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Trigonometry A Unit Circle Approach 10th Edition Sullivan
SOLUTIONS MANUAL
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edition-sullivan-solutions-manual/
Trigonometry A Unit Circle Approach 10th Edition Sullivan TEST BANK
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edition-sullivan-test-bank/
Section 2.1
Chapter 2
Trigonometric Functions
16.
1. C = 2πr ; A = πr2
17.
2. A =
1
2π(2) = 2π
2
3. standard position
4. central angle
5. d
6. rθ ;
1
r2
θ
2
18.
19.
7. b
8.
s
;
θ 20.
t t
9. True
10. False; υ = rω
1
1
.

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 
12.
13.
21.
22.
 1 1 1 º
14.
_
15.
23.
24.
40º10'25" = 40 +10⋅
60
+ 25⋅
60
⋅
60

≈ (40 + 0.1667 + 0.00694)º
≈ 40.17º
61º 42'21" =

61+ 42⋅
1
+ 21⋅
1
⋅
1 º

60 60 60

 
≈ (61+ 0.7000 + 0.00583)º
≈ 61.71º

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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
25. 50º14'20" = 50 +14⋅
1
+ 20⋅
1
⋅
1 º
60 60 60
32. 29.411º = 29º + 0.411º
= 29º + 0.411(60')
≈ (50 + 0.2333 + 0.00556)º
≈ 50.24º
= 29º +24.66'
= 29º +24'+ 0.66'
= 29º +0.66(60")
 1 1 1 º
26. 73º 40'40" = 73+ 40⋅ + 40⋅ ⋅  = 29º +24'+ 39.6"
 60 60 60 
≈ (73+ 0.6667 + 0.0111)º
≈ 73.68º
 1 1 1 º
≈ 29º 24'40"
33. 19.99º =19º + 0.99º
=19º + 0.99(60')
27. 9º9'9" = 9 + 9⋅ + 9⋅ ⋅  =19º +59.4'
 60 60 60 
= (9 + 0.15 + 0.0025)º
≈ 9.15º
=19º +59'+ 0.4'
=19º +59'+ 0.4(60")
=19º +59'+ 24"
 1 1 1 º =19º59'24"28. 98º 22'45" = 98 + 22⋅ + 45⋅ ⋅ 
 60 60 60 
≈ (98 + 0.3667 + 0.0125)º
≈ 98.38º
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"
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"
= 44º0'36"
35. 30° = 30⋅
π
radian =
π
radian
180 6
30. 61.24º = 61º + 0.24º
= 61º +0.24(60')
36. 120° =120⋅
π
radian =
2π
radians
180 3
= 61º +14.4'
37. 240° = 240⋅
π
radian =
4π
radians
= 61º +14'+ 0.4'
= 61º +14'+ 0.4(60")
= 61º +14'+ 24"
= 61º14'24"
38.
180 3
330° = 330⋅
π
radian =
11π
radians
180 6
31. 18.255º =18º + 0.255º
=18º + 0.255(60')
39. −60° = −60⋅
π
radian
180
= −
π
radian
3
=18º +15.3'
=18º +15'+ 0.3'
=18º +15'+ 0.3(60")
40. −30° = −30⋅
π
radian = −
π
radian
180 6
π
=18º +15'+18"
=18º15'18"
41. 180° =180⋅ radian = π radians
180
42. 270° = 270⋅
π
radian =
3π
radians

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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
180 2

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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
= −
= −
π
43.
44.
−135° = −135⋅
π
radian = −
3π
radians
180 4
−225° = −225⋅
π
radian = −
5π
radians
180 4
60. 73° = 73⋅
π
radian
180
73π
= radians
180
≈1.27 radians
45.
46.
47.
−90° = −90⋅
π
radian = −
π
radians
180 2
−180° = −180⋅
π
radian = −π radians
180
π
=
π
⋅
180
degrees = 60°
61. −40° = −40⋅
π
radian
180
2π
radian
9
≈ −0.70 radian
π
3 3 π 62. −51° = −51⋅
180
radian
48.
5π
=
5π
⋅
180
degrees = 150°
17π
radian
6 6 π 60
≈ −0.89 radian
49.
5π 5π 180
− = − ⋅ degrees = − 225°
4 4 π
63. 125° =125⋅ π
180
radian
50. −
2π
= −
2π
⋅
180
degrees = −120° 25π
3 3 π = radians
36
51.
π
=
π
⋅
180
degrees = 90°
≈ 2.18 radians
2 2 π
64. 350° = 350⋅ π
180
radian
52. 4π = 4π⋅
180
degrees = 720°
π =
35π
18
radians
53.
π
=
π
⋅
180
degrees =15°
≈ 6.11 radians
12 12 π
65. 3.14 radians = 3.14⋅
180
degrees ≈179.91º
54.
5π
=
5π
⋅
180
degrees = 75°
π
12 12 π
66. 0.75 radian = 0.75⋅
180
degrees ≈ 42.97º
55. −
π
= −
π
⋅
180
degrees = −90°
π
2 2 π
67. 2 radians = 2⋅
180
degrees ≈114.59º
56.
57.
−π = −π⋅
180
degrees = −180°
π
−
π
= −
π
⋅
180
degrees = −30°
68.
π
3 radians = 3⋅
180
degrees ≈171.89º
π
6 6 π
69. 6.32 radians = 6.32⋅
180
degrees ≈ 362.11º
58.
3π 3π 180
− = − ⋅ degrees = −135°
4 4 π
70. 2 radians = 2 ⋅
180
degrees ≈ 81.03º
59. 17° =17⋅
π
radian =
17π
radian ≈ 0.30 radian π
180 180

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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
3
4
2
6
2
2
71. r =10 meters;θ =
1
radian;
2
s = rθ =10⋅
1
= 5 meters
2
81. θ =
1
radian;
3
A =
1
r2
θ
2
2 =
1
r2  1 
A = 2 ft2
72.
73.
r = 6 feet; θ = 2 radian;
θ =
1
radian; s = 2 feet;
3
s = rθ = 6⋅2 =12 feet  
 
2 =
1
r2
6
2
s = rθ
r =
s
=
2
= 6 feet
12 = r
r = 12 = 2 3 ≈ 3.464 feet
θ (1/ 3)
82. θ =
1
radian;
4
A = 6 cm2
74. θ =
1
radian; s = 6 cm;
4
s = rθ
A =
1
2
r2
θ
6 =
1
r2  1 
r =
s
=
θ
6
(1/ 4)
= 24 cm  
 
6 =
1
r2
75. r = 5 miles; s = 3 miles; 8
2
s = rθ
θ =
s
=
3
= 0.6 radian
48 = r
r = 48 = 4 3 ≈ 6.928 cm
r 5
83. r = 5 miles; A = 3 mi2
76. r = 6 meters; s = 8 meters; 1 2
s = rθ
θ =
s
=
8
=
4
≈1.333 radians
A = r θ
2
3 =
1
(5)
2
θ
r 6 3 2
3 =
25
θ
77. r = 2 inches; θ = 30º = 30⋅
π
=
π
radian;
180 6
s = rθ = 2⋅
π
=
π
≈1.047 inches
6 3
2
θ =
6
= 0.24 radian
25
78. r = 3 meters; θ =120º =120⋅
π
=
2π
radians
180 3
84. r = 6 meters;
A =
1
r2
θ
2
A = 8 m2
79.
s = rθ = 3⋅
2π
= 2π ≈ 6.283 meters
3
r =10 meters; θ =
1
radian
2
A =
1
r2
θ =
1
(10)
2  1 
=
100
=25 m2
8 =
1
(6)
2
θ
2
8 =18θ
θ =
8
=
4
≈ 0.444 radian
18 9
 
2 2   4
85. r = 2 inches; θ = 30º = 30⋅
π
=
π
180 6
radian
80. r = 6 feet; θ = 2 radians
A =
1
r2
θ =
1
2
2  π 
=
π
≈1.047 in2
A =
1
r2
θ =
1
(6)
2
(2)=36 ft2
2 2
( )  
2 2   3

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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
4
3
3
18
18
86. r = 3 meters; θ =120º =120⋅
π
=
2π
radians
180 3
92. r = 40 inches; θ = 20º =
π
radian
9
A =
1
r2
θ =
1
(3)
2  2π 
=3π ≈ 9.425 m2
s = rθ = 40⋅
π
=
40π
≈13.96 inches
2 2

3

9 9
87. r = 2 feet; θ =
π
3
 
radians 93. r = 4 m; θ = 45º = 45⋅
π
=
π
radian
180 4
s = rθ = 2⋅
π
=
2π
≈ 2.094 feet A =
1
r2
θ =
1
(4)
2  π 
= 2π ≈ 6.28 m2
3 3
A =
1
r2
θ =
1
(2)
2  π 
= =
2π
≈ 2.094 ft2
 
2 2  
 
2 2   3 94. r = 3 cm; θ = 60º = 60⋅
π
=
π
radians
88. r = 4 meters; θ =
π
radian
180 3
A =
1
r2
θ =
1
(3)
2  π 
=
3π
≈ 4.71 cm2
6
s = rθ = 4⋅
π
=
2π
≈ 2.094 meters
 
2 2   2
36 3 95. r = 30 feet; θ =135º =135⋅
π
=
π
radians
A =
1
r2
θ =
1
(4)
2  π 
=
4π
≈ 4.189 m2
180 4
2 2

6

3 1 2 1 2  3π 
 675π 2
  A = r θ = (30)   = ≈ 1060.29 ft
89. r =12 yards; θ = 70º = 70⋅
π
=
7π
radians
2 2  4  2
2
180 18
s = rθ =12⋅
7π
≈14.661 yards
18
96. r =15 yards;
A =
1
r2
θ
2
A =100 yd
A =
1
r2
θ =
1
(12)
2  7π 
= 28π ≈ 87.965 yd2 100 =
1
(15)
2
θ
 
2 2   2
100 =112.5θ
90. r = 9 cm; θ = 50º = 50⋅
π
=
5π
radian θ =
100
=
8
≈ 0.89 radian
180 18
s = rθ = 9⋅
5π
≈ 7.854 cm
112.5 9
°
or
8
⋅
180
=
160 
≈ 50.93°
18 9 π 
π 
 
A =
1
r2
θ =
1
(9)
2  5π 
=
45π
≈ 35.343 cm2
 
2 2   4 97. A =
1
r2
θ −
1
r 2
θ θ = 120° =
2π
2
1
2
2
3
91. r = 6 inches 1 2π
= (34)2 1 2π
− (9)2
In 15 minutes,
θ =
15
rev =
1
⋅360º = 90º =
π
radians
2 3 2 3
60 4 2
=
1
(1156)
2π
−
1
(81)
2π
s = rθ = 6⋅
π
= 3π ≈ 9.42 inches
2
2 3 2 3
π π
In 25 minutes,
= (1156) − (81)
3 3
θ =
25
rev =
5 5π
⋅360º =150º = radians =
1156π
−
81π
60 12 6
s = rθ = 6⋅
5π
= 5π ≈15.71 inches
3 3
1075π 2
6 =
3
≈ 1125.74 in

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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
98. A =
1
r2
θ −
1
r 2
θ θ = 125° =
25π 104. r = 5 m; ω = 5400 rev/min = 10800π rad/min
2
1
2
2
36 v = rω = (5)⋅10800π m/min = 54000π cm/min
1
= (30)2 25π 1
− (6)2 25π
v = 54000π
cm
⋅
1m
⋅
1km
⋅
60min
2 36 2 36 min 100cm 1000m 1hr
≈ 101.8 km/hr
1
= (900)
25π 1
− (36)
25π
2 36 2 36 105. d = 26 inches; r = 13 inches; v = 35 mi/hr
= (450)
25π
− (18)
25π v =
35mi
⋅
5280ft
⋅
12in.
⋅
1hr
36 36 hr mi ft 60 min
= 36,960 in./min
=
11250π
−
450π
36 36 ω =
v
=
36,960in./min
r 13 in.10800π 2
= = 300π ≈ 942.48 in
36
≈ 2843.08 radians/min
≈
2843.08rad
⋅
1rev
99. r = 5 cm; t = 20 seconds; θ =
1
radian
3
min 2π rad
≈ 452.5 rev/min
θ (1/3) 1 1 1
ω = = = ⋅ = radian/sec 106. r =15 inches; ω = 3 rev/sec = 6π rad/sec
t 20 3 20 60
s rθ 5⋅(1/3) 5 1 1
v = rω =15⋅6π in./sec = 90π ≈ 282.7 in/sec
v = = = = ⋅ = cm/sec in. 1ft 1mi 3600sec
t t 20 3 20 12 v = 90π ⋅ ⋅ ⋅ ≈16.1 mi/hr
sec 12in. 5280ft 1hr
100. r = 2 meters; t = 20 seconds; s = 5 meters
107. r = 3960 miles
θ (s /r) (5/2) 5 1 1
ω = = = = ⋅ = radian/sec θ = 35º9'− 29º57'
t t 20 2 20 8 = 5º12'
v =
s
=
5 1
= m/sec = 5.2º
t 20 4
= 5.2⋅
π
101. r = 25 feet; ω = 13 rev/min = 26π rad/min 180
v = rω = 25⋅26π ft./min = 650π ≈ 2042.0 ft/min ≈ 0.09076 radian
s = rθ = 3960(0.09076) ≈ 359 miles
v = 650π
ft. 1mi 60min
⋅ ⋅ ≈ 23.2 mi/hr
min 5280ft 1hr
108. r = 3960 miles
θ = 38º 21'− 30º 20'
102. r = 6.5 m; ω = 22 rev/min = 44π rad/min = 8º1'
v = rω = (6.5)⋅44π m/min = 286π ≈ 898.5 m/min ≈ 8.017º
v = 286π
m 1km 60min
⋅ ⋅ ≈ 53.9 km/hr = 8.017⋅
π
103. r = 4 m;
min 1000m 1hr
ω = 8000 rev/min = 16000π rad/min
180
≈ 0.1399 radian
s = rθ = 3960(0.1399) ≈ 554 miles
v = rω = (4)⋅16000π m/min = 64000π cm/min
v = 64000π
cm
⋅
1m
⋅
1km
⋅
60min
109. r = 3429.5 miles
ω =1 rev/day = 2π radians/day =
π
radians/hr
min 100cm
≈ 120.6 km/hr
1000m 1hr 12
v = rω = 3429.5⋅
π
≈ 898 miles/hr
12

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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
( )
( )
110. r = 3033.5 miles
ω =1 rev/day = 2π radians/day =
π
12
radians/hr
115. r = 4 feet;
v = rω
= 4⋅20π
ω =10 rev/min = 20π radians/min
111.
v = rω = 3033.5⋅
π
≈ 794 miles/hr
12
r = 2.39×105
miles
ω =1 rev/27.3 days
= 2π radians/27.3 days
= 80π
ft
min
80π ft 1mi 60min
= ⋅ ⋅
min 5280 ft hr
≈ 2.86 mi/hr
=
π
radians/hr
116. d = 26 inches; r =13 inches;
112.
12⋅27.3
v = rω = 2.39×105
⋅
π
≈ 2292 miles/hr
327.6
r = 9.29×107
miles
ω =1 rev/365 days
ω = 480 rev/min = 960π radians/min
v = rω
=13⋅960π
=12480π
in
min
12480π in 1 ft 1 mi 60 min
= 2π radians/365 days
=
min
⋅ ⋅ ⋅
12 in 5280 ft hr
=
π
radians/hr
12⋅365
v = rω = 9.29×107
⋅
π
≈ 66,633 miles/hr
4380
≈ 37.13 mi/hr
ω =
v
r
80mi/hr 12in 5280ft 1hr 1rev
= ⋅ ⋅ ⋅ ⋅
13 in 1 ft 1 mi 60 min 2π rad
113. r1 = 2 inches; r2 = 8 inches; ≈1034.26 rev/min
ω1 = 3 rev/min = 6π radians/min
Find ω2
: 117. d = 8.5 feet; r = 4.25 feet; v = 9.55 mi/hr
v1 = v2 ω =
v
=
9.55mi/hr
r1ω1 = r2ω2
r 4.25 ft
9.55 mi 1 5280 ft 1 hr 1 rev
2(6π) = 8ω2
12π
= ⋅ ⋅
hr 4.25 ft mi
⋅ ⋅
60 min 2π
ω2 =
8
≈ 31.47 rev/min
114.
=1.5π radians/min
1.5π
= rev/min
2π
=
3
rev/min
4
r = 30 feet
118. Let t represent the time for the earth to rotate 90
miles.
t
=
24
90 2π(3559)
t =
90(24)
≈ 0.0966 hours ≈ 5.8 minutes
2π(3559)
ω =
1rev 2π π
= = ≈ 0.09 radian/sec
70 sec 70 sec 35
v = rω = 30 feet ⋅
π rad
=
6π ft
≈ 2.69 feet/sec
35 sec 7 sec

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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
119. A = πr2
= π(9)2
= 81π
We need ¾ of this area.
3
81π =
243
π . Now
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
4 4
we calculate the small area.
A = πr2
= π(3)2
= 9π
7.2° = 7.2°⋅
π
=
π
radian . Therefore,
180° 25
s = rθ
500 = r ⋅
π
25
We need ¼ of the small area.
1
9π =
9
π r =
25
⋅500 =
12,500
≈ 3979 miles
4 4
So the total area is:
243
π +
9
π =
252
π = 63π
4 4 4
square feet.
120. First we find the radius of the circle.
C = 2πr
8π = 2πr
4 = r
π π
C = 2π r = 2π ⋅
12,500
= 25,000 miles.
π
The radius of Earth is approximately 3979 miles,
and the circumference is approximately 25,000
miles.
124. a. The length of the outfield fence is the arc
length subtended by a central angle θ = 96°
with r = 200 feet.
The area of the circle is A = πr2
= π(4)2
= 16π . s = r ⋅θ = 200⋅96°⋅
π
≈ 335.10 feet
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
So the area of the rectangle that is outside of the
circle is 44 − 4π u2
.
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.
121. The earth makes one full rotation in 24 hours. A =
1
R2
θ −
1
r2
θ =
θ
(R2
− r2
)
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:
v =
s
=
2π(3960)
≈1037 miles/hr
2 2 2
=
96°
⋅
π
(2002
−1902
)2 180°
=
4π
(3900) ≈ 3267.26
15
The area of the warning track is about
3267.26 square feet.
t 24
122. Find s, when r = 3960 miles and θ =1'.
θ =1'⋅
1degree
⋅
π radians
60 min 180 degrees
≈ 0.00029 radian
s = rθ = 3960(0.00029) ≈1.15 miles
Thus, 1 nautical mile is approximately 1.15
statute miles.

11. 136
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Section 2.1: Angles and Their MeasureChapter 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:
135. x2
− 9 cannot be zero so the domain is:
{x | x ≠ ±3}
v1 = v2
136. Shift to the left 3 units would give y = x + 3 .
r1ω1 = r2ω2
r1ω1
=
r2ω2
Reflecting about the x-axis would give
y = − x + 3 . Shifting down 4 units would result
r2
ω1
r2
ω1
in y = − x + 3 − 4 .
r1
=
ω2
r2 ω1
126. Answers will vary.
127. If the radius of a circle is r and the length of the
137.
x1+ x2
,
y1+ y2
2 2
3+ (−4) 6 + 2
= ,
2 2
arc subtended by the central angle is also r, then
=
−1
,
8 1
= − ,4
the measure of the angle is 1 radian. Also,
1 radian =
180
degrees .
π
2 2 2
1° =
1
360
revolution
Section 2.2
128. Note that 1° =1°⋅
 π radians 
≈ 0.017 radian 2 2 2

180°
 1. c = a + b 
and 1 radian ⋅
 180° 
≈ 57.296° .
π radians


2. f (5) = 3(5)− 7 = 15 − 7 = 8
Therefore, an angle whose measure is 1 radian is
larger than an angle whose measure is 1 degree.
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.
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
3. True
4. equal; proportional
 1 3 
5. −
2
,
2 
 
6. −
1
2
7. b
8. (0,1)
 2 2 
can be rewritten as follows: s = rθ =
π
rθ .
180
9.

 2
,
2 
131 – 133. Answers will vary.
10. a
134. f (x) = 3x + 7
11.
y
;
x
r r
0 = 3x + 7 3x = −7 → x = −
7
3

12. 137
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Section 2.1: Angles and Their MeasureChapter 2: Trigonometric Functions
12. False

13. 138
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138
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Copyright © 2016 Pearson Education, Inc.
Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
= −
5
2
cott   ⋅
3
13.
 3 1  3 1
P =  ,   x = , y = 15.
 2 21 
P = − ,   x
2
, y =
21
 2 2  2 2  5 5  5 5
sint = y =
1
2
cost = x =
3
2
sint = y =
21
5
cost = x = −
2
5
 1   21 
         
   
tant =
y
=
 2 
=
1
⋅
2
=
1
⋅
3
=
3
tant =
y
=
 5 
=
21 5 21
− = −
x  3  2 3 3 3 3 x 
−
2  5  2  2
 2 
 
5

 
csct =
1
=
1
=1⋅
2
= 2
 
csct =
1
=
1
=1⋅
5
=
5
⋅
21
=
5 21
y  1  1 y  21  21 21 21 21
 2 
 
5

   
sect =
1
=
1
=1

−
5 
= −
5
sect =
1
=
1
=1⋅
2
=
2
⋅
3
=
2 3
2
 
x  3 

3 3 3 3 x 
−
  2  2
 
 2   5 
 
 2 
 3  − 
x   2 5
    
 cott = = = − ⋅
y  
x  2  3 2 21 5 21
cott = = = ⋅ = 3  y  1  2 1

2

 
 5 
= −
2
⋅
21
= −
2 21
14.
 1 3  1 3
P =  ,−   x = , y = −
21 21 21
 2 2  2 2
sint = y = −
3
 1 2 6  1 2 62 16. P = − ,   x = − , y =
cost = x =
1  5 5  5 5
2
 3 
− 

sint = y =
2 6
5
tant =
y
=
 2 
= −
3
⋅
2
= − 3 cost = x = −
1
x  1  2 1
 
 
5
 2 6 
 
 
csct =
1
=
1
= −
2
= −
2
⋅
3
= −
2 3 tant =
y
=
 5 
=
2 6
−
5
= − 2 6
y  3 

3 3 3 3 x 
−
1  5  1 
− 
 
5

 2   
sect =
1
=
1
=1⋅
2
= 2 csct =
1
=
1
=1⋅
5
=
5
⋅
6
=
5 6
x  1  1

2

y  2 6 
 
2 6 2 6 6 12
 
 1 
x 

2

= =
1 2
= − = −
1
⋅
3
= −
 5 
y  3 
− 
2 3 3 3 3
 2 

14. 139
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Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 
y  2
−
2 2
3
−
2
2
−
 
sect =
1
=
1
=1

−
5 
= −5 csct =
1
=
1
= 1⋅
2
=
2
⋅
2
= 2
x  1 
 
1

− 
  y  2  2 2 2
 5   2 
 

−
1 

  
 
sect =
1
=
1
=1⋅
2
=
2
⋅
2
= 2
cott =
x
=

 5  = −
1

5
y  2 6  5  2 6 
x  2 
 

2 2 2

5
  2 
 
= −
=1
⋅
6
= −
6
 
 2 
 2 6 6 12 x 2 cott = = =1
y  2 
17.
 2 2  2 2
P = − ,  x = − , y =
 2 
   
 2 2  2 2
sint =
2
19.
 2 2 1  2 2 1
P = , −  x = , y = −
 
3 3 3 3
cost = x = −
2
2
 2 
 
tant = =
 
= −1
 
sint = y = −
1
3
cost = x =
2 2
3
x  2 
−   1 
  2 
tant =
y
=
 3 = −
1
⋅
3
csct =
1
=
1
=1⋅
2
=
2
⋅
2
= 2
x  2 2  3 2 2
y  2 

2 2 2

3

 

2

 
= −
1
⋅
2
= −
2
sect =
1
=
1
=1

−
2 
= −
2
⋅
2
= − 2
2 2 2 4
x  2 

2

2 2 1 1  3  
−  csct = = =1−  = −3
 2  y  1 
   1 
 2 
− 

 3 
cott =
x
=
 2 
= −1 sect =
1
=
1
=1
 3 
=
3
⋅
2
=
3 2
y  2  x  2 2   
2 2 2 4

2
  
   3 
18.
 2 2  2 2
P = ,  x = , y =
 2 2 
 
x   2 2  3 
  cott = =
y
= −  = −2 2
1 3 1 2 2  2 2
   

3

sint = y =
2
2

15. 140
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Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 
   5 2  5 2
cost = x =
2 20. P = − , −
3 3
 x = − , y
3 3
2
 2 
y  

 
= −
 
sint = y = −
2
3
tant = = =1 5x  2  cost = x = −
 2  3
 

16. 141
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Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 
 
2
3
2 2
 
 2 
   
 

−
2 
25. csc
11π 
= csc
 3π
+
8π 
   
 
2
 
2 2

tant =
y
=  3 2 3
= − −     
x  5 
− 

3  5 
= csc
 3π
+ 4π

 3   2 
=
2
⋅
5
=
2 5 = csc
 3π
+ 2⋅2π

5 5 5  2 
csct =
1
=
1
=1

−
3 
= −
3 = csc
 3π 
y  2 
 
2

2
 
− 
 
 
= −1
sect =
1
=
1
=1

−
3  26. sec(8π) = sec(0 + 8π )
x  5 

5

 
−
3  = sec(0 + 4⋅2π ) = sec(0) =1
 
3 5 3 5  3π   3π 
= − ⋅ = −
5 5 5
27. cos−  = cos 
   
 5 
= cos
 π
−
4π 
− 
 
2 2

x  3 
 5  3  5  
cott = = = − −  =  π y 
−
2  3  2  2 = cos
 + (−1)⋅2π 
3

 2 
21.
 
sin
11π 
= sin
 3π
+
8π  = cos
 π 
 2 

2
 
2 2

    = 0
= sin
 3π
+ 4π


2
 28. sin(−3π) = −sin(3π) 
= sin
 3π
+ 2⋅2π


( ) ( )

2
 = −sin π + 2π = −sin π = 0
22.
23.
24.
 
= sin
 3π 
 
= −1
cos(7π) = cos(π + 6π )
= cos(π + 3⋅2π ) = cos(π) = −1
tan (6π) = tan(0 + 6π) = tan (0) = 0
cot
 7π 
= cot
 π
+
6π 
2 2 2
29.
30.
31.
32.
sec(−π) = sec(π) = −1
tan(−3π) = − tan(3π)
= − tan(0 + 3π ) = − tan(0) = 0
sin 45º + cos60º =
2
+
1
=
1+ 2
2 2 2
sin30º − cos45º =
1
−
2
=
1− 2
2 2 2
   
= cot
 π
+ 3π

= cot
 π 
= 0
33. sin90º + tan 45º =1+1 = 2

2
 
2

34. cos180º − sin180º = −1− 0 = −1   
35. sin 45º cos45º =
2
⋅
2
=
2
=
1
2 2 4 2

17. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 2  
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2
3
tan  
2
 
2
−
3
= −
= −
36. tan 45º cos30º =1⋅
3
=
3 cos
2π
=
1
=1

−
2 
= −2
2 2 3  1 
 
1

37. csc45º tan 60º = 2 ⋅ 3 = 6
− 
 
 
 1 
 
cot
2π
=  2 = −
1
⋅
2
= −
1
⋅
3
= −
3
38. sec30º cot 45º =
2 3
⋅1 =
2 3 3  3  2 3 3 3 3
3 3 
2

39. 4sin90º −3tan180º = 4⋅1−3⋅0 = 4
 
48. The point on the unit circle that corresponds to
θ =
5π
=150º is

−
3
,
1 
.40. 5cos90º −8sin 270º = 5⋅0 −8(−1) = 8  
6 2 2
41. 2sin
π
− 3tan
π
= 2⋅
3
−3⋅
3
=
3 6 2 3
3 − 3 = 0
 
sin
5π
=
1
6 2
cos
5π
= −
42. 2sin
π
+ 3tan
π
= 2⋅
2
+ 3⋅1 =
4 4 2
2 + 3
6 2
 1 
5π 

2

 1  2
= = ⋅ −
 3 3
⋅ = −
 
 
43. 2sec
π
+ 4cot
π
= 2⋅ 2 + 4⋅
3
= 2 2 +
4 3
6 3 2 
− 
3  3 3
4 3 3 3  2 
csc
5π
=
1
=1⋅
2
= 2
44. 3csc
π
+ cot
π
= 3⋅
2 3
+1 = 2 3 +1
3 4 3
6  1  1
 
 
sec
5π
=
1
=1⋅

−
2  3 2 3
⋅ = −
45. csc
π
+ cot
π
=1+ 0 =1
6   
3

3 3
2 2 − 
 
 2 
46. secπ −csc
π
= −1−1 = −2 
π −
3 
2
cot
5
=  2  = −
3
⋅
2
= − 3
47. The point on the unit circle that corresponds to
θ =
2π
=120º is

−
1
,
3 
.
3  2 2 
6  1  2 1
 
 
49. The point on the unit circle that corresponds to
sin
2π
=
3
θ = 210º =
7π  3 1 
is − , − .
3 2 6

2 2

cos
2π 1  
sin 210º
1
3 2
 3 
2
cos210º = −
= 3
π   
   2

18. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 2  
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 1
tan
2
=
 2 
=
3
⋅−
2
 = − 3
3 
−
1  2  1   1 

2

−    
tan 210º =  2 = −
1
⋅−
2 3 3
⋅ =
csc
2π
=
1
=
2
⋅
3
=
2 3  3 
− 

2  3  3 3
3  3  3 3 3  2 

2

1  2 
 
csc210º =
 1 
=1⋅−
−
 = −2


19. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 2  
140
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 
2
 
= −
2
sec210º =
1
=1⋅

−
2  3 2 3
⋅ = −
csc
3π
=
1
=1⋅
2
⋅
2
=
2 2
= 2
 3 
 
3

3 3 4  

2 2 2  2
−    2 
 2  
 3 
sec
3π
=
1
=1⋅

−
2 
⋅
2
= −
2 2
= − 2

−


4  2 
 
2

2 2
 2 
 3  2   
− 
cot 210º = = − ⋅−  = 3  2 
−
1  2  1 

2

 2  
− 
cot
3π
=  2  = −
2
⋅
2
= −150. The point on the unit circle that corresponds to
4  2  2 2
θ = 240º =
4π  1 3 
is − , − .

2

   
3
3sin 240º = −
2
 2 2 
52. The point on the unit circle that corresponds to
θ =
11π
= 495º is

−
2
,
2 
.
cos240º
1
2
4
11π 2
 2 2 
 3  sin =
4 2
−  
  
tan 240º =  2  = −
3
⋅−
2
 = 3 11π 2

−
1  2  1  cos = −
4 2

2

   2 
csc240º =
1
=1⋅

−
2 
⋅
3
= −
2 3
tan
11π  
=
2 
=
2
⋅

−
2 
= −1
 3 
 
3

3 3   
− 
 4  2  2  2 
−
 2  
2

sec240º =
1
=1⋅

−
2 
= − 2 11π  
1 2 2 2 2
 1 
 
1
 csc
4
= = 1⋅ ⋅ = = 2
  2− 
 

 
2 2 2

2


−
1 
11π
 
1  2  2 2 2        sec = = 1⋅− ⋅ = − = − 2
cot 240º =  2  = −
1
⋅−
2
⋅
3
=
3
4  

2 2 2
 3  2  3  3 3 2  
− 
− 
 
 2 
 2 
− 51. The point on the unit circle that corresponds to
cot
11π
=
 2 
= −
2
⋅
2
= −1
θ =
3π
=135º is

−
2
,
2 
.
4  2  2 2
 4  2 2 
sin
3π
=
2
4 2

20. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 2  
141
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= −
2

2

53. The point on the unit circle that corresponds to
θ =
8π
= 480º is

−
1
,
3 
.
cos
3π
= −
2  
3 2 2
4 2
 2   
sin
8π
=
3
π     3 2
tan
3
= 2 
=
2
⋅−
2
 = −1
4  2  2  2  cos
8π 1
−  3 2 

21. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 2  
142
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2
−
2
 2
 
 
 3  55. The point on the unit circle that corresponds to
π   
  

θ 405º
9π
is
 2
,
2 
.tan
8
=
 2 
=
3
⋅−
2
 = − 3 = =  
3 
−
1  2  1  4  2 2 

2

 
csc
8π
=
1
= 1⋅
2
⋅
3
=
2 3
sin 405º =
2
2
3  3  3 3 3
cos405º =
2

2
 2 
sec
8π
=
1
= 1⋅

−
2 
= −2
 2 
 
3  1 
 
1

tan 405º = 2  =
2
⋅
2
=1− 
 
   2  2 2
 
 1 
   2 
cot
8π
=
 2
= −
1
⋅
2
⋅
3
= −
3
csc405º =
1
=1⋅
2
⋅
2
= 23  3  2 3 3 3  2  2 2

2


2
   
sec405º =
1
=1⋅
2
⋅
2
= 2
 2  2 2
54. The point on the unit circle that corresponds to  
θ =
13π
= 390º is
 3
,
1 
.
 2 
   2 
6
sin
13π
=
1
6 2
 2 2   
cot 405º =
2 
=1
 2 
 
cos
13π
=
3
6 2
 2 
56. The point on the unit circle that corresponds to
 1  13π  3 1 
  θ = 390º = is 
 ,  .
tan
13π
=
2 =
1
⋅
2
⋅
3
=
3 6  2 2 
6  3  2 3 3 3 1
 2  sin390º =
2 
csc
13π
=
1
= 1⋅
2
= 2 cos390º =
3
6  1  1
 
 
2
 1 
 
tan390º =
 
=
1
⋅
2
⋅
3
=
3
sec
13π
=
1
= 1⋅
2
⋅
3
=
2 3
  2 3 3 3
6  3  3 3 3 
3

 2   2 
 
 3 
13π  2  3 2
csc390º =
1
=1⋅
2
= 2
 1  1
 
 2 
cot = = ⋅ = 3
6  1  2 1 1 2 3 2 3
2

sec390º = =1⋅ ⋅ =
3
   3  3 3

2

 
 3 

22. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 2  
143
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cot390º = 2  =
3
⋅
2
= 3
 1  2 1

23. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 2  
144
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−
  ⋅
3
57. The point on the unit circle that corresponds to
θ
π
= −30º is
 3
, −
1 
.
59. The point on the unit circle that corresponds to
θ = −135º = −
3π
is

−
2
, −
2 
.= −    
6
sin

−
π 
= −
1
 2 2  4  2 2 
2

6

2
sin(−135º) = −
 
cos

−
π 
=
3
2
cos(−135º) = −
2

6

2 2 
 1 
 
  2 
− 
 
tan

−
π 
=  2 1 2
= − ⋅
3
= − tan(−135º) =
 2 
= −
2
⋅ −
2
 =1
 6   3  2 3 3 3  2 
−
2  2 

2
 
2

 
csc

−
π 
=
1
− 2
 
csc(−135º) =
1
=1⋅

−
2

⋅
2
= − 2

6

 1 

    
− 
 2
− 
  2  2
 2   2 
sec

−
π 
=
1
=
2
⋅
3
=
2 3 1  2  2

6

 3 

3 3 3
sec(−135º) =

=1⋅−
2 
2
⋅
2
= 2
 

2

 
−
2

   
 π 
  3 
       
  2 
− 
cot−  =
 2 
= 
 3 2
⋅ −  = − 3 cot(−135º) =  2  =1
 6 

 1 
  2   1 
−     2 
−
 2 
58. The point on the unit circle that corresponds to

2

 
60. The point on the unit circle that corresponds to
θ
π
= −60º is
 1
, −
3 
. 4π 
1 3 = −  
θ = −240º = − is − ,  .3
sin

−
π 
= −
3
 2 2  3  2 2 
3

3

2
sin(−240º) =
2
 
cos

−
π 
=
1
cos −240º = −
1

3

2
( )
 
 3 
2
 3 
 π 
 −    
2

3  2 
tan−  =  2  = −
3 2
⋅ = − 3 tan(−240º) =   = ⋅−  = − 3
 3   1  2 1 
−
1  2  1 

2

 
2

 
csc

−
π 
=
1
=1⋅

−
2  3 2 3
⋅ = −
 
csc(−240º) =
1
=1⋅
 2  3 2 3
⋅ =

24. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
 2  
145
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2 2
−
2
 

3


 3  
3

3 3
 3  
3

3 3
   
 
 
− 
 
 2 
sec

−
π 
=
1
=1⋅
2
= 2 sec(−240º) =
1
=1⋅

−
2 
= −2 3 
 1  1  1 
 
1

 
  − 
 

 
 1   1 
        
cot −240º =  2 = −
1
⋅
2 3 3
⋅ = −
cot

−
π 
=
 2 
=
1
⋅

−
2 
⋅ 3
= −
3 ( )

3



3  2

 3

3 3
 3  2  3  3 3
   
−  
2

 2   

25. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions

 
−
 
2
cot  
2
2
2
 14π  

2


2
2
61. The point on the unit circle that corresponds to 64. The point on the unit circle that corresponds to
θ =
5π
= 450º is (0,1). θ
13π
= −390° is
 3 1 
2
= −
6  ,− .
2 2
 
sin
5π
=1
2
csc
5π
=
1
=1
2 1 sin

−
13π  1
 = − cos

−
13π  3
 =
cos
5π
= 0
2
sec
5π
=
1
= undefined
2 0
 6  2
 1 
 
 6  2
tan
5π
=
1
= undefined cot
5π
=
0
= 0 tan

−
13π 
=
 2
= −
1
⋅
2
⋅
3
= −
3
2 0 2 1  6   3  2 3 3 3

2

62. The point on the unit circle that corresponds to  13π 

 
1
θ = 5π = 900º is (−1, 0) . csc−
6
=
 1 
− 2
sin5π = 0 csc5π =
1
= undefined
 
− 
 
0
 13π  1 2 3 2 3
cos5π = −1 sec5π =
1
= −1
sec−
6
 =
 3 
= ⋅ =
3 3 3
 
−1 
2

tan5π =
0
= 0
−1
cot5π =
−1
= undefined
0
 
 3 
 13π 

2
  3   2 
− = =  ⋅ − = − 3
63. The point on the unit circle that corresponds to 
6
  1   2  
1

θ
14π
= −840° is

−
1
, −
3 
.
 
− 
 
   
= −  3  2 2  65. Set the calculator to degree mode:
sin

−
14π 
= −
3
cos

−
14π 
= −
1 sin 28º ≈ 0.47 .

3

2

3

2   
 3 
− 
tan

−
14π 
=
 2 
= −
3
⋅

−
2 
= 3
3

 1 

2

1

 
− 
 
  66. Set the calculator to degree mode:
cos14º ≈ 0.97 .
csc

−
14π 
=
1
=1⋅

−
2  3 2 3
⋅ = −

3

 3 
 
3

3 3   
− 
 
sec

−
14π 
=
1
=1⋅

−
2 
= −2
3

 1 
 
1

 
− 
 
  67. Set the calculator to degree mode:
sec21º =
1
≈1.07 .
−
1 
cot −  
= −
1
⋅

−
2  3 3
⋅ =
cos21°

3

 3 

2

3

3 3   
− 
 

26. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
= −
= −
68. Set the calculator to degree mode:
cot 70º =
1
≈ 0.36 .
tan 70º
74. Set the calculator to radian mode: tan 1 ≈1.56 .
69. Set the calculator to radian mode: tan
π
≈ 0.32 .
10
75. Set the calculator to degree mode: sin1º ≈ 0.02 .
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 = x2
+ y2
= 9 +16 = 25 = 5
sinθ =
4
cscθ =
5
5 4
71. Set the calculator to radian mode: cosθ
3
secθ
5
cot
π
=
1
≈ 3.73 .
= − = −
5 3
4 3
12 tan
π tanθ
3
cotθ
4
= − = −
12
78. For the point (5,−12) , x = 5 , y = −12 ,
r =
sinθ
x2
+ y2
=
12
25 +144 =
cscθ
169 =13
13
72. Set the calculator to radian mode:
= − = −
13 12
csc
5π
=
1
≈1.07 . cosθ =
5
secθ =
13
13 sin
5π 13 5
13 tanθ
12
cotθ = −
5
5 12
79. For the point (2,−3) , x = 2 , y = −3 ,
r = x2
+ y2
= 4 + 9 = 13
73. Set the calculator to radian mode: sin 1 ≈ 0.84 . sinθ =
−3
⋅
13 3 13
cscθ = −
=13
13 13 13 3
cosθ =
2
⋅
13
=
2 13
secθ =
13
tanθ
13 13
3
= −
13 2
cotθ = −
2
2 3

27. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
3 4 3 4
80. For the point (−1,−2) , x = −1, y = −2 , 84. For the point (0.3, 0.4) , x =0.3 , y = 0.4 ,
r = x2
+ y2
= 1+ 4 = 5 r = x2
+ y2
= 0.09 + 0.16 = 0.25 = 0.5
sinθ =
−2
⋅
5
= −
2 5
cscθ =
5
= −
5 sinθ =
0.4
=
4
cscθ =
0.5
=
5
5 5 5 −2 2 0.5 5 0.4 4
cosθ =
0.3
=
3
secθ =
0.5
=
5
cosθ =
−1
⋅
5
= −
5
secθ =
5
= − 5 0.5 5 0.3 3
5 5 5 −1
0.4 4 0.3 3
tanθ =
−2
= 2 cotθ =
−1
=
1 tanθ = =
0.3 3
cotθ = =
0.4 4
−1
81. For the point (−2,−2) , x = −2 ,
−2 2
y = −2 ,
85. sin 45º +sin135º +sin 225º +sin315º
2 2  2   2 
r = x2
+ y2
= 4 + 4 = 8 = 2 2 =
2
+
2
+ −
2  + −
2 
− 2 2 2 2 2
   
= 0sinθ = ⋅ = −
2 2 2 2
cscθ = = − 2
− 2
− 2 2 2 2 2 86. tan 60º + tan150º =
 3 
3 + −cosθ = ⋅ = − secθ = = − 2  
2 2 2 2 −2  3 
− 2 − 2 3 3 − 3 2 3
tanθ = = 1 cotθ = = 1 = =
− 2 −2 3 3
82. For the point (−1,1) , x = −1, y =1, 87. sin 40º +sin130º +sin 220º +sin310º
r = x2
+ y2
= 1+1 = 2 = sin 40º +sin130º +sin(40º +180º)+
sin(130º +180º)
sinθ =
1
⋅
2
=
2
cscθ =
2
= 2
2 2 2 1 = sin 40º +sin130º −sin 40º −sin130º
= 0
cosθ =
−1
⋅
2
= −
2
secθ =
2
= − 2
2 2 2 −1 88. tan 40º + tan140º = tan 40º + tan(180º −40º)
tanθ =
1
= −1 cotθ =
−1
= −1 = tan 40º − tan 40º
−1 1
= 0
83. For the point
 1
,
1 
, x =
1
, y =
1
, 89. If f (θ ) = sinθ = 0.1, then 
 
f (θ +π ) = sin(θ + π) = −0.1 .
r = x2
+ y2
=
1
+
1
=
25 5
=
9 16 144 12
1 5
90. If f (θ ) = cosθ = 0.3, then
sinθ = 4 1 12 3
= ⋅ = cscθ = 12 =
5 4 5
⋅ =
f (θ +π ) = cos(θ + π) = −0.3 .
5 4 5 5
12
1 12 1 3
4 91. If f (θ ) = tanθ = 3 , then
1
cosθ = 3 1 12 4
= ⋅ =
5
secθ = 12 =
5 3 5
⋅ =
f (θ +π ) = tan(θ + π) = 3.
5 3 5 5
12
1 12 1 4
3
92. If f (θ ) = cotθ = −2 , then
1
tanθ = 4 =
1
⋅
3
=
3
1 4 1 4
3
1
cotθ = 3 =
1
⋅
4
=
4
1 3 1 3
4
f (θ +π ) = cot(θ + π) = −2 .

28. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
5
3
4 4

2


   
93. If sinθ =
1
, then cscθ =
1
=1⋅
5
= 5 . 107. f h
π
= sin 2
π
5  1  1 6 6
 
  π 3
94. If cosθ =
2
, then secθ =
1
=1⋅
3
=
3
.
= sin =
3 2
95.
3
f (60º) = sin (60º) =
3
2
 2  2 2
 
  108. g (p(60°)) = cos
60°
2
= cos30° =
3
2
cos315°
96.
97.
g (60º) = cos (60º) =
1
2
f
 60º 
= sin
 60º 
= sin(30º) =
1
109. p(g(315°)) =
=
2
1
cos315°
2

2
 
2

2 1 2 2
98.
   
g
 60º 
= cos
 60º 
= cos(30º) =
3
= ⋅ =
2 2 4
5π 5π 1
2
 
2

2 110. h f = 2 sin = 2⋅ = 1   
6 6 2
2
2 2  3  3
99.  f (60º) = (sin 60º) =
 
=
111. a. f
 π 
= sin
 π 
=
2
 2  4
2
   
    2
 π 2 
2 2  1  1 The point  ,  is on the graph of f.
100. g (60º) = (cos60º) =   =
 2  4
 4 2 
 2 π 
b. The point  ,  is on the graph of f −1
.
101. f (2⋅60º) = sin(2⋅60º) = sin(120º) =
3
2
 2 4 
102.
103.
g (2⋅60º) = cos(2⋅60º) = cos(120º) = −
1
2
2 f (60º) = 2sin(60º) = 2⋅
3
= 3
2
c. f
 π
+
π 
−3 = f
 π 
−3
 4 4   2 
= sin
 π 
−3
 
=1−3
= −2
 π
−

is on the graph of
104. 2g (60º) = 2cos(60º) = 2⋅
1
= 1
2
The point  , 2
 4 
y = f

x +
π 
−3 .
4

105.
 
f (−60º) = sin(−60º) = sin(300º) = −
3
2
106. g (−60º) = cos(−60º) = cos(300º) =
1
2

29. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
θ cosθ −1 cosθ −1
θ
0.5
0.4
0.2
0.1
0.01
0.001
0.0001
0.00001
−0.1224
−0.0789
−0.0199
−0.0050
−0.00005
0.0000
0.0000
0.0000
−0.2448
−0.1973
−0.0997
−0.0050
−0.0050
−0.0005
−0.00005
−0.000005
θ sinθ sinθ
θ
0.5
0.4
0.2
0.1
0.01
0.001
0.0001
0.00001
0.4794
0.3894
0.1987
0.0998
0.0100
0.0010
0.0001
0.00001
0.9589
0.9735
0.9933
0.9983
1.0000
1.0000
1.0000
1.0000
6 6

6 6



0
0
0
2
0
0
0
112. a. g
 π 
= cos
 π 
=
3
   
    2
116.
 π 3 
The point  ,  is on the graph of g.
 6 2 
 3 π 
b. The point  ,  is on the graph of g−1
.
 2 6 
c. 2g
 π
−
π 
= 2g(0)
 
= 2cos(0)
= 2⋅1
= 2
Thus, the point
 π
, 2 is on the graph of
g (θ ) =
cosθ −1
approaches 0 as θ
θ
approaches 0.
6

 
y = 2g

x −
π 
.
117. Use the formula R(θ ) =
=
v 2=
sin(
2θ )
g
with

6

 
g = 32.2ft/sec2
; θ = 45º ; v =100 ft/sec :
113. Answers will vary. One set of possible answers
is −
11π
, −
5π
,
π
,
7π
,
13π
.
3 3 3 3 3
R(45º) (100)2
sin(2⋅45º )
= ≈ 310.56 feet
32.2
v 2
sinθ
2
114. Answers will vary. One ser of possible answers Use the formula H (θ ) = 0 ( )
2g
with
115.
is −
13π
, −
5π
,
3π
,
11π
,
19π
4 4 4 4 4
g = 32.2ft/sec2
; θ = 45º ; v =100 ft/sec :
1002
(sin 45º )2
H (45º) = ≈ 77.64 feet
2(32.2)
118. Use the formula R( )
( )v sin 2θ
θ =
0
g
with
g = 9.8 m/sec2
; θ = 30º ; v =150 m/sec :
R(30º) = 1502
sin(2⋅30º )
9.8
≈1988.32 m
2 2
v (sinθ )Use the formula H (θ ) = 0
2g
with
g = 9.8 m/sec2
; θ = 30º ; v =150 m/sec :
f (θ ) =
sinθ
θ
approaches 1 as θ approaches 0. 1502
(sin30º )2
H (30º) = ≈ 286.99 m
2(9.8)
v 2
sin(2θ )
119. Use the formula R( ) =
0
with
θ
g
g = 9.8 m/sec2
; θ = 25º ; v = 500 m/sec :

30. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
2
R(25º) =
500 sin(2⋅25º )
≈ 19,541.95 m
9.8

31. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
0
2
0
0
0
o
2 2
v (sinθ )Use the formula H (θ ) = 0
2g
with 123. Note: time on road =
distanceonroad
rate
on road
g = 9.8 m/sec2
; θ = 25º ; v = 500 m/sec : =
8− 2x
8
5002
(sin 25º )2
H (25º) = ≈ 2278.14 m
2(9.8) =1−
x
4
1
120. Use the formula R(θ ) =
v sin(2θ )
with
g
g = 32.2ft/sec2
; θ = 50º ; v = 200 ft/sec :
2002
sin(2⋅50º )
R(50º) = ≈1223.36 ft
=1−
=1−
tan=θ
4
1
4 tanθ
32.2
2 2
a. T(30º) =1+
2
−
1
3sin30º 4 tan30º
v (sinθ )
Use the formula H (θ ) =
0
with 2 1
2g
g = 32.2ft/sec2
; θ = 50º ; v = 200 ft/sec :
=1+
1
−
1
3⋅ 4⋅
2 3
2002
(sin50º )2
H (50º) = ≈ 364.49 ft
2(32.2) =1+
4
−
3
≈1.9 hr
3 4
Sally is on the paved road for
121. Use the formula t (θ ) =
2a
g sinθ cosθ
with
1−
1
≈ 0.57 hr .
4 tan30º
g = 32 ft/sec2
and a =10 feet :
2 1
a. t (30) =
2(10) ≈1.20 seconds
32sin30º⋅cos30º
b. T(45º ) =1+ −
3sin 45º 4 tan 45º
2 1
b. t (45) =
2(10) ≈1.12 seconds
32sin 45º⋅cos45º
=1+
3⋅
1
2
−
4⋅1
c. t (60) =
2(10) ≈1.20 seconds
32sin 60º⋅cos60º
=1+
2 2
−
1
≈1.69 hr
3 4
Sally is on the paved road for
122. Use the formula 1−
1
= 0.75 hr .
x(θ ) = cosθ + 16 + 0.5cos(2θ) .
4 tan 45
T(60º ) =1+
2
−
1
x(30) = cos(30º)+ 16 + 0.5cos(2⋅30º ) c.
3sin 60o
4 tan 60o
= cos(30º)+ 16 + 0.5cos(60º) =1+
2
−
1
≈ 4.90 cm
x(45) = cos(45º)+
= cos(45º)+
≈ 4.71 cm
16 + 0.5cos(2⋅45º )
16 + 0.5cos(90º)
3⋅
3 4⋅ 3
2
=1+
4
−
1
3 3 4 3
≈1.63 hr
Sally is on the paved road for
1−
1
≈ 0.86 hr .
4 tan 60o

32. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
3
3
d. T(90º ) =1+
2
−
1
.
3sin90º 4 tan90º
But tan90º 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
time would be:
2
+1 =
5
≈1.67 hours. The
3 3
path would be to leave the first house
127. tan
θ
=
H
2 2D
tan
20°
=
H
2 2(200)
H = 400 tan10°
H ≈ 71
The tree is approximately 71 feet tall.
walking 1 mile in the sand straight to the θ H
road. Then turn and walk 8 miles on the
road. Finally, turn and walk 1 mile in the
sand to the second house.
124. When θ = 30º :
V (30º) =
1
π(2)3 (1+ sec30º)
≈ 251.42 cm3
128. tan =
2 2D
tan
0.52°
=
H
2 2(384400)
H = 768800 tan 0.26°
H = 3488
3
When θ = 45º :
(tan 30º)2
The moon has a radius of 1744 km.
V (45º) =
1
π (2)3 (1+ sec45º)
≈117.88 cm3 129. cosθ + sin2
θ =
41
;cos
49
2
θ + sin2
θ = 1
3 (tan 45º)2
Substitute x = cosθ; y = sinθ and solve these
When θ = 60º :
3
simultaneous equations for y.
x + y2
=
41
; x2
+ y2
= 1
V (60º) =
1
π (2)3 (1+ sec60º)
≈ 75.40 cm3
49
3
125. tan
θ
=
H
(tan 60º)
2
y2
= 1− x2
x + (1− x2
) =
41
49
2 2D
x2
− x −
8
= 0tan
22°
=
6
492 2D
2D tan11° = 6
D =
3
= 15.4
Using the quadratic formula:
a = 1,b = −1,c = −
8
49
tan11°
−(−1) ± (−1)2
− 4(1)(− 8
)
Arletha is 15.4 feet from the car.
x =
49
2
1± 1+ 32
1± 81
1± 9
8 149 49 7
126. tan
θ
=
H
2 2D
= = =
2 2
= or −
2 7 7
1
tan
8°
=
555
2 2D
Since the point is in quadrant III then x = −
7
2
2D tan 4° = 555 and y2
= 1− −
1
= 1−
1
=
48
D =
555
= 3968
2 tan 4°
The tourist is 3968 feet from the monument.
y = −
7 49 49
48 4 3
= −
49 7

33. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
[ ]
2
130. cos2
θ − sinθ = −
1
;cos2
θ + sin2
θ = 1
9
c. Using the MAXIMUM feature, we find:
20
Substitute x = cosθ; y = sinθ and solve these
simultaneous equations for y.
x2
− y = −
1
; x2
+ y2
= 1
9
x2
= 1− y2
45°
0
90°
(1− y2
) − y = −
1
9
y2
+ y −
10
= 0
9
9y2
+ 9y −10 = 0
Using the quadratic formula:
132.
R is largest when θ = 67.5º .
Slope of M =
sinθ − 0
=
sinθ
= tanθ .
cosθ − 0 cosθ
Since L is parallel to M, the slope of L is equal to
the slope of M. Thus, the slope of L = tanθ .
a = 9,b = 9,c = −10 133. a. When t = 1, the coordinate on the unit circle
is approximately (0.5, 0.8) . Thus,
−(9) ±
y =
(9)2
− 4(9)(−10)
18 sin1 ≈ 0.8 csc1 ≈
1
≈1.3
0.8
=
−9± 81+ 360
=
−9± 441
=
−9± 21
18 18 18
Since the point is in quadrant II then
cos1 ≈ 0.5
0.8
sec1 ≈
1
= 2.0
0.5
0.5
y =
−3+ 21
=
12
=
2
18 18 3
and
tan1 ≈ = 1.6
0.5
cot1 ≈ ≈ 0.6
0.8
x2
= 1−
2
= 1−
4
=
5
Set the calculator on RADIAN mode:
3 9 9
x = −
5
= −
= 5
9 3
2 b. When t = 5.1, the coordinate on the unit
131. a. R(60) =
32 2
[sin (2 ⋅60º)− cos(2⋅60º)−1]
32
2
circle is approximately (0.4,−0.9) . Thus,
132 2
= sin (120º)− cos(120º)−1
32
sin5.1 ≈ −0.9 csc5.1 ≈ ≈ −1.1
−0.9
 3  1   cos5.1 ≈ 0.4 sec5.1 ≈
1
= 2.5
≈ 32 2  − −  −1 0.4
 2
≈ 16.56 ft
 2  
tan5.1 ≈
−0.9
0.4
≈ −2.3 cot5.1 ≈
0.4
≈ −0.4
−0.9
b. Let Y1 =
20
322
2
32
sin (2x)− cos(2x)−1 Set the calculator on RADIAN mode:
45° 90°
0

34. Section 2.2: Trigonometric Functions: Unit Circle ApproachChapter 2: Trigonometric Functions
134. a. When t = 2 , the coordinate on the unit
circle is approximately (−0.4, 0.9) . Thus,
138.
sin 2 ≈ 0.9 csc2 ≈
1
0.9
≈1.1
cos 2 ≈ −0.4
tan 2 ≈
0.9
−0.4
= −2.3
sec2 ≈
1
= −2.5
−0.4
cot 2 ≈
−0.4
≈ −0.4
0.9
Set the calculator on RADIAN mode:
b. When t = 4 , the coordinate on the unit
circle is approximately (−0.7,−0.8) . Thus,
sin 4 ≈ −0.8
cos 4 ≈ −0.7
tan 4 ≈
−0.8
≈1.1
−0.7
csc4 ≈
1
≈ −1.3
−0.8
sec4 ≈
1
≈ −1.4
−0.7
cot 4 ≈
−0.7
≈ 0.9
−0.8 Answers will vary.
Set the calculator on RADIAN mode:
139. y =
2
x − 7
135 – 137. Answers will vary.
x =
2
y − 7
x(y − 7) = 2
xy − 7x = 2
xy = 7x + 2
y =
7x + 2
= 7 +
2
x x
f 1
(x) = 7 +
2
x
140. 180
−13π
3
= −780° is
141. f (x + 3) =
(x + 3) −1
(
x + 3)2
+ 2
x + 2
=
x2
+ 6x + 9 + 2
x + 2
=
x2
+ 6x +11

35. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
142. d = (1− 3)2
+ (0 − (−6))2
19. cos
33π
= cos
 π
+8π

= cos
 π
+ 4⋅2π

4

4
 
4

= (−2)2
+ (6)2    
=
π
= 4 + 36 = 40 = 2 10 cos
4
=
2
2
Section 2.3 20. sin
9π
= sin
 π
+ 2π

= sin
π
=
2
4

4

4 2
1. All real numbers except −
1
;
 

x | x
1
 ≠ − 
2  2 21. tan(21π) = tan(0 + 21π) = tan(0) = 0
2. even 9π  π   π 
22. csc = csc + 4π = csc 2
+ 2⋅2π
3. False
4. True
2  2   
= csc
π
2
5. 2π , π
6. All real number, except odd multiples of
π 23.
=1
sec
17π
= sec
 π
+ 4π

= sec
 π
+ 2⋅2π

2 4

4
 
4

7. b.
   
= sec
π
4
8. a
9. 1
17π
= 2
 π   π 
24. cot = cot + 4π = cot 4
+ 2⋅2π
10. False; secθ =
1
cosθ
4  4   
= cot
π
4
11. sin 405º = sin(360º + 45º) = sin 45º =
2
2
25.
=1
tan
19π
= tan
 π
+ 3π

= tan
π
=
3
12. cos420º = cos(360º + 60º) = cos60º =
1 6

6

6 3
 
2
25π  π   π 
13. tan 405º = tan(180º + 180º + 45º) = tan 45º =1 26. sec = sec + 4π = sec + 2⋅2π
14. sin390º = sin(360º + 30º) = sin30º =
1
2
6  6   6 
= sec
π
6
2 3
15. csc450º = csc(360º + 90º) = csc90º =1 =
3
16. sec540º = sec(360º + 180º ) = sec180º = −1
17. cot390º = cot(180º + 180º + 30º) = cot 30º = 3
18. sec420º = sec(360º + 60º) = sec60º = 2
27. Since sin θ > 0 for points in quadrants I and II,
and cosθ < 0 for points in quadrants II and III,
the angle θ lies in quadrant II.
28. Since sin θ < 0 for points in quadrants III and
IV, and cosθ > 0 for points in quadrants I and
IV, the angle θ lies in quadrant IV.

36. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
5
= −
 
5
2
5
29. Since sinθ < 0 for points in quadrants III and
secθ =
1
=
1
= −
5
IV, and tanθ < 0 for points in quadrants II and cosθ  3  3
IV, the angle θ lies in quadrant IV.
30. Since cosθ > 0 for points in quadrants I and IV,
and tanθ > 0 for points in quadrants I and III,
the angle θ lies in quadrant I.
− 
 
cotθ =
1
= −
3
tanθ 4
2 5 5
31. Since cosθ > 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.
37. sinθ = , cosθ =
5 5
 2 5 
 
θ =
sinθ
=  5  =
2 5
⋅
5
=32. Since cosθ < 0 for points in quadrants II and III,
tan
cosθ
2
 5  5 5
and tanθ > 0 for points in quadrants I and III,
the angle θ lies in quadrant III.
 5 
 
cscθ =
1
=
1
=1⋅
5
⋅
5
=
5
33. Since secθ < 0 for points in quadrants II and III,
and sinθ > 0 for points in quadrants I and II, the
sinθ  2 5 
 

2 5 5 2
angle θ lies in quadrant II.  5 
secθ =
1
=
1
=
5
⋅
5
= 534. Since cscθ > 0 for points in quadrants I and II,
and cosθ < 0 for points in quadrants II and III,
cosθ  5  5 5
 
the angle θ lies in quadrant II.  5 
cotθ =
1
=
1
35. sinθ
3
, cosθ =
4
5 5
tanθ 2

−
3  5 2 5
sinθ 
5
 3 5 3
tanθ = = = − ⋅ = −
38. sinθ = − , cosθ = −
5 5
cosθ  4 
 
 
5 4 4
sinθ
 5 
− 
 5   5   5  1
cscθ =
1
=
1
= −
5 tanθ = =
cosθ 
= −
2 5 
⋅−  =
5 2 5 2
sinθ 
−
3  3 − 
5    

5
  
 
cscθ =
1
=
1
= 1⋅

−
5
 5
⋅ = − 5
secθ =
1
=
1
=
5
sinθ  
cosθ  4  4  5  
−
5  5

5
 
5

 
cotθ =
1
= −
4  
secθ =
1
=
1
=

−
5 
⋅
5
= −
5
tanθ 3 cosθ  2 5 
 
2 5

5 2 
− 
 
36. sinθ =
4
, cosθ = −
3
5 5
 4 
 
cotθ =
1
=
1
=1⋅
2
= 2 tan
θ  1  1
 
 

37. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
5
5
tanθ =
sinθ
=
5 4  5  4
= ⋅ − = −
cosθ  3 

5

3

3
− 
 
 
cscθ =
1
=
1
=
5
sinθ  4  4
 
 

38. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
tan  
2
−
2
2
2
1−
−
= −
tan  
= −
2
39. sinθ =
1
, cosθ =
3
secθ =
1
=
1
=
3
⋅
2
=
3 2
2 2 cosθ  2 2  2 2 2 4
 1   3 
sinθ 

2

θ = =
=
1
⋅
2
⋅
3
=
3
 
1 1 4 2
cosθ  3  2 3 3 3 cotθ = = = − ⋅ = −2 2
 2  tanθ  2  2 2
−
   4 
 
cscθ =
1
=
1
=1⋅
2
= 2
sinθ  1  1
 
  42. sinθ =
2 2
, cosθ = −
1
3 3
secθ =
1
=
1
=
2
⋅
3
=
2 3  2 2 
cosθ  3  3 3 3
sinθ  3 
2 2 3
 2  tanθ = =   = ⋅− = −2 2
  cosθ  1  3 1
 
cotθ =
1
=
1
=
3
⋅
3
= 3
 3 
tanθ  3  3 3 1 1 3 2 3 2
 3  cscθ = = =1⋅ ⋅ =
sinθ  2 2  2 2 2 4
 
 3 
40. sinθ =
3
, cosθ =
1
 
1 1  3 
2 2
 3 
secθ = =
cosθ 
−
=1⋅−  = −3
1   1 
3

sinθ  
 

 
3 2
tanθ = = = ⋅ = 3
cotθ =
1
=
1
= −
1
⋅
2
= −
2
cosθ  1  2 1
 
 
tanθ −2 2 2 2 2 4
cscθ =
1
=
1
=1⋅
2
⋅
3
=
2 3
43. sinθ =
12
, θ in quadrant II
sinθ  3  3 3 3 13
 2 
 
secθ =
1
=
1
=1⋅
2
= 2
Solve for cosθ : sin2
θ + cos2
θ =1
cos2
θ =1−sin2
θ
cosθ  1  1
 
  Since θ
cosθ = ± 1−sin2
θ
is in quadrant II, cosθ < 0 .
cotθ =
1
=
1 1 3 3
= ⋅ = cosθ = − 1− sin2
θ
tanθ 3 3 3 3
12 
  = − 1−
144
= −
25
= −
5
41. sinθ 1
, cosθ =
2 2
3 3
 1 
 

13 
12 
sinθ 

13

θ = =
169 169 13
12  13  12
= ⋅−  = −
tanθ =
sinθ
=  3
= −
1
⋅
3
⋅
2
= −
2 cosθ 
−
5  13  5  5
cosθ  2 2 

3 2 2 2 4

13

 3 
 
1 1 13  cscθ = = =
cscθ =
1
=
1
=1⋅

−
3 
= −3
sinθ 12  12
 

39. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
3
sinθ  1 
 
1
 13 
− 
 
 

40. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
4
1−
 
5

5 −
,
12
sinθ  5 3  5  3
1−
2
= −
2
2
secθ =
1
=
1
= −
13
cscθ =
1
=
1
= −
5
cosθ 
−
5  5 sinθ 
−
3  3

13
 
5

 
cotθ =
1
=
1
= −
5
 
secθ =
1
=
1
= −
5
tanθ 
−
12  12 cosθ 
−
4  4
 5 
 
5

   
cotθ =
1
=
1
=
4
44. cosθ =
3
,
5
θ in quadrant IV tanθ  3  3
 
 
Solve for sinθ : sin2
θ + cos2
θ =1
sin2
θ = 1− cos2
θ 46. sinθ = −
5
, θ in quadrant III
Since θ
sinθ = ± 1− cos2
θ
is in quadrant IV, sinθ < 0 .
2
13
Solve for cosθ : sin2
θ + cos2
θ =1
cos2
θ = 1− sin2
θ
sinθ = − 1− cos2
θ = −
 3 
 
 5 
16 4
= − = −
25 5 Since θ
cosθ = ± 1− sin2
θ
is in quadrant III, cosθ < 0 .
cosθ = − 1− sin2
θ = − 1−

−
5 

−
4  
13

sinθ 
5
 4 5 4
 
tanθ = = = − ⋅ = −
cosθ  3 
 
 
5 3 3 144 12
= − = −
169 13

−
5 
cscθ =
1
=
1
= −
5
sinθ  
5  13  5 13 sinθ 
−
4  4 tanθ = = = − ⋅−  =

5
 cosθ  12  13  12  12
  −
 13

secθ =
1
=
1
=
5
1 1 13
cosθ  3  3 cscθ = = = −
  sinθ
   5  5
 
 13
cotθ =
1
=
1
= −
3
1 1 13
tanθ 
−
4  4 secθ = =
cosθ 
= −
12  12

3
 −
  
13

45. cosθ
4
θ in quadrant III
 
cotθ =
1
=
1
=
12
5
Solve for sinθ : sin2
θ + cos2
θ =1
sin2
θ = 1− cos2
θ
tanθ
5
 5  5
 
 
sinθ = ± 1− cos2
θ
47. sinθ = , 90º < θ <180º, θ
13
in quadrant II
Since θ is in quadrant III, sinθ < 0 . Solve for cosθ : sin2
θ + cos2
θ =1
2 2
sinθ = − 1− cos2
θ cos θ =1− sin θ
= − 1−

−
4 
= − 1−
16
= −
9
= −
3
cosθ = ± 1− sin2
θ

5

25 25 5
Since θ is in quadrant II, cosθ < 0 .
 

−
3 
tanθ = =   = − ⋅ − =
cosθ = − 1− sin2
θ = −  5 
 
13 

41. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
cosθ  4 

5

4

4
− 
 
= − 1−
25
= −
144 12
= −
 5  169 169 13

42. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
 
3
3
,
−
5
5
−
,
−
= −
 5 
sinθ 
13

tanθ = =
5  13  5
= ⋅ − = −
sinθ = 1− cos2
θ
2
cosθ  12  13

 12

12 = −

−
1 

1
= − =
8 2 2
=
− 
  1 
3
 1
9 9 3
 13   

 
cscθ =
1
=
1
=
13 2 2
 
sinθ  5  5
tanθ =
sinθ
=
 3 
=
2 2
⋅

−
3 
= −2 2
13

cosθ  1 

3

1

 
secθ =
1
=
1
= −
13
− 
 
 
cosθ 
−
12  12 1 1 3 2 3 2

13
 cscθ = = = ⋅ =
  sinθ  2 2 
 
 2 2 2 4
cotθ =
1
=
1
= −
12
 3 
tanθ 
−
5  5 1 1
12

secθ = = = −3
48.
 
cosθ =
4
, 270º < θ < 360º; θ in quadrant IV
cosθ  1 
 
 
5 cotθ =
1
=
1
⋅
2
= −
2
Solve for sinθ : sin2
θ + cos2
θ =1
sinθ = ± 1− cos2
θ 50. sinθ
tanθ
2
= −
−2 2 2
π < θ <
3π
,
4
θ in quadrant III
Since θ is in quadrant IV, sinθ < 0 .
2
3 2
Solve for cosθ : sin2
θ + cos2
θ =1
sinθ = − 1− cos2
θ = − 1−
 4  9 3
= − = − 2
 
 
 3 
 

25 5
Since θ
cosθ = ± 1−sin θ
is in quadrant III, cosθ < 0 .
2
tanθ =
sinθ
=  5 = −
3
⋅
5
= −
3 cosθ = − 1− sin2
θ = − 1−

−
2 
cosθ  4 

5 4 4

3

 
 
cscθ =
1
=
1
= −
5
 
= −
5
= −
5
9 3
sinθ 
−
3  3 
−
2 

5
  
 
tanθ =
sinθ
=  3
secθ =
1
=
1
=
5 cosθ  5 
−
cosθ  4  4  3 

5
  
 
2  3  5 2 5
= − ⋅ − ⋅ =
cotθ =
1
=
1
= −
4
3  5  5 5
tanθ 
−
3  3
1 1 3
4

cscθ = = = − 
sinθ  2  2
 
49. cosθ
1 π
<θ < π, θ in quadrant II  3 
3 2 1 1 3 5 3 5
Solve for sinθ : sin2
θ + cos2
θ =1 secθ =
cosθ
= = − ⋅ = −
  5

43. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
sin2
θ =1− cos2
θ 5 5 5

−
3 
sinθ = ± 1− cos2
θ
 
1 1 5 5 5
Since θ is in quadrant II, sinθ > 0 . cotθ = = = ⋅ =
tanθ  2 5  2 5 5 2
 5 
 

44. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
= −
3
2
3
2
= −
15
2
3
 4 
2
 
 
51. sinθ =
2
, tanθ < 0, so θ is in quadrant II secθ =
1
=
1 4
= −4
3 cosθ 
−
1  1
Solve for cosθ : sin2
θ + cos2
θ =1
cos2
θ =1−sin2
θ

4

 
cotθ =
1
=
1
⋅
15
=
15
cosθ = ± 1−sin2
θ tanθ 15 15 15
Since θ is in quadrant II, cosθ < 0 .
53. secθ = 2, sinθ < 0, so θ is in quadrant IV
cosθ = − 1− sin2
θ
Solve for cosθ : cosθ =
1
=
1
= − 1−
 2 
= − 1−
4
= −
5
= −
5 secθ 2
  9 9 3 Solve for sinθ : sin2
θ + cos2
θ =1
2 2
 2 
 
sin θ =1− cos θ
tanθ =
sinθ
=
 3
=
2
⋅

−
3  2 5
= −
sinθ = ± 1− cos2
θ
cosθ 

5  3

5

5 
−
3 
Since θ is in quadrant IV, sinθ < 0 .
2
 
cscθ =
1
=
1
=
3
sinθ = − 1− cos θ
 1 
1 3 3
sinθ  2  2
 
 
= − 1− 
 2 

= − 1− = − = −
4 4 2
3 
secθ =
1
=
1
= −
=3 3 5
sinθ − 
2 3 2
= −
= =
 
= − ⋅ = −cosθ
 5  5 5
−  tanθ
cosθ
3
 1  2 1
 3   
cotθ =
1
=
1
= −
5
= − 5
cscθ =
1
=
1
= −
2
= −
2 3
tanθ 
− 2 5  2 5 2 sinθ  3  3 3

5

−    2 
cotθ =
1
=
1
= −
3
52. cosθ
1
, tanθ > 0
4
tanθ − 3 3
Since tanθ =
sinθ
cosθ
> 0 and cosθ < 0 , sinθ < 0 .
54. cscθ = 3, cotθ < 0, so θ is in quadrant II
Solve for sinθ : sin2
θ + cos2
θ =1
sinθ = ± 1− cos2
θ
sinθ = − 1− cos2
θ
2
Solve for sinθ : sinθ =
1
=
1
cscθ 3
Solve for cosθ : sin2
θ + cos2
θ =1
2
= − 1−

−
1 
= − 1−
1 cosθ = ± 1− sin θ

4

16 Since θ is in quadrant II, cosθ < 0 . 
15
= − = −
cosθ = − 1− sin2
θ
16 4  1  1 8 2 2
 15 
− 

= − 1− 
 
= − 1− = − = −
9 9 3
tanθ =
sinθ 15  4 
= = − ⋅ − = 15
 1 
cosθ  1 

4

1

sinθ  
3


45. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
4
3
− 
 
 
tanθ = =
cosθ 
 
2 2 
cscθ =
1
=
1
= −
4
⋅
15
= −
4 15
− 
 
sinθ  15  15 15 15 1  3  2 2
−  = ⋅− ⋅ = −
 4  3  2 2  2 4

46. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
3 2

2
4
1 1
4
3
3
4
2 = −

 
 
2
θ
−
2
 
secθ =
1
=
1
= − ⋅
3 sinθ = − 1− cos2
θ
cosθ  2 2 
− 

2 2 2 4
= − 1−

−
4  16 9 3
= − 1− = − = −
 3   5  25 25 5
cotθ =
1
=
1
= −
4
⋅
2
= −2 2 cscθ =
1
=
1
= −
5
tanθ  2  2 2 sinθ 
−
3  3
− 
 
5

 4 
 
1
55. tanθ =
3
, sinθ < 0, so θ is in quadrant III
4
57. tanθ = − , sinθ > 0, so θ is in quadrant II
3
Solve for secθ : sec2
θ =1+ tan2
θ Solve for secθ : sec2
θ =1+ tan2
θ
2
secθ = ± 1+ tan2
θ secθ = ± 1+ tan θ
Since θ is in quadrant III, secθ < 0 . Since θ is in quadrant II, secθ < 0 .
2
secθ = − 1+ tan2
θ secθ = − 1+ tan θ
2
2
= − 1+

−
1 

= − 1+
1
= −
10 10
= −
= − 1+
 3 
 = − 1+
9 25 5
= − = − 
16 16 4  3  9 9 3
cosθ =
1
= −
4 3 3 10
cosθ = = = − = −
secθ 5 secθ  10  10 10
− 
sinθ = − 1− cos2
θ
= − 1−

−
4 
= − 1−
16
= −
9 3
= −
sinθ =
 3 
1− cos2
θ

5

25 25 5 2
  
= 1−−
3 10 
 = 1−
90
cscθ =
1
sinθ =
1

−
3 
= −
5
3
 10  100

5
 10 10
 
cotθ =
1
=
1
=
4
= =
100 10
1 1
tanθ  3  3 cscθ = = = 10
  sin   10 
 
 10 
56. cotθ =
4
, cosθ < 0, so θ is in quadrant III cotθ =
1
=
1
= −3
3
tanθ =
1
=
1
=
3
tanθ  1 
 
 
cotθ  4  4
 
  58. secθ = −2, tanθ > 0, so θ is in quadrant III
Solve for secθ : sec2
θ =1+ tan2
θ
secθ = ± 1+ tan2
θ
Solve for tanθ : sec2
θ =1+ tan2
θ
tanθ = ± sec2
θ −1
Since θ is in quadrant III, secθ < 0 . tanθ = sec2
θ −1 = (−2)2
−1 = 4 −1 = 3
secθ = − 1+ tan2
θ
cosθ =
1
= −
1
= − 1+
 3 
= − 1+
9
= −
25
= −
5 secθ 2
  16 16 4 cotθ =
1
=
1
⋅
3
=
3

47. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
cosθ =
1
= −
4 tanθ 3 3 3
secθ 5

48. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions


12
 
12


2
sinθ = − 1− cos2
θ 75. sec

−
π  π
 = sec =
2 3
= − 1−

−
1 
= − 1−
1
= −
3
= −
3
 6  6 3

2

4 4 2 2 3
   π  π
76.
−  = − = −
csc csc
cscθ =
1
=
1
= −
2
⋅
3
= −
2 3  3  3 3
sinθ  3  3 3 3
−  77. sin2
(40º)+ cos2
(40º) =1
 2 
78. sec2
(18º)− tan2
(18º) =1
59.
60.
61.
62.
sin(−60º ) = −sin 60º = −
3
2
cos(−30º) = cos30º =
3
2
tan(−30º) = − tan30º = −
3
3
sin(−135º ) = −sin135º = −
2
2
79.
80.
81.
sin(80º)csc(80º) = sin(80º)⋅
1
=1
sin(80º)
tan(10º)cot(10º) = tan(10º)⋅
1
=1
tan (10º)
sin(40º)tan(40º)− = tan(40º)− tan(40º) = 0
cos(40º)
cos(20º)
63. sec(−60º) = sec60º = 2
64. csc(−30º ) = −csc30º = −2
65. sin(−90º ) = −sin90º = −1
66. cos(−270º) = cos270º = 0
67. tan

−
π 
= − tan
π
= −1
82.
83.
84.
cot(20º)− = cot(20º)− cot(20º) = 0
sin(20º)
cos(400º)⋅sec(40º) = cos(40º +360º)⋅sec(40º)
= cos(40º)⋅sec(40º)
= cos(40º)⋅
1
=1
cos(40º)
tan(200º)⋅cot(20º) = tan(20º +180º)⋅cot(20º)

4

4 = tan(20º)⋅cot(20º) 
= tan 20º
1
⋅ =1
68. sin(−π) = −sin π = 0
( )
tan(20º)
69. cos

−
π 
= cos
π
=
2

4

4 2 85.
 π   25π   π   25π 
  sin− csc  = −sin  csc 
 12   12  12   12 
70. sin

−
π 
= −sin
π
= −
3 = −sin
 π 
csc
 π
+
24π 

3

3 2

12
 
12 12

     

= −sin
 π 
csc
 π
+ 2π

71. tan(−π) = − tan π = 0

12
 
12

72. sin

−
3π 
= −sin
3π
= −(−1) =1
   
= −sin
 π 
csc
 π 
   

2

2   π  1

49. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
 
= −sin ⋅ = −1
73. csc

−
π 
= −csc
π
= − 2
12 
sin
 π 
12
 4  4
 
74.
 
sec(−π) = secπ = −1

50. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
18  18 
86. sec

−
π 
cos
 37π 
= sec
 π 
cos
 37π  92. If cotθ = −2 , then

18
 
18
 
18
 
18

       
= sec
 π 
cos
 π
+
36π 
cotθ + cot(θ − π)+ cot(θ − 2π)
= −2 + (−2)+ (−2) = −6

18
 
18 18

   
= sec
 π 
cos
 π
+ 2π

93. sin1º +sin 2º +sin3º +...+ sin357º

18
 
18
 +sin358º +sin359º
   
= sec
 π 
cos
 π 
   
= sec
 π 
⋅
1
=1
 
  
= sin1º +sin 2º +sin3º + ⋅⋅⋅ + sin(360º −3º )
+sin(360º −2º ) + sin(360º −1º)
= sin1º +sin 2º +sin3º + ⋅⋅⋅+sin(−3º )
+ sin(−2º ) + sin(−1º)
87.
sin(−20º) + tan(200º)cos(380º)
18 
sec
π

18  = sin1º +sin 2º +sin3º + ⋅⋅⋅−sin 3º −sin 2º −sin1º
= sin(180º) = 0
94. cos1º +cos2º +cos3º + ⋅⋅⋅+cos357º
+ cos358º +cos359º
88.
−sin(20º)= + tan(20º +180º)cos(20º +360º)
−sin(20º)= + tan(20º)cos(20º)
= − tan(20º)+ tan(20º) = 0
sin(70º) + tan(−70º)cos(−430º)
sin(70º)= − tan(70º)cos(430º)
sin(70º)= − tan(70º)cos(70º +360º)
sin(70º)= − tan(70º)cos(70º)
= tan(70º)− tan(70º) = 0
= cos1º +cos2º +cos3º +...+ cos(360º −3º )
+ cos(360º −2º ) + cos(360º −1º)
= cos1º +cos2º +cos3º +...+ cos(−3º)
+ cos(−2º ) + cos(−1º)
= cos1º +cos2º +cos3º +... + cos3º
+ cos2º +cos1º
= 2cos1º +2cos2º +2cos3º +...+ 2cos178º
+ 2cos179º +cos180º
= 2cos1º +2cos2º +2cos3º +...+ 2cos(180º −2º )
+ 2cos(180º −1º) + cos(180º)
= 2cos1º +2cos2º +2cos3º +...− 2cos2º
− 2cos1º +cos180º
= cos180º = −1
95. The domain of the sine function is the set of all
real numbers.
89. If sinθ = 0.3 , then
sinθ + sin(θ + 2π)+ sin(θ + 4π)
96. The domain of the cosine function is the set of
all real numbers.
= 0.3+ 0.3+ 0.3 = 0.9 97. f (θ) = tanθ is not defined for numbers that are
90. If cosθ = 0.2 , then
cosθ + cos(θ + 2π)+ cos(θ + 4π)
odd multiples of
π
.
2
= −0.2 + 0.2 + 0.2 = 0.6 98. f (θ) = cotθ is not defined for numbers that are
91. If tanθ = 3, then
tanθ + tan(θ + π)+ tan(θ + 2π)
multiples of π .
= 3+ 3+ 3 = 9
99. f (θ) = secθ is not defined for numbers that are
odd multiples of
π
.
2

51. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
100. f (θ) = cscθ is not defined for numbers that are
114. a. f (−a) = f (a) =
1
multiples of π .
101. The range of the sine function is the set of all
real numbers between −1 and 1, inclusive.
102. The range of the cosine function is the set of all
real numbers between −1 and 1, inclusive.
4
b. f (a) + f (a + 2π) + f (a − 2π)
= f (a) + f (a) + f (a)
1 1 1
= + +
4 4 4
3
=
103. The range of the tangent function is the set of all 4
real numbers.
104. The range of the cotangent function is the set of
all real numbers.
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 .
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) = −sinθ . Its graph is symmetric with
respect to the origin.
108. The cosine function is even because
cos(−θ) = cosθ . Its graph is symmetric with
respect to the y-axis.
109. The tangent function is odd because
tan(−θ) = − tanθ . Its graph is symmetric with
respect to the origin.
110. The cotangent function is odd because
cot(−θ) = −cotθ . Its graph is symmetric with
respect to the origin.
111. The secant function is even because
115. a.
b.
116. a.
b.
117. a.
b.
118. a.
b.
f (−a) = − f (a) = −2
f (a) + f (a + π) + f (a + 2π)
= f (a) + f (a) + f (a)
= 2 + 2 + 2 = 6
f (−a) = − f (a) = −(−3) = 3
f (a) + f (a + π) + f (a + 4π)
= f (a) + f (a) + f (a)
= −3+ (−3) + (−3)
= −9
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
sec(−θ) = secθ . Its graph is symmetric with 119. Since tanθ =
500 1 y
= = , then
respect to the y-axis.
112. The cosecant function is odd because
csc(−θ) = −cscθ . Its graph is symmetric with
respect to the origin.
1500 3 x
r2
= x2
+ y2
= 9 +1 =10
r = 10
sinθ =
1
=
1
.
1+ 9 10
113. a. f (−a) = − f (a) = −
1 5 5
3 T = 5 −

+
1   1 
b. f (a) + f (a + 2π) + f (a + 4π) 3⋅
3
 
10

= f (a) + f (a) + f (a)
1 1 1
= + + = 1
3 3 3
   
= 5 − 5 + 5 10
= 5 10 ≈15.8 minutes

52. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
   
120. a. tanθ =
1
=
y
for 0 < θ <
π
.
123. Suppose there is a number p, 0 < p < 2π for
4 x 2 which sin(θ + p) = sinθ for all θ . If θ = 0 ,
r2
= x2
+ y2
=16 +1 =17 then sin (0 + p) = sin p = sin 0 = 0 ; so that
r = 17 π  π   π 
p = π . If θ = then sin
 + p = sin  .
Thus, sinθ =
1
.
2  2   2 
17
T(θ) =1+
2
−
1
But p = π . Thus, sin
 3π 
= −1 = sin
 π 
=1,
 2   2 

3⋅
1  
4⋅
1 

17
  4 
or −1 =1. This is impossible. The smallest
positive number p for which sin(θ + p) = sinθ   
for all θ must then be p = 2π .
=1+
2 17
−1 =
2 17
≈ 2.75 hours
3 3
124. Suppose there is a number p, 0 < p < 2π , for
b. Since tanθ =
1
, x = 4 . Sally heads
4
directly across the sand to the bridge,
which cos(θ + p) = cosθ for all θ . If θ =
π
,
2
crosses the bridge, and heads directly across
the sand to the other house.
c. θ must be larger than 14º , or the road will
not be reached and she cannot get across the
river.
then cos
 π
+ p

= cos
 π 
= 0 ; so that p = π .
2
 
2
   
If θ = 0 , then cos(0 + p) = cos(0) . But
p = π . Thus cos(π) = −1 = cos(0) =1, or
−1 = 1. This is impossible. The smallest
positive number p for which cos(θ + p) = cosθ
121. Let P = (x, y) be the point on the unit circle that
corresponds to an angle t. Consider the equation
tant =
y
= a . Then y = ax . Now x2
+ y2
=1,
for all θ must then be
1
p = 2π .
x
1
so x2
+ a2
x2
=1 . Thus, x = ± and
1+ a2
a
y = ± . That is, for any real number a ,
1+ a2
there is a point P = (x, y) on the unit circle for
125.
126.
secθ = : Since cosθ has period 2π , so
cosθ
does secθ.
cscθ =
1
: Since sinθ has period 2π , so
sinθ
does cscθ.
which tant = a . In other words, 127. If P = (a,b) is the point on the unit circle
−∞ < tant < ∞ , and the range of the tangent
function is the set of all real numbers.
122. Let P = (x, y) be the point on the unit circle that
corresponds to an angle t. Consider the equation
corresponding to θ , then Q = (−a,−b) is the
point on the unit circle corresponding to θ + π .
Thus, tan(θ + π) =
−b
=
b
= tanθ . If there
−a a
cott =
x
= a . Then x = ay . Now x2
+ y2
=1,
exists a number p, 0 < p < π, for which
y tan(θ + p) = tanθ for all θ , then if θ = 0 ,
so a2
y2
+ y2
=1. Thus,
1
y = ± and
tan( p) = tan(0) = 0. But this means that p is a
1+ a2
a
x = ± . That is, for any real number a ,
multiple of π . Since no multiple of π exists in
the interval (0,π), this is impossible. Therefore,
1+ a2
there is a point P = (x, y) on the unit circle for
which cott = a . In other words, −∞ < cott < ∞ ,
and the range of the tangent function is the set of
all real numbers.
the fundamental period of f (θ ) = tanθ is π .

53. Section 2.3: Properties of the Trigonometric FunctionsChapter 2: Trigonometric Functions
a
128. cotθ =
1
: Since tanθ has period π , so
tanθ
does cotθ .
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:
129. Let P = (a,b) be the point on the unit circle
corresponding to θ . Then cscθ =
1
=
1
secθ =
1
=
1
b sinθ
a cosθ
cotθ =
a
=
1
=
1
b  b 
 
 
tanθ
130. Let P = (a,b) be the point on the unit circle
corresponding to θ . Then
tanθ =
b
=
sinθ
139. f (25) = 3 25 − 9 + 6
= 3 16 + 6 = 12 + 6 = 18
a cosθ
So the point (25,18) is on the graph.
cotθ =
a
=
cosθ
b sinθ
140. y = (0)3
− 9(0)2
+ 3(0) − 27
131. (sinθ cosφ)2
+ (sinθ sinφ)2
+ cos2
θ
= sin2
θ cos2
φ + sin2
θ sin2
φ + cos2
θ
= sin2
θ(cos2
φ + sin2
φ) + cos2
θ
= sin2
θ + cos2
θ
=1
= −27
Thus the y-intercept is (0, −27) .
132 – 136. Answers will vary. Section 2.4
(−x)4
+ 3 x4
+ 3 2
137. f (−x) = = = f (x) . Thus,
(−x)2
− 5 x2
− 5
f (x) is even.
1. y = 3x
Using the graph of y = x2
, vertically stretch the
138. We need to use completing the square to put the
function in the form
f (x) = a(x − h)2
+ k
f (x) = −2x2
+12x −13
= −2(x2
− 6x) −13
graph by a factor of 3.
= −2 x2
− 6x + 144
4( 2)2
−13+ 2 144
4( 2)2
= −2(x2
− 6x + 9) −13+18
= −2(x2
− 6x + 9) + 5
= −2(x − 3)2
+ 5

54. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions
 
, .
= − ω
2. y = 2x g. The x-intercepts of sin x are
Using the graph of y = x , compress {x | x = kπ, k an integer}
horizontally by a factor of
1
. 12. a. The graph of y = cos x crosses the y-axis at
2 the point (0, 1), so the y-intercept is 1.
b. The graph of
0 < x < π .
y = cos x is decreasing for
c. The smallest value of y = cos x is −1 .
d. cos x = 0 when x =
π
,
3π
2 2
3. 1;
π
e. cos x =1 when x = −2π, 0, 2π;
cos x = −1 when x = −π, π.
3 11π π π 11π
2 f. cos x = when x = − ,− , ,
4. 3; π
2 6 6 6 6
g. The x-intercepts of cos x are
5. 3;
2π
=
π

x | x =
(2k +1)π
2
, k an integer


6 3
13.
 
y = 2sin x
6. True This is in the form y = Asin(ωx) where A = 2
7. False; The period is
2π
= 2 .
and ω =1. Thus, the amplitude is
2 2
A = 2 = 2
π
8. True
14.
and the period is T =
π
=
ω
y = 3cos x
π
= 2π .
1
9. d This is in the form y = Acos(ωx) where A = 3
10. d
and ω =1. Thus, the amplitude is A = 3 = 3
2π 2π
= = = π .11. a. The graph of y = sin x crosses the y-axis at
and the period is T 2
ω 1
the point (0, 0), so the y-intercept is 0.
15. y = −4cos(2x)
b. The graph of y = sin x is increasing for This is in the form y = Acos(ωx) where
−
π
< x <
π
. A = −4 and ω = 2 . Thus, the amplitude is
2 2
A = −4 = 4 and the period is
c. The largest value of y = sin x is 1.
T =
2π
=
2π
= π .
ω 2
d. sin x = 0 when x = 0, π, 2π .
16. y = −sin
 1
x

e. sin x =1 when x = −
3π
,
π
;
 2 
sin x = −1 when x
2 2
π 3π
= −
This is in the form
and
1
y = Asin(ωx) where A = −1
ω = . Thus, the amplitude is
2 2 2
T =
2π
=
2π
= 4π .
A = −1 =1
f. sin x
1
when x = −
5π
,−
π
,
7π
,
11π
2 6 6 6 6
and the period is 1
2

55. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions
   x
cos x

ω 3
ω
17. y = 6sin(π x)
22. y =
9
cos

−
3π
x

=
9
cos
 3π 
This is in the form y = Asin(ωx) where A = 6 5  2  5  2 
and ω = π . Thus, the amplitude is
and the period is T =
2π
=
2π
= 2 .
A = 6 = 6 This is in the form y = Acos(ωx) where A =
9
5
18.
ω π
y = −3cos(3x)
and ω =
3π
. Thus, the amplitude is
2
9 9
This is in the form y = Acos(ωx) where A = −3
A = = and the period is
5 5
and ω = 3. Thus, the amplitude is A = −3 = 3
T =
2π
=
2π
=
4
.
and the period is T =
2π
=
2π
.
ω 3π 3
19.
ω 3
1  3 
y = −  
2
23. F
2  2  24. E
This is in the form y = Acos(ωx) where
25. A1
A = −
2
and ω =
3
. Thus, the amplitude is
2 26. I
A = −
1
=
1
and the period is
2 2
T =
2π
=
2π
=
4π
.
27. H
28. B
3
2
29. C
20. y =
4
sin
 2
x

3

3
 30. G 
This is in the form y = Asin(ωx) where A =
4
3
31. J
and ω =
2
. Thus, the amplitude is A =
4
=
4 32. D
3 3 3
33. Comparing y = 4cos x to y = Acos(ωx) , we
and the period is T =
2π
=
2π
= 3π . find A = 4 and ω =1. Therefore, the amplitude2
3
is 4 = 4 and the period is
2π
= 2π . Because
21. y =
5
sin

−
2π
x

= −
5
sin
 2π
x
 1
3

3

3

3
 the amplitude is 4, the graph of y = 4cos x will
   
lie between −4 and 4 on the y-axis. Because the
This is in the form y = Asin(ωx) where
5
A = −
3
period is 2π , one cycle will begin at x = 0 and
end at x = 2π . We divide the interval [0,2π ]
and ω =
2π
. Thus, the amplitude is
3 into four subintervals, each of length
2π
=
π
by
4 2
A = −
5
=
5
and the period is finding the following values:
3 3
T =
2π
=
2π
= 3.
0,
π
2
, π ,
3π
, and 2π
2
ω 2π
3
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
coordinates of the five key points for
y = 4 cos x , we multiply the y-coordinates of the
five key points for
key points are
y = cos x by A = 4 . The five

56. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions
,0

,3
,0(0,4) ,
 π 
, (π,−4) ,
 3π 
, (2π,4)
direction to obtain the graph shown below.

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.
From the graph we can determine that the
domain is all real numbers, (−∞,∞) and the
range is [−3,3].
From the graph we can determine that the 35. Comparing y = −4sin x to y = Asin (ωx) , we
domain is all real numbers, (−∞,∞) and the
range is [−4,4].
find A = −4 and ω =1. Therefore, the amplitude
2π
is −4 = 4 and the period is = 2π . Because
1
34. Comparing y = 3sin x to y = Asin(ωx) , we the amplitude is 4, the graph of y = −4sin x will
find A = 3 and ω =1. Therefore, the amplitude lie between −4 and 4 on the y-axis. Because the
is 3 = 3 and the period is
2π
= 2π . Because
1
period is 2π , one cycle will begin at x = 0 and
end at x = 2π . We divide the interval [0,2π ]
the amplitude is 3, the graph of y = 3sin x will 2π π
lie between −3 and 3 on the y-axis. Because the
period is 2π , one cycle will begin at x = 0 and
into four subintervals, each of length
π
= by
4 2
3π
end at x = 2π . We divide the interval [0,2π ] finding the following values: 0, , π ,
2
, 2π
2
into four subintervals, each of length
2π
=
π
by
4 2
finding the following values:
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
coordinates of the five key points for
y = −4sin x , we multiply the y-coordinates of
0,
π
, π ,
3π
, and 2π the five key points for y = sin x by A = −4 . The
2 2
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
five key points are
(0,0) ,
 π
,−4

, (π,0),
 3π
,4 , (2π,0)
coordinates of the five key points for y = 3sin x ,

2
 
2

   
we multiply the y-coordinates of the five key We plot these five points and fill in the graph of
points for y = sin x by A = 3. The five key the curve. We then extend the graph in either
points are (0,0) ,
 π
 , (π,0),
 3π
,−3

,
direction to obtain the graph shown below.

2
 
2

(2π,0)
   
We plot these five points and fill in the graph of
the curve. We then extend the graph in either

57. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions

–––
From the graph we can determine that the 37. Comparing y = cos(4x) to y = Acos(ωx) , we
domain is all real numbers, (−∞,∞) and the find A =1 and ω = 4 . Therefore, the amplitude
range is [−4,4].
is 1 =1 and the period is
2π
=
π
. Because the
4 2
36. Comparing y = −3cos x to y = Acos(ωx) , we amplitude is 1, the graph of y = cos(4x) will lie
find A = −3 and ω =1. Therefore, the amplitude between −1 and 1 on the y-axis. Because the
is −3 = 3 and the period is
2π
= 2π . Because
1
period is
π
, one cycle will begin at x = 0 and
2
the amplitude is 3, the graph of y = −3cos x will
end at x =
π
. We divide the interval

0,
π 
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
2  2  
into four subintervals, each of length
π / 2
=
π
4 8
by finding the following values:
0,
π
,
π
,
3π
, and
π
finding the following values:
0,
π
, π ,
3π
, and 2π
8 4 8 2
These values of x determine the x-coordinates of
2 2 the five key points on the graph. The five key
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
points are
 π   π
−

,
 3π
,0 ,
 π
,1

coordinates of the five key points for (0,1) , 
8
,0 , 
4
, 1 
8
 
2

       
y = −3cos x , we multiply the y-coordinates of We plot these five points and fill in the graph of
the five key points for
five key points are


y = cos x by
 3π 

A = −3 . The the curve. We then extend the graph in either
direction to obtain the graph shown below.
(0,−3) ,
 π
,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
From the graph we can determine that the
(0, 3)
5
(2 , 3)
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].

58. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions
,0
38. Comparing y = sin(3x) to y = Asin (ωx) , we y-axis. Because the period isπ , one cycle will
find A =1 and ω = 3. Therefore, the amplitude begin at x = 0 and end at x = π . We divide the
interval [0,π ] into four subintervals, each of
is 1 =1 and the period is
2π
. Because the
3
length
π
by finding the following values:
amplitude is 1, the graph of y = sin(3x) will lie 4
between −1 and 1 on the y-axis. Because the
period is
2π
, one cycle will begin at x = 0 and
0,
π
,
π
,
3π
, and π
4 2 4
3
end at x =
2π
. We divide the interval

0,
2π 
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
coordinates of the five key points for
3  3 
  y = −sin (2x) , we multiply the y-coordinates of
into four subintervals, each of length
2π /3
=
π
4 6
the five key points for
five key points are
y = sin x by A = −1 .The
by finding the following values:  π   π   3π 
0,
π
,
π
,
π
, and
2π (0,0) , 
4
,−1 , 
2
,0 ,  ,1 , (π,0)
4
     
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
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
(0,0) ,
 π
,1

,
 π 
,
 π
,−1

,
 2π 
,0
2
3

6
 
3
    

––– ( , 0)     2   3  ( –––, 1)4
( 4
, 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. 2
(0, 0)
2
––
2 x
(–– , 0)2
( 4
, 1)
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(2x).
From the graph we can determine that the
Comparing y = cos(2x) to y = Acos(ωx) , we
domain is all real numbers, (−∞,∞) and the find A =1 and ω = 2 . Therefore, the amplitude
2π
range is [−1,1] .
39. Since sine is an odd function, we can plot the
is 1 =1 and the period is
amplitude is 1, the graph of
= π . Because the
2
y = cos(2x) will lie
equivalent form y = −sin(2x). between −1 and 1 on the y-axis. Because the
period isπ , one cycle will begin at x = 0 and
Comparing y = −sin(2x) to y = Asin(ωx) , we end at x = π . We divide the interval [0,π ] into
find A = −1 and ω = 2 . Therefore, the π
amplitude is −1 =1 and the period is
2π
= π .
2
four subintervals, each of length
4
the following values:
by finding
Because the amplitude is 1, the graph of π π 3π

59. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions
y = −sin(2x) will lie between −1 and 1 on the 0, , ,
4 2
, and π
4

60. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions
,0 ,0
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
the five key points for
five key points are
y = sin x by A = 2 . The
coordinates of the five key points for
y = cos(2x), we multiply the y-coordinates of
(0,0) , (π,2), (2π,0), (3π,−2), (4π,0)
We plot these five points and fill in the graph of
the five key points for
five key points are
y = cos x by A = 1.The the curve. We then extend the graph in either
direction to obtain the graph shown below.
(0,1) ,
 π 
,
 π
,−1

,
 3π 
, (π,1)
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.
From the graph we can determine that the
domain is all real numbers, (−∞,∞) and the
range is [−2,2].
42. Comparing y = 2cos
 1
x

to y = Acos(ωx) ,
From the graph we can determine that the

4

domain is all real numbers, (−∞,∞) and the
range is [−1,1] .
we find
 
A = 2 and ω =
1
. Therefore, the
4
41. Comparing y = 2sin
 1
x

to y = Asin(ωx) ,
amplitude is 2 = 2 and the period is
2π
= 8π .
1/ 4

2
 Because the amplitude is 2, the graph of 
y = 2cos
 1
x

will lie between −2 and 2 on
we find A = 2 and ω =
1
. Therefore, the 
4

2
amplitude is 2 = 2 and the period is
2π
= 4π .
1/ 2
Because the amplitude is 2, the graph of
 
the y-axis. Because the period is8π , one cycle
will begin at x = 0 and end at x = 8π . We
divide the interval [0,8π ] into four subintervals,
y = 2sin
 1
x

will lie between −2 and 2 on the each of length
8π
= 2π by finding the following
2
 4 
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
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 y-
of length
4π
= π
4
by finding the following
coordinates of the five key points for
y = 2cos
 1
x

, we multiply the y-coordinates
values: 
4

0, π , 2π , 3π , and 4π
 
of the five key points for y = cos x by
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
coordinates of the five key points for
y = 2sin
 1
x

, we multiply the y-coordinates of
2

A = 2 .The five key points are
(0,2) , (2π,0), (4π,−2) , (6π,0), (8π,2)
We plot these five points and fill in the graph of
the curve. We then extend the graph in either
 

61. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions
= −
= −
8
= −
 
= −
 
direction to obtain the graph shown below.
y
direction to obtain the graph shown below.
( 2 , 0)
2
(0, 2) (8 , 2)
( 8 , 2) (2 , 0)
(6 , 0)
x
8 4 4 8
2
( 4 , 2) (4 , 2)
From the graph we can determine that the
domain is all real numbers, (−∞,∞) and the
range is [−2,2].
From the graph we can determine that the
domain is all real numbers, (−∞,∞) and the
 1 1
43. Comparing y
1
cos(2x) to y = Acos(ωx) , range is − , 
we find
= −
2
A
1
and ω = 2 . Therefore, the
.
 2 2
1
2 44. Comparing y = −4sin
 
x

to y = Asin(ωx) ,
8

amplitude is −
1
=
1
and the period is
2π
= π .
2 2 2 we find A = −4 and ω =
1
. Therefore, the
8
Because the amplitude is
1
, the graph of
2 amplitude is −4 = 4 and the period is
y
1
cos(2x) will lie between −
1
and
1
on
2π
=16π . Because the amplitude is 4, the
2 2 2 1/8
the y-axis. Because the period isπ , one cycle  1 
will begin at x = 0 and end at x = π . We divide graph of y = −4sin x will lie between −4
 
the interval [0,π ] into four subintervals, each of
length
π
by finding the following values:
4
0,
π
,
π
,
3π
, and π
and 4 on the y-axis. Because the period is16π ,
one cycle will begin at x = 0 and end at
x =16π . We divide the interval [0,16π ] into
four subintervals, each of length
16π
= 4π by
4 2 4 4
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
coordinates of the five key points for
y
1
cos(2x), we multiply the y-coordinates
2
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 y-
coordinates of the five key points for
of the five key points for y = cos x by
y = −4sin
 1
x

, we multiply the y-coordinates
A
1
.The five key points are
 8 
2
  π 1 

 3π   1 
 of the five key points for
The five key points are
y = sin x by A = −4 .

0,−
1 
,
 π
,0 , , , ,0 , π,−
2
 
4
 
2 2
 
4
 
2

          (0,0) , (4π,−4) , (8π,0) , (12π,4) , (16π,0)
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

62. Section 2.4: Graphs of the Sine and Cosine FunctionsChapter 2: Trigonometric Functions
,5
direction to obtain the graph shown below.
y
5 (12 , 4)
direction to obtain the graph shown below.
16 8
(0, 0)
x
(16 , 0)
(8 , 0)
5 (4 , 4)
From the graph we can determine that the
domain is all real numbers, (−∞,∞) and the
range is [−4,4]. From the graph we can determine that the
domain is all real numbers, (−∞,∞) and the
45. We begin by considering y = 2sin x . Comparing range is [1,5].
y = 2sin x to y = Asin(ωx) , 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 = Acos(ωx) , we find A = 3
and the period is
2π
= 2π . Because the
1 and ω =1. Therefore, the amplitude is 3 = 3
amplitude is 2, the graph of y = 2sin x will lie
and the period is
2π
= 2π . Because the
between −2 and 2 on the y-axis. Because the
period is 2π , one cycle will begin at x = 0 and
1
amplitude is 3, the graph of y = 3cos x will lie
end at x = 2π . We divide the interval [0,2π ]
into four subintervals, each of length
2π
=
π
by
4 2
finding the following values:
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 ]
2π π
0,
π
, π ,
3π
, and 2π
into four subintervals, each of length = by
4 2
2 2 finding the following values:
These values of x determine the x-coordinates of π
, π ,
3π
, and 2πthe five key points on the graph. To obtain the y- 0,
2 2
coordinates of the five key points for
y = 2sin x + 3 , we multiply the y-coordinates of
These values of x determine the x-coordinates of
the five key points on the graph. To obtain the y-
the five key points for y = sin x by A = 2 and coordinates of the five key points for
then add 3 units. Thus, the graph of y = 3cos x + 2 , we multiply the y-coordinates of
y = 2sin x + 3 will lie between 1 and 5 on the y- the five key points for y = cos x by A = 3 and
axis. The five key points are then add 2 units. Thus, the graph of
(0,3),
 π
 , (π,3) ,
 3π
,1

, (2π,3)
y = 3cos x + 2 will lie between −1 and 5 on the

2
 
2

y-axis. The five key points are   
  3π We plot these five points and fill in the graph of (0,5),
 π
,2 , (π,−1), ,2 , (2π,5)
the curve. We then extend the graph in either

2
 
2

   
We plot these five points and fill in the graph of
the curve. We then extend the graph in either