4 3π 180 degrees = 135° = ⋅ 4 π

π 6

=

 3π  − cos    4

 = cos  − 5π + 2π  − cos  3π         4   4  3π 3π = cos   − cos   = 0  4   4 

9. cos ( −120° ) = cos (120° ) = −

1 2

10. tan 330° = tan (150° + 180° ) = tan (150° ) = − 11. sin

π 2

16. Set the calculator to radian mode: 28π 1 cot = ≈ 2.747 28π 9 tan 9

1 2

5π 

8. cos  −  4

− tan

2π ≈ 0.309 5

⋅1 radian

9π 9π radian = ⋅1 radian 2 2 9π 180 degrees = 810° = ⋅ 2 π

7. sin

14. Set the calculator to radian mode: cos

3 3

19π π  3π  = sin − tan  + 4π  4 2  4  = sin

π 2

 3π − tan   4

17. To remember the sign of each trig function, we primarily need to remember that sin θ is positive in quadrants I and II, while cos θ is positive in quadrants I and IV. The sign of the other four trig functions can be determined directly from sine and cosine by knowing sin θ 1 1 tan θ = , sec θ = , csc θ = , and cos θ cos θ sin θ cos θ . cot θ = sin θ

  = 1 − ( −1) = 2 

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Chapter 2: Trigonometric Functions

θ in QI θ in QII θ in QIII θ in QIV

sin θ cos θ tan θ sec θ csc θ cot θ + + + + + + + − − − + − + − − − − + + − − + − −

18. Because f ( x) = sin x is an odd function and

since f (a) = sin a =

3 , then 5

1 1 3 5 3 5 = 5 =− ⋅ =− sin θ − 3 5 5 5

sec θ =

1 1 3 = = cos θ 23 2

cot θ =

1 1 2 5 2 5 = =− ⋅ =− tan θ − 25 5 5 5

12 π and < θ < π (in quadrant II) 2 5 Using the Pythagorean Identities: 2 144 169  12  sec 2 θ = tan 2 θ + 1 =  −  + 1 = +1 = 25 25  5

21. tan θ = −

3 f (−a ) = sin(−a ) = − sin a = − . 5

5 and θ in quadrant II. 7 Using the Pythagorean Identities: 2 25 24 5 cos 2 θ = 1 − sin 2 θ = 1 −   = 1 − = 49 49 7

19. sin θ =

169 13 =± 25 5 Note that sec θ must be negative since θ lies in 13 quadrant II. Thus, sec θ = − . 5 1 1 5 cos θ = = =− sec θ − 135 13 sec θ = ±

24 2 6 =± 49 7 Note that cos θ must be negative because θ lies cos θ = ±

in quadrant II. Thus, cos θ = −

csc θ =

2 6 . 7

tan θ =

sin θ , so cos θ 12  5  12 −  = 5  13  13

tan θ =

5 sin θ 5 7  6 5 6 = 27 6 =  − =− ⋅ cos θ − 7 7 2 6  6 12

csc θ =

1 1 7 = = sin θ 75 5

csc θ =

1 1 13 = 12 = sin θ 13 12

sec θ =

1 1 7 6 7 6 = =− ⋅ =− cos θ − 2 7 6 12 2 6 6

cot θ =

1 1 5 = =− tan θ − 125 12

cot θ =

1 1 12 6 2 6 = 5 6 =− ⋅ =− tan θ − 12 5 5 6 6

sin θ = ( tan θ )( cos θ ) = −

22. The point ( 2, 7 ) lies in quadrant I with x = 2

and y = 7 . Since x 2 + y 2 = r 2 , we have

2 3π and < θ < 2π (in quadrant IV). 2 3 Using the Pythagorean Identities: 2 4 5 2 2 2 sin θ = 1 − cos θ = 1 −   = 1 − = 9 9 3

r = 22 + 7 2 = 53 . So, y 7 7 53 7 53 sin θ = = = ⋅ = . 53 r 53 53 53

20. cos θ =

23. The point ( −5,11) lies in quadrant II with x = −5 and y = 11 . Since x 2 + y 2 = r 2 , we

5 5 =± 9 3 Note that sin θ must be negative because θ lies sin θ = ±

in quadrant IV. Thus, sin θ = − tan θ =

have r =

5 . 3

cos θ =

( −5 )2 + 112

= 146 . So,

−5 −5 146 5 146 x . = = ⋅ =− 146 r 146 146 146

sin θ − 35 5 3 5 = 2 =− ⋅ =− cos θ 3 2 2 3

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Chapter 2 Test 24. The point ( 6, −3) lies in quadrant IV with x = 6 2

2

2

and y = −3 . Since x + y = r , we have 2

r = 62 + ( −3) = 45 = 3 5 . So, tan θ =

π

y −3 1 = =− x 6 2

y = A sin (ω x − φ ) , we see that 1 π , and φ = . The graph is a sine 6 3

ω

=

π  y = tan  x +  . Next, reflect this graph about 4  π  the y-axis to obtain the graph of y = tan  − x +  .  4 Finally, shift the graph up 2 units to obtain the π  graph of y = tan  − x +  + 2 . 4  π  y = tan  − x +  + 2 4 

curve with amplitude A = 2 , period T=

units to the left to obtain the graph of

4

x π 25. Comparing y = 2sin  −  to 3 6

A=2, ω =

π  26. y = tan  − x +  + 2  4 Begin with the graph of y = tan x , and shift it

y

2π = 6π , and phase shift 1/ 3

4

π φ π x π = 6 = . The graph of y = 2sin  −  ω 1/ 3 2 3 6 will lie between −2 and 2 on the y-axis. One φ π period will begin at x = = and end at ω 2 2π φ π 13π x= + = 6π + = . We divide the 2 2 ω ω  π 13π  interval  ,  into four subintervals, each of 2 2  6π 3π = . length

=

4 2 7π   7π π     13π   2 , 2π  ,  2π , 2  ,  2 ,5π  , 5π , 2          The five key points on the graph are π   7π   13π  , 0  , ( 5π , −2 ) ,  ,0  , 0  , ( 2π , 2 ) ,  2   2   2  We plot these five points and fill in the graph of the sine function. The graph can then be extended in both directions.

2

−2π

(⫺4␲, 2)

A , the period is given by

11␲ , (⫺ –––– 2 0(

5␲ , (⫺ ––– 2 0(

(⫺␲, ⫺2)

ω

, and the phase

φ . Therefore, we have A = −3 , ω π 3π ω = 3 , and φ = 3  −  = − . The equation  4

4

3π  for the graph is y = −3sin  3 x + 4 

 . 

28. The area of the walk is the difference between the area of the larger sector and the area of the smaller shaded sector.

(2␲, 2)

( ––␲2 , 0( 7␲ , ( ––– 2 0(

shift is given by

W a x

x

27. For a sinusoidal graph of the form y = A sin (ω x − φ ) , the amplitude is given by

y 3 2 1

π

−π

l

k

50° 3 ft

(5␲, ⫺2)

The area of the walk is given by 1 1 A = R 2θ − r 2θ , 2 2

θ

(

)

R2 − r 2 2 where R is the radius of the larger sector and r is =

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Chapter 2: Trigonometric Functions

the radius of the smaller sector. The larger radius is 3 feet longer than the smaller radius because the walk is to be 3 feet wide. Therefore, R = r + 3 , and A= = =

( r + 3) 2(

θ

θ

2

θ

(r

2

2

− r2

)

+ 6r + 9 − r

2

Chapter 2 Cumulative Review 2 x2 + x − 1 = 0

1.

( 2 x − 1)( x + 1) = 0 x=

)

1 or x = −1 2

 1 The solution set is −1,  .  2

( 6r + 9 ) 2

The shaded sector has an arc length of 25 feet 5π radians . The and a central angle of 50° = 18 s 25 90 feet . radius of this sector is r = = 5π =

θ

18

2. Slope = −3 , containing (–2,5) Using y − y1 = m( x − x1 ) y − 5 = −3 ( x − (−2) ) y − 5 = −3( x + 2) y − 5 = −3 x − 6 y = −3 x − 1

π

Thus, the area of the walk is given by 5π   90   5π  540  A = 18  6   + 9  = + 9  2   π   36  π  5π 2 = 75 + ft ≈ 78.93 ft 2 4

3. radius = 4, center (0,–2)

Using ( x − h ) + ( y − k ) = r 2 2

( x − 0)

29. To throw the hammer 83.19 meters, we need v2 s= 0 g v0 2 83.19 m = 9.8 m/s 2 2 v0 = 815.262 m 2 / s 2 v0 = 28.553 m/s Linear speed and angular speed are related according to the formula v = r ⋅ ω . The radius is r = 190 cm = 1.9 m . Thus, we have 28.553 = r ⋅ ω 28.553 = (1.9 ) ω ω = 15.028 radians per second radians 60 sec 1 revolution ω = 15.028 ⋅ ⋅ sec 1 min 2π radians ≈ 143.5 revolutions per minute (rpm) To throw the hammer 83.19 meters, Adrian must have been swinging it at a rate of 143.5 rpm upon release.

2

2

+ ( y − ( −2 ) ) = 42 2

x 2 + ( y + 2 ) = 16 2

4. 2 x − 3 y = 12 This equation yields a line. 2 x − 3 y = 12 −3 y = −2 x + 12 2 y = x−4 3 2 The slope is m = and the y-intercept is −4 . 3 Let y = 0 : 2 x − 3(0) = 12 2 x = 12 x=6 The x-intercept is 6.

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Chapter 2 Cumulative Review

5. x 2 + y 2 − 2 x + 4 y − 4 = 0

b.

y = x3

c.

y = sin x

d.

y = tan x

x2 − 2 x + 1 + y 2 + 4 y + 4 = 4 + 1 + 4

( x − 1) + ( y + 2 ) = 9 2 2 ( x − 1) + ( y + 2 ) = 32 2

2

This equation yields a circle with radius 3 and center (1,–2).

6. y = ( x − 3) 2 + 2

Using the graph of y = x 2 , horizontally shift to the right 3 units, and vertically shift up 2 units.

8. 7. a.

y = x2

f ( x) = 3x − 2 y = 3x − 2

x = 3y − 2

Inverse

x + 2 = 3y x+2 =y 3 x+2 1 f −1 ( x) = = ( x + 2) 3 3

9. Since ( sin θ ) + ( cos θ ) = 1 , then 2

2

( sin14 ) + ( cos14 ) o

2

o

2

− 3 = 1 − 3 = −2

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Chapter 2: Trigonometric Functions 10. y = 3sin(2 x) Amplitude:

11. tan

A = 3 =3

π 4

− 3cos

π 6

+ csc

2π Period: T= =π 2 φ 0 Phase Shift: = =0 ω 2

 3 = 1 − 3   + 2 6  2 

π

3 3 2 6−3 3 = 2 = 3−

12. The graph is a cosine graph with amplitude 3 and period 12. 2π Find ω : 12 =

ω

12ω = 2π

ω=

2π π = 12 6

π The equation is: y = 3cos  6

 x . 

Chapter 2 Projects Project I – Internet Based Project Project II 1. November 15: High tide: 11:18 am and 11:15 pm November 19: low tide: 7:17 am and 8:38 pm 2. The low tide was below sea level. It is measured against calm water at sea level. 3.

Nov

14 0-24 15 24-48 16 48-72 17 72-96 18 96-120 19 120-144 20 144-168

Low Tide

Low Tide

High Tide

High Tide

Time

Ht (ft) t

Time

Ht (ft) t

Time

Ht (ft)

t

Time

6:26a

2.0

6.43

4:38p

1.4 16.63

9:29a

2.2

9.48

11:14p 2.8

23.23

6:22a

1.6

30.37

5:34p

1.8 41.57

11:18a 2.4

35.3

11:15p 2.6

47.25

6:28a

1.2

54.47

6:25p

2.0 66.42

12:37p 2.6

60.62

11:16p 2.6

71.27

6:40a

0.8

78.67

7:12p

2.4 91.2

1:38p

2.8

85.63

11:16p 2.6

95.27

6:56a

0.4

102.93

7:57p

2.6 115.95

2:27p

3.0

110.45

11:14p 2.8

119.23

7:17a

0.0

127.28

8:38p

2.6 140.63

3:10p

3.2

135.17

11:05p 2.8

143.08

7:43a -0.2

151.72

3:52p

3.4

159.87

Ht (ft) t

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Chapter 2 Projects

Low tides of 1.49 feet when t = 178.2 and t = 190.3.

4. The data seems to take on a sinusoidal shape (oscillates). The period is approximately 12 hours. The amplitude varies each day: Nov 14: 0.1, 0.7 Nov 15: 0.4, 0.4 Nov 16: 0.7, 0.3 Nov 17: 1.0, 0.1 Nov 18: 1.3, 0.1 Nov 19: 1.6, 0.1 Nov 20: 1.8 5. Average of the amplitudes: 0.66 Period : 12 Average of vertical shifts: 2.15 (approximately) There is no phase shift. However, keeping in mind the vertical shift, the amplitude y = A sin ( Bx ) + D A = 0.66

12 = B=

2π B

High tides of 2.81 feet occur when t = 172.2 and t = 184.3.

Looking at the graph for the equation in part (6) and using MAX/MIN for values between t = 168 and t = 192: A low tide of 1.38 feet occurs when t = 175.7 and t = 187.8 .

D = 2.15

π

≈ 0.52 6 Thus, y = 0.66sin ( 0.52 x ) + 2.15

(Answers may vary)

A high tide of 3.08 feet occurs when t = 169.8 and t = 181.9 .

6. y = 0.848sin ( 0.52 x + 1.25 ) + 2.23

The two functions are not the same, but they are similar.

8. The low and high tides vary because of the moon phase. The moon has a gravitational pull on the water on Earth. 7. Find the high and low tides on November 21 which are the min and max that lie between t = 168 and t = 192 . Looking at the graph of the equation for part (5) and using MAX/MIN for values between t = 168 and t = 192 :

Project III 1. s (t ) = 1sin ( 2π f 0 t ) 2. T0 =

2π 1 = 2π f 0 f0

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Chapter 2: Trigonometric Functions

t

1 4 f0

s (t ) 0 1

1 2 f0

3 4 f0

1 f0

0

−1

0

2. s = rθ s 65 θ= = = 0.0164 r 3960

4. Let f 0 = 1 =1. Let 0 ≤ x ≤ 12 , with Δx = 0.5 . Label the graph as 0 ≤ x ≤ 12T0 , and each tick

mark is at Δx = 2

1 . 2 f0

h = 0.396 miles 0.396 × 5280 = 2090 feet

0

12

4. Maui:

12 12T0 = __ f0

Oahu

Oahu

s

θ

Maui

6. M = 0 1 0 → P = 0 π 0

10,023 ft

396

i 0m

s 110 = = 0.0278 r 3960 3960 = cos(0.278) 3960 + h 3960 = 0.9996(3960 + h) h = 1.584 miles h = 1.584 × 5280 = 8364 feet

8. [0, 4 T0 ] S0 [4 T0 , 8 T0 ] S1 [ 8 T0 , 12 T0 ] S0 2 12

Hawaii: Oahu

θ

Project IV

13,796 ft

396

Oahu

s

Peak of Lanaihale

s=

mi

i

θ

}

39 60

Lanai

m 65

39 60 m

i

Oahu

3,370 ft

396

i 0m

Lanai

0

mi

s 190 = = 0.0480 r 3960 3960 = cos(0.480) 3960 + h 3960 = 0.9988(3960 + h) h = 4.752 miles h = 4.752 × 5280 = 25, 091 feet

θ=

1. Lanai:

}

39 60

Hawaii

mi

39 60 m i

s

Peak of Mauna Kea i 0m 19

θ

Oahu

s=

−2

θ

Maui

θ=

7. S0 (t ) = 1sin(2π f 0 t + 0) , S1 (t ) = 1sin(2π f 0 t + π )

0

}

θ

mi

1 5 9 45 , t= , t= ,…, t= 4 f0 4 f0 4 f0 4 f0

Peak of Haleakala

0 11 s=

39 60 m i

−2

5. t =

3960 = cos(0.164) 3960 + h 3960 = 0.9999(3960 + h)

3.

39 60 m i

3.

0

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Hawaii


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Chapter 2 Projects

Molokai: Oahu

s i

θ

}

39 60 m

Molokai

mi 40

θ

Peak of Kamakou

s=

39 60 m i

Oahu

4,961 ft

396

i 0m

Molokai

s 40 = = 0.0101 r 3960 3960 = cos(0.0101) 3960 + h 3960 = 0.9999(3960 + h) h = 0.346 miles h = 0.346 x5280 = 2090 feet

θ=

5. Kamakou, Haleakala, and Lanaihale are all visible from Oahu.

Project V

Answers will vary.

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P r o j e c t

a t

M o t o r o l a

Digital Transmission over the Air Digital communications is a revolutionary technology of the century. For many years, Motorola has been one of the leading companies to employ digital communication in wireless devices, such as cell phones. Figure 1 shows a simplified overview of a digital communication transmission over the air. The information source to be transmitted can be audio, video, or data. The information source may be formatted into a digital sequence of symbols from a finite set Ean F={0, 1}. So 0110100 is an example of a digital sequence. The period of the symbols is denoted by T. The principle of digital communication systems is that, during the finite interval of time T, the information symbol is represented by one digital waveform from a finite set of digital waveforms before it is sent. This technique is called modulation. Modulation techniques use a carrier that is modulated by the information to be transmitted.The modulated carrier is transformed into an electromagnetic field and propagated in the air through an antenna.The unmodulated carrier can be represented in its general form by a sinusoidal function s(t)=A sin Av0 t+f B, where A is the amplitude, v0 is the radian frequency, and f is the phase. Let’s assume that A=1, f=0, and v0=2pf0 radian, where f0 is the frequency of the unmodulated carrier. 1. Write s(t) using these assumptions. 2. What is the period, T0 , of the unmodulated carrier?

3. Evaluate s(t) for t=0, 1/A4f0 B, 1/A2f0 B, 3/A4f0 B, and 1/f0 . 4. Graph s(t) for 0 ⱕ t ⱕ 12T0 . That is, graph 12 cycles of the function. 5. For what values of t does the function reach its maximum value? [Hint: Express t in terms of f0]. Three modulation techniques are used for transmission over the air: amplitude modulation, frequency modulation and phase modulation. In this project, we are interested in phase modulation. Figure 2 illustrates this process. An information symbol is mapped onto a phase that modulates the carrier. The modulated carrier is expressed by Si(t)=sin A2pf0 t+ci B. Let’s assume the following mapping scheme: Ean F

S

Ecn F

0

c0 = 0

1

c1 = p

6. Map the binary sequence M=010 into a phase sequence P. 7. What is the expression of the modulated carrier S0(t) for ci=c0 and S1(t) for ci=c1 ? 8. Let’s assume that in the sequence M the period of each symbol is T=4T0 . For each of the three intervals C0, 4T0 D, C4T0 , 8T0 D, and C8T0 , 12T0 D, indicate which of S0(t) or S1(t) is the modulated carrier. On the same graph, illustrate M, P, and the modulated carrier for 0 ⱕ t ⱕ 12T0 . Carrier s (t )

Information source

Figure 1

Format

Digital symbols

Modulate

Digital waveform (Modulated carrier) Si (t )

Transmit

Simplified Overview of a Digital Communication Transmission Carrier s (t ) ⫽ sin (2␲f 0t )

Digital symbols {␣n }

Mapping

Figure 2

Phase {␺n }

Modulate

Digital waveforms Si (t ) ⫽ sin (2␲f 0t ⫹ ␺i )

Principle of Phase Modulation

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Trigonometry A Unit Circle Approach 10th Edition Sullivan Solutions Manual  

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Trigonometry A Unit Circle Approach 10th Edition Sullivan Solutions Manual  

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