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TRANSCRIPT
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12.
13.
21. 22.
1 1 1 º14.
_
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 Measure Chapter 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 º26. 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. 9º 9 ' 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 = 19º 59 ' 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 "
= 44º 0 ' 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 Measure Chapter 2: Trigonometric Functions
180 2
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Section 2.1: Angles and Their Measure Chapter 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
radian6 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 Measure Chapter 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
r 2θ
2
2 = 1
r 2 1
A = 2 ft 2
72.
73.
r = 6 feet; θ = 2 radian;
θ = 1
radian; s = 2 feet; 3
s = rθ = 6 ⋅ 2 = 12 feet
2 = 1
r 2
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
r 2θ
6 = 1
r 2 1
r = s
= θ
6
(1/ 4) = 24 cm
6 = 1
r 2
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
θ77. 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
r 2θ
2
A = 8 m2
79.
s = rθ = 3 ⋅ 2π
= 2π ≈ 6.283 meters 3
r = 10 meters; θ = 1
radian 2
A = 1
r 2θ =
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
r 2θ =
1 2
2 π =
π ≈ 1.047 in
2
A = 1
r 2θ = 1
(6)2
(2) =36 ft 2
2 2
( ) 2 2 3
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Section 2.1: Angles and Their Measure Chapter 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
r 2θ =
1 (3)
2 2π =3π ≈ 9.425 m
2
s = rθ = 40 ⋅ π
= 40π
≈ 13.96 inches2 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
r 2θ =
1 (4)2 π
= 2π ≈ 6.28 m2
3 3
A = 1
r 2θ =
1 (2)2 π
= = 2π
≈ 2.094 ft 2
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
r 2θ =
1 (3)2 π
= 3π
≈ 4.71 cm2
6
s = rθ = 4 ⋅ π
= 2π
≈ 2.094 meters
2 2 2
3 6 3 95. r = 30 feet; θ = 135º = 135 ⋅ π
= π radians
A = 1
r 2θ =
1 (4)
2 π =
4π ≈ 4.189 m
2 180 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
r 2θ
2
A = 100 yd
A = 1
r 2θ =
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 radian180 18
s = rθ = 9 ⋅ 5π
≈ 7.854 cm
112.5 9 °
or 8
⋅ 180
= 160
≈ 50.93°18 9 π
π
A = 1
r 2θ =
1 (9)2 5π
= 45π
≈ 35.343 cm2
2 2 4
97. A = 1
r 2θ − 1
r 2θ 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 Measure Chapter 2: Trigonometric Functions
98. A = 1
r 2θ −
1 r
2θ 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 ⋅ 60 min
2 36 2 36 min 100cm 1000m 1hr
≈ 101.8 km/hr1
= (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 = 35 mi
⋅ 5280 ft
⋅ 12 in.
⋅ 1 hr
36 36 hr mi ft 60 min
= 36, 960 in./min=
11250π −
450π
36 36 ω = v
= 36, 960 in./min
r 13 in.10800π 2
= = 300π ≈ 942.48 in 36
≈ 2843.08 radians/min
≈ 2843.08 rad
⋅ 1 rev
99. r = 5 cm; t = 20 seconds; θ = 1
radian 3
min 2π rad ≈ 452.5 rev/min
θ (1/ 3) 1 1 1ω = = = ⋅ = 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 3600 sect 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 milesθ ( s / r ) ( 5 / 2) 5 1 1
ω = = = = ⋅ = radian/sec θ = 35º 9 '− 29º 57 't t 20 2 20 8 = 5º12 '
v = s
= 5 1
= m/sec = 5.2ºt 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 60 min
⋅ ⋅ ≈ 23.2 mi/hrmin 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 60 min
⋅ ⋅ ≈ 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
⋅ 60 min
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 Measure Chapter 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 1 mi 60 min = ⋅ ⋅
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
80 mi/hr 12 in 5280 ft 1 hr 1 rev = ⋅ ⋅ ⋅ ⋅
13 in 1 ft 1 mi 60 min 2π rad113. r
1 = 2 inches; r
2 = 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
= 9.55 mi/hr
r1ω
1 = r
2ω
2 r 4.25 ft
9.55 mi 1 5280 ft 1 hr 1 rev2(6π) = 8ω
2
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 = 1 rev 2π π
= = ≈ 0.09 radian/sec70 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 Measure Chapter 2: Trigonometric Functions
119. A = π r 2
= 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 = π r 2
= 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 = 2π r
8π = 2π r
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 = π r 2 = 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 u
2 .
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
R2θ −
1 r 2θ =
θ (R2 − r
2 )
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 (200
2 −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'⋅ 1 degree
⋅ 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.
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Section 2.1: Angles and Their Measure Chapter 2: Trigonometric Functions
125. r1
rotates at ω1
rev/min , so v1
= r1ω
1 .
r2
rotates at ω2
rev/min , so v2
= r2ω
2 .
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 .
r1ω
1 = r
2ω
2
r1ω
1 = r2ω
2
Reflecting about the x-axis would give
y = − x + 3 . Shifting down 4 units would result
r2ω
1 r2ω
1 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. x
1 + x
2 ,
y1 + y
2
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
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Section 2.1: Angles and Their Measure Chapter 2: Trigonometric Functions
12. False
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
= −
5
2
cot t ⋅
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
sin t = y = 1
2
cos t = x = 3
2
sin t = y = 21
5
cos t = x = − 2
5
1 21
tan t = y
= 2 =
1 ⋅
2 = 1
⋅ 3
= 3 tan t =
y = 5
= 21 5 21 − = −
x 3 2 3 3 3 3 x −
2 5 2 2
2
5
csc t = 1
= 1
= 1⋅ 2
= 2
csc t = 1
= 1
= 1⋅ 5
= 5
⋅ 21
= 5 21
y 1 1 y 21 21 21 21 21
2
5
sec t = 1
= 1
= 1
− 5
= − 5
sec t = 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
cot t = = = − ⋅ y
x 2 3 2 21 5 21cot t = = = ⋅ = 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
sin t = y = − 3
1 2 6
1 2 6
2 16. P = − , x = − , y =
cos t = x = 1
5 5 5 5
2
3 −
sin t = y = 2 6
5
tan t = y
= 2
= − 3
⋅ 2
= − 3 cos t = x = − 1
x 1 2 1
5
2 6
csc t =
1 =
1 = −
2 = −
2 ⋅
3 = −
2 3 tan t = y
= 5
= 2 6
− 5 = − 2 6
y 3
3 3 3 3 x
− 1 5 1
−
5
2
sec t = 1
= 1
= 1⋅ 2
= 2 csc t =
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
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
y 2
−
2 2
3
−
2
2
−
sec t = 1
= 1
= 1
− 5
= −5
csc t = 1
= 1
= 1⋅ 2
= 2
⋅ 2
= 2x 1
1
− y 2 2 2 2
5 2
− 1
sec t = 1
= 1
= 1⋅ 2
= 2
⋅ 2
= 2
cot t = 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 cot t = = = 1 y 2
17. 2 2 2 2
P = − , x = − , y = 2
2 2 2 2
sin t = 2
19.
2 2 1 2 2 1
P = , − x = , y = −
3 3 3 3
cos t = x = − 2
2
2
tan t = =
= −1
sin t = y = − 1
3
cos t = x = 2 2
3
x 2 − 1
2 tan t =
y =
3 = − 1
⋅ 3
csc t = 1
= 1
= 1⋅ 2
= 2
⋅ 2
= 2 x 2 2 3 2 2
y 2
2 2 2
3
2
= −
1 ⋅
2 = −
2
sec t = 1
= 1
= 1
− 2
= − 2
⋅ 2
= − 2 2 2 2 4
x 2
2
2 2
1 1 3 − csc t = = = 1 −
= −3 2 y 1
1
2 −
3
cot t = x
= 2
= −1
sec t = 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
cot t = = y
= − = −2 2 1 3 1 2 2 2 2
3
sin t = y = 2
2
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
5 2 5 2
cos t = x = 2
20.
P = − , −
3 3 x = − , y
3 32
2
y
= −
sin t = y = − 2
3
tan t = = = 1 5x 2 cos t = x = − 2
3
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
3
2 2
2
−
2
25.
csc 11π
= csc 3π
+ 8π
2
2 2
tan t = 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
csc t = 1
= 1
= 1
− 3
= − 3
= csc 3π
y 2
2
2
−
= −1
sec t = 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
cot t = = = − − =
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 + cos 60º = 2
+ 1
= 1 + 2
2 2 2
sin 30º − cos 45º = 1
− 2
= 1 − 2
2 2 2
= cot π
+ 3π
= cot π
= 0
33. sin 90º + tan 45º = 1 +1 = 2
2
2 34. cos180º − sin 180º = −1 − 0 = −1
35.
sin 45º cos 45º = 2
⋅ 2
= 2
= 1
2 2 4 2
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
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2
3
tan
2
2
−
3
= −
= −
36.
tan 45º cos 30º = 1⋅ 3
= 3
cos 2π
= 1
= 1
− 2
= −22 2 3
1
1
37. csc 45º tan 60º =
2 ⋅ 3 = 6
−
1
cot 2π
= 2 = − 1
⋅ 2 = −
1 ⋅ 3
= − 3
38.
sec 30º cot 45º = 2 3
⋅1 = 2 3 3 3 2 3 3 3 3
3 3 2
39. 4 sin 90º − 3 tan 180º = 4 ⋅1 − 3 ⋅ 0 = 4
48. The point on the unit circle that corresponds to
θ = 5π
= 150º is
− 3
, 1
.40. 5 cos 90º − 8 sin 270º = 5 ⋅ 0 − 8(−1) = 8 6 2 2
41. 2sin
π − 3 tan
π = 2 ⋅
3 − 3 ⋅
3 =
3 6 2 3
3 − 3 = 0
sin 5π
= 1
6 2
cos 5π
= −
42. 2sin π
+ 3 tan π
= 2 ⋅ 2
+ 3 ⋅1 = 4 4 2
2 + 3 6 2
1
5π 2 1 2
= = ⋅ −
3 3
⋅ = −
43. 2sec π
+ 4 cot 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. sec π − 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
cos 210º = − = 3
π 2
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 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
csc 210º =
1 = 1⋅ −
−
= − 2
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
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2
= −
2
sec 210º = 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
3 sin 240º = − 2
2 2
52. The point on the unit circle that corresponds to
θ = 11π
= 495º is
− 2
, 2
.
cos 240º 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
csc 240º = 1
= 1⋅
− 2
⋅ 3
= − 2 3
tan 11π
= 2
= 2
⋅
− 2
= −1 3
3
3 3 −
4 2 2 2
−
2 2
sec 240º = 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
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
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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 2tan
3 = 2 =
2 ⋅ −
2 = −1
4 2 2 2 cos 8π 1
− 3 2
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
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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 cos 405º =
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 csc 405º =
1 = 1⋅
2 ⋅
2 = 2
3 3 2 3 3 3 2 2 2
2
2
sec 405º = 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
sin 390º = 2
csc 13π
= 1
= 1⋅ 2
= 2 cos 390º =
3
6 1 1
2
1
tan 390º =
= 1
⋅ 2
⋅ 3
= 3
sec 13π
= 1
= 1⋅ 2
⋅ 3
= 2 3
2 3 3 36 3 3 3 3
3
2 2
3
13π 2
3 2
csc 390º = 1
= 1⋅ 2
= 2 1 1 2
cot = = ⋅ = 36 1 2 1
1 2 3 2 3 2 sec 390º = = 1⋅ ⋅ =
3
3 3 3
2
3
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
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cot 390º = 2 = 3
⋅ 2
= 3
1 2 1
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
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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
⋅ =
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
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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
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 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
sin 5π = 0
csc 5π = 1
= undefined
−
0 13π 1 2 3 2 3
cos 5π = −1
sec 5π = 1
= −1 sec −
6 =
3 = ⋅ =
3 3 3
−1
2
tan 5π = 0
= 0 −1
cot 5π = −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:
sec 21º = 1
≈ 1.07 . −
1
cot −
= − 1
⋅
− 2
3 3
⋅ =
cos 21°
3
3
2
3
3 3 −
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 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: sin 1º ≈ 0.02 .
76. Set the calculator to degree mode: tan 1º ≈ 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 + y 2 = 9 +16 = 25 = 5
sin θ = 4 cscθ =
5
5 4
71. Set the calculator to radian mode:
cosθ 3
secθ 5
cot π
= 1
≈ 3.73 .
= − = − 5 3
4 312
tan π
tan θ
3 cot θ
4
= − = −
12
78. For the point (5, −12) , x = 5 , y = −12 ,
r =
sin θ
x2 + y
2 =
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 tan θ 12 cot θ = −
5
5 12
79. For the point (2, −3) , x = 2 , y = −3 ,
r = x2 + y 2 = 4 + 9 = 13
73. Set the calculator to radian mode: sin 1 ≈ 0.84 . sin θ = −3
⋅ 13 3 13
cscθ = − = 13
13 13 13 3
cosθ = 2
⋅ 13
= 2 13 secθ =
13
tan θ
13 13
3 = −
13 2
cot θ = − 2
2 3
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 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 + y 2 = 1 + 4 = 5 r = x2 + y 2 = 0.09 + 0.16 = 0.25 = 0.5
sin θ = − 2
⋅ 5
= − 2 5
cscθ = 5
= − 5 sin θ =
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 35 5 5 −1
0.4 4 0.3 3
tan θ = − 2
= 2 cot θ = −1
= 1
tan θ = = 0.3 3
cot θ = = 0.4 4
−1
81. For the point (−2, −2) , x = − 2 ,
− 2 2
y = − 2 ,
85. sin 45º + sin 135º + sin 225º + sin 315º
2 2 2 2
r = x2 + y 2 = 4 + 4 = 8 = 2 2 = 2
+ 2
+
− 2
+
− 2
− 2 2 2 2 2
= 0sin θ = ⋅ = − 2 2 2 2
csc θ = = − 2 − 2
− 2 2 2 2 2
86.
tan 60º + tan150º = 3
3 + −cos θ = ⋅ = − sec θ = = − 2
2 2 2 2 − 2 3
− 2 − 2 3 3 − 3 2 3tan θ = = 1 cot θ = = 1 = =
− 2 − 2 3 3
82. For the point (−1, 1) , x = −1 , y = 1 , 87. sin 40º + sin 130º + sin 220º + sin 310º
r = x2 + y 2 = 1 +1 = 2 = sin 40º + sin 130º + sin (40º +180º ) +
sin (130º +180º )sin θ =
1 ⋅
2 =
2 cscθ =
2 = 2
2 2 2 1 = sin 40º + sin 130º − sin 40º − sin 130º
= 0cosθ =
−1 ⋅
2 = −
2 secθ = 2
= − 22 2 2 −1 88. tan 40º + tan 140º = tan 40º + tan (180º −40º )
tan θ = 1 = −1 cot θ =
−1 = −1 = tan 40º − tan 40º
−1 1 = 0
83. For the point 1
, 1
, x = 1
, y =
1 ,
89. If
f (θ ) = sin θ = 0.1 , then
f (θ + π ) = sin(θ + π) = − 0.1 .
r = x2 + y 2 =
1 +
1 = 25 5
=9 16 144 12
1 5 90. If f (θ ) = cosθ = 0.3 , then
sin θ = 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 ) = tan θ = 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 ) = cot θ = − 2 , then
1
tan θ = 4 = 1
⋅ 3
= 3
1 4 1 4
3
1
cot θ = 3 = 1
⋅ 4
= 4
1 3 1 3
4
f (θ + π ) = cot(θ + π) = −2 .
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
5
3
4 4
2
93. If sin θ = 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
= cos 30° = 3
2
cos 315°
96.
97.
g (60º ) = cos (60º ) = 1
2
f 60º
= sin 60º
= sin (30º ) = 1
109. p (g (315°)) =
=
2
1 cos 315°
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 3 399. 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 ) = (cos 60º ) = =
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 ) = 2 sin (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. 2 g (60º ) = 2 cos (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
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 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 sin θ sin θ 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. 2 g π
− π
= 2g (0)
= 2 cos(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 sin θ 2
114. Answers will vary. One ser of possible answers
Use the formula H (θ ) = 0 ( ) 2 g
with
115.
is − 13π
, − 5π
, 3π
, 11π
, 19π
4 4 4 4 4 g = 32.2ft/sec
2 ; 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 2v ( sin θ )
Use the formula H (θ ) = 0 2 g
with
g = 9.8 m/sec2
; θ = 30º ; v = 150 m/sec :
f (θ ) = sin θ
θ approaches 1 as θ approaches 0.
1502 (sin 30º )
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 :
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
2
R (25º ) = 500 sin(2 ⋅ 25º )
≈ 19, 541.95 m 9.8
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
0
2
0
0
0
o
2 2v ( sin θ )
Use the formula H (θ ) = 0 2 g
with 123. Note: time on road = distance on road
rate on road
g = 9.8 m/sec2
; θ = 25º ; v = 500 m/sec : = 8 − 2 x
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 tan θ
32.2
2 2 a. T (30º ) = 1 +
2 −
1
3 sin 30º 4 tan 30ºv ( sin θ )
Use the formula H (θ ) = 0
with
2 1 2 g
g = 32.2ft/sec2
; θ = 50º ; v = 200 ft/sec :
= 1 + 1
− 1
3 ⋅ 4 ⋅ 2 3
2002 (sin 50º )
2
H (50º ) = ≈ 364.49 ft 2(32.2)
= 1 + 4
− 3
≈ 1.9 hr 3 4
Sally is on the paved road for121. Use the formula t (θ ) =
2a
g sin θ cosθ
with
1 − 1
≈ 0.57 hr . 4 tan 30º
g = 32 ft/sec2
and a = 10 feet : 2 1
a. t (30) = 2(10)
≈ 1.20 seconds 32 sin 30º ⋅cos 30º
b. T (45º ) = 1 + − 3sin 45º 4 tan 45º
2 1
b. t (45) = 2(10)
≈ 1.12 seconds 32 sin 45º ⋅cos 45º
= 1 +
3 ⋅ 1
2
− 4 ⋅1
c. t (60) = 2(10)
≈ 1.20 seconds 32 sin 60º ⋅cos 60º
= 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 + 0.5 cos(2θ ) . 4 tan 45
T (60º ) = 1 + 2
− 1
x (30) = cos (30º ) + 16 + 0.5 cos(2 ⋅ 30º ) c. 3sin 60
o
4 tan 60o
= cos (30º ) + 16 + 0.5 cos (60º ) = 1 + 2
− 1
≈ 4.90 cm
x (45) = cos (45º ) +
= cos (45º ) +
≈ 4.71 cm
16 + 0.5 cos(2 ⋅ 45º )
16 + 0.5 cos (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 60
o
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
3
3
d. T (90º ) = 1 + 2
− 1
. 3 sin 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
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 tan 10°
H ≈ 71
The tree is approximately 71 feet tall.
walking 1 mile in the sand straight to the θ Hroad. 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 + sec 30º ) ≈ 251.42 cm
3
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 + sec 45º )
≈ 117.88 cm3
129. cos θ + sin 2
θ = 41
; cos 49
2 θ + sin 2
θ = 1
3 ( tan 45º )2
Substitute x = cos θ; y = sin θ and solve these
When θ = 60º : 3
simultaneous equations for y.
x + y 2 =
41 ; x
2 + y2 = 1V (60º ) =
1 π (2)
3 (1 + sec 60º )
≈ 75.40 cm3
493
125. tan θ
= H
( tan 60º )2
y 2 = 1 − x2
x + (1 − x2 ) =
41
492 2D
x2 − x −
8 = 0
tan 22°
= 6
492 2D
2D tan 11° = 6
D = 3
= 15.4
Using the quadratic formula:
a = 1, b = −1, c = − 8
49
tan 11° −(−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 1 49 49 7
126. tan θ
= H
2 2D
= = = 2 2
= or − 2 7 7
1tan
8° =
555
2 2D
Since the point is in quadrant III then x = − 7
2
2D tan 4° = 555
and y 2 = 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
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 2: Trigonometric Functions
[ ]
2
130. cos2 θ − sin θ = −
1 ; cos
2 θ + sin 2 θ = 1
9
c. Using the MAXIMUM feature, we find: 20
Substitute x = cos θ; y = sin θ and solve these
simultaneous equations for y.
x2 − y = −
1 ; x
2 + y 2 = 1
9
x2 = 1 − y 2
45° 0
90°
(1 − y 2 ) − y = −
1
9
y 2 + y −
10 = 0
9
9 y 2 + 9 y −10 = 0
Using the quadratic formula:
132.
R is largest when θ = 67.5º .
Slope of M = sin θ − 0
= sin θ
= tan θ . 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 = tan θ .
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
sin 1 ≈ 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 tan 1 ≈ = 1.6
0.5 cot 1 ≈ ≈ 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 unit131. 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 sin 5.1 ≈ −0.9 csc 5.1 ≈ ≈ −1.1
−0.9
3 1 cos 5.1 ≈ 0.4 sec 5.1 ≈ 1 = 2.5
≈ 32 2 − − − 1 0.4
2
≈ 16 .56 ft
2
tan 5.1 ≈ −0.9
0.4
≈ −2.3
cot 5.1 ≈ 0.4
≈ −0.4 −0.9
b. Let Y1
=
20
322
2
32 sin (2 x ) − cos (2x ) −1
Set the calculator on RADIAN mode:
45° 90° 0
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Section 2.2: Trigonometric Functions: Unit Circle Approach Chapter 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 csc 2 ≈ 1
0.9
≈ 1.1
cos 2 ≈ −0.4
tan 2 ≈ 0.9
−0.4
= −2.3
sec 2 ≈ 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
csc 4 ≈ 1
≈ −1.3 −0.8
sec 4 ≈ 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 − 7 x = 2
xy = 7 x + 2
y = 7 x + 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 + 6 x + 9 + 2 x + 2
= x2 + 6 x + 11
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Section 2.3: Properties of the Trigonometric Functions Chapter 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. cos 420º = cos(360º + 60º ) = cos 60º = 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.
sin 390º = sin(360º + 30º ) = sin 30º = 1
2
6 6 6
= sec π 6
2 315. csc 450º = csc(360º + 90º ) = csc 90º = 1 =
3
16. sec 540º = sec(360º + 180º ) = sec180º = −1
17. cot 390º = cot(180º + 180º + 30º ) = cot 30º = 3
18. sec 420º = sec(360º + 60º ) = sec 60º = 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.
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
5
= −
5
2
5
29. Since sin θ < 0 for points in quadrants III and secθ =
1 =
1 = −
5IV, and tan θ < 0 for points in quadrants II and cosθ 3 3IV, the angle θ lies in quadrant IV.
30. Since cosθ > 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.
−
cot θ =
1 = −
3
tan θ 4
2 5 531. 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. sin θ = , cosθ = 5 5
2 5
θ = sin θ
= 5 =
2 5 ⋅
5 =
32. Since cosθ < 0 for points in quadrants II and III, tan
cosθ 2
5 5 5and tan θ > 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 sin θ > 0 for points in quadrants I and II, the
sin θ 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
cot θ = 1
= 1
35. sin θ 3
, cosθ = 4
5 5
tan θ 2
−
3
5 2 5
sin θ 5
3 5 3 tan θ = = = − ⋅ = −
38. sin θ = − , cosθ = − 5 5
cosθ 4
5 4 4
sin θ 5 − 5
5
5 1
cscθ = 1
= 1
= − 5 tan θ = =
cosθ
= −
2 5 ⋅ − =
5 2 5 2
sin θ −
3 3 − 5
5
cscθ = 1
= 1
= 1⋅
− 5
5
⋅ = − 5
secθ = 1
= 1
= 5
sin θ
cosθ 4 4 5
−
5 5
5
5
cot θ = 1
= − 4
secθ = 1
= 1
= −
5 ⋅
5 = −
5
tan θ 3
cosθ
2 5
2 5
5 2 −
36. sin θ = 4
, cosθ = − 3
5 5
4
cot θ =
1 =
1 = 1⋅
2 = 2 tan
θ 1 1
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
5
5
tan θ = sin θ
= 5 4 5 4
= ⋅ − = −
cosθ
3
5
3
3 −
cscθ = 1
= 1
= 5
sin θ 4 4
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
tan
2
−
2
2
2
1 −
−
= −
tan
= −
2
39. sin θ = 1
, cosθ = 3 secθ =
1 =
1 =
3 ⋅
2 =
3 2
2 2 cosθ 2 2 2 2 2 4
1 3
sin θ 2
θ = =
= 1
⋅ 2
⋅ 3
= 3
1 1 4 2
cosθ 3 2 3 3 3 cot θ = = = − ⋅ = − 2 2
2 tan θ 2 2 2
−
4
cscθ = 1
= 1
= 1⋅ 2
= 2sin θ 1 1
42. sin θ =
2 2 , cosθ = −
1
3 3
secθ = 1
= 1
= 2 ⋅
3 =
2 3 2 2 cosθ 3 3 3 3
sin θ 3
2 2 3 2
tan θ = = = ⋅ − = − 2 2 cosθ 1 3 1
cot θ =
1 =
1 =
3 ⋅ 3
= 3 3
tan θ 3 3 3 1 1 3 2 3 2 3
cscθ = = = 1⋅ ⋅ =
sin θ 2 2 2 2 2 4
3
40.
sin θ = 3
, cosθ = 1
1 1 3
2 2
3
secθ = = cosθ
−
= 1⋅ − = −3 1 1
3
sin θ
3 2tan θ = = = ⋅ = 3
cot θ = 1
= 1
= − 1
⋅ 2
= − 2
cosθ 1 2 1
tan θ − 2 2 2 2 2 4
cscθ = 1
= 1
= 1⋅ 2
⋅ 3
= 2 3
43.
sin θ = 12
, θ in quadrant II
sin θ 3 3 3 3 13 2
secθ = 1
= 1
= 1⋅ 2
= 2
Solve for cosθ : sin 2 θ + cos2 θ = 1
cos2 θ = 1 − sin 2 θ
cosθ 1 1
Since θ cosθ = ± 1 − sin 2 θ
is in quadrant II, cosθ < 0 .
cot θ = 1
= 1 1 3 3
= ⋅ = cosθ = − 1 − sin 2 θ
tan θ 3 3 3 3
12 = −
1 − 144
= −
25 = −
5
41. sin θ 1 , cosθ =
2 2
3 3
1
13
12
sin θ 13
θ = =
169 169 13
12 13 12 = ⋅ − = −
tan θ = sin θ
= 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 sin θ 12 12
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
3
sin θ
1
1 13
−
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
4
1 −
5
5 −
,
12
sin θ 5 3 5 3
1 −
2
= −
2
2
secθ = 1
= 1
= − 13 cscθ =
1 =
1 = −
5
cosθ −
5 5 sin θ −
3 3
13
5
cot θ = 1
= 1
= − 5
secθ = 1
= 1
= − 5
tan θ −
12 12 cosθ −
4 4
5
5
cot θ = 1
= 1
= 4
44. cosθ = 3 ,
5 θ in quadrant IV
tan θ 3 3
Solve for sin θ : sin 2 θ + cos2 θ = 1
sin 2 θ = 1 − cos
2 θ 46.
sin θ = − 5
, θ in quadrant III
Since θ
sin θ = ± 1 − cos2 θ
is in quadrant IV, sin θ < 0 . 2
13
Solve for cosθ :
sin 2 θ + cos
2 θ = 1
cos2 θ = 1 − sin 2 θ
sin θ = − 1 − cos2 θ = −
3 5
16 4 = − = −
25 5
Since θ cosθ = ± 1 − sin 2 θ
is in quadrant III, cosθ < 0 .
cosθ = − 1 − sin 2 θ = −
1 −
− 5
−
4 13
sin θ 5
4 5 4
tan θ = = = − ⋅ = −cosθ 3
5 3 3 144 12 = − = −
169 13
−
5 cscθ =
1 =
1 = −
5 sin θ
5 13 5 13 sin θ −
4 4 tan θ = = = − ⋅ − = 5 cosθ 12 13
12 12 −
13
secθ = 1
= 1
= 5
1 1 13cosθ 3 3 cscθ = = = −
sin θ
5 5 13
cot θ = 1
= 1
= − 3
1 1 13tan θ
− 4 4 secθ = =
cosθ
= − 12 12
3 −
13
45. cosθ 4 θ in quadrant III
cot θ = 1
= 1
= 12
5
Solve for sin θ :
sin 2 θ + cos
2 θ = 1
sin 2 θ = 1 − cos
2 θ
tan θ 5
5 5
sin θ = ± 1 − cos2 θ 47. sin θ = , 90º < θ < 180º , θ
13 in quadrant II
Since θ is in quadrant III, sin θ < 0 . Solve for cosθ : sin 2 θ + cos2 θ = 1 2 2
sin θ = − 1 − cos2 θ cos θ = 1 − sin θ
= − 1 −
− 4
= − 1 − 16
= −
9 = −
3 cosθ = ± 1 − sin 2 θ
5
25 25 5 Since θ is in quadrant II, cosθ < 0 .
−
3
tan θ = = = − ⋅ − =
cosθ = − 1 − sin 2 θ = −
5 13
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
cosθ
4
5
4
4 −
= − 1 −
25 = −
144 12 = −
5 169 169 13
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
3
3
,
−
5
5
− ,
−
= −
5
sin θ 13
tan θ = =
5 13 5
= ⋅ − = −
sin θ = 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
sin θ 5 5 tan θ =
sin θ = 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 = = = ⋅ =
sin θ 2 2
2 2 2 4
cot θ = 1
= 1
= − 12 3
tan θ −
5 5 1 1
12 secθ = = = −3
48.
cosθ = 4
, 270º < θ < 360º ;
θ in quadrant IV
cosθ 1
5 cot θ = 1
= 1
⋅ 2
= − 2
Solve for sin θ : sin 2 θ + cos2 θ = 1
sin θ = ± 1 − cos2 θ
50.
sin θ
tan θ
2 = −
− 2 2 2
π < θ < 3π
,
4
θ in quadrant III
Since θ is in quadrant IV, sin θ < 0 .
2
3 2
Solve for cosθ : sin 2 θ + cos
2 θ = 1
sin θ = − 1 − cos2 θ = − 1 −
4 9 3
= − = − 2
3
25 5 Since θ
cosθ = ± 1 − sin θ is in quadrant III, cosθ < 0 .
2
tan θ = sin θ
= 5 = −
3 ⋅
5 = −
3 cosθ = − 1 − sin 2 θ = −
1 −
−
2
cosθ
4
5 4 4
3
cscθ = 1
= 1
= − 5
= − 5
= − 5
9 3
sin θ −
3 3 −
2
5
tan θ = sin θ
= 3
secθ = 1
= 1
= 5 cosθ 5
−cosθ
4 4
3
5
2 3 5 2 5
= − ⋅ −
⋅ =cot θ =
1 =
1 = −
4 3 5 5 5tan θ
− 3 3
1 1 3 4
cscθ =
= = − sin θ 2 2
49. cosθ 1 π
< θ < π , θ in quadrant II 3
3 2 1 1 3 5 3 5Solve for sin θ : sin
2 θ + cos2 θ = 1 secθ =
cosθ
= = − ⋅ = − 5
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
sin 2 θ = 1 − cos
2 θ 5 5 5
−
3
sin θ = ± 1 − cos2 θ
1 1 5 5 5Since θ is in quadrant II, sin θ > 0 . cot θ = = = ⋅ =
tan θ 2 5 2 5 5 2
5
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
= −
3
2
3
2
= −
15
2
3
4
2
51. sin θ = 2
, tan θ < 0, so θ is in quadrant II secθ = 1
= 1 4
= − 43 cosθ
− 1 1
Solve for cosθ : sin 2 θ + cos
2 θ = 1
cos2 θ = 1 − sin 2 θ
4
cot θ = 1
= 1
⋅ 15
= 15
cosθ = ± 1 − sin 2 θ tan θ 15 15 15
Since θ is in quadrant II, cosθ < 0 . 53. secθ = 2, sin θ < 0, so θ is in quadrant IV
cosθ = − 1 − sin 2 θ Solve for cosθ : cosθ =
1 =
1
= − 1 − 2
= −
1 − 4
= − 5
= − 5 secθ 2
9 9 3 Solve for sin θ : sin 2 θ + cos2 θ = 1
2 2
2
sin θ = 1 − cos θ
tan θ = sin θ
= 3
= 2
⋅
− 3 2 5
= − sin θ = ± 1 − cos2 θ
cosθ
5 3
5
5
− 3
Since θ is in quadrant IV, sin θ < 0 .
2
cscθ = 1
= 1
= 3
sin θ = − 1 − cos θ
1
1 3 3
sin θ 2 2
= − 1 − 2
= − 1 − = − = − 4 4 2
3
secθ = 1
= 1
= − =3 3 5
sin θ
− 2
3 2
= − = =
= − ⋅ = −cosθ
5 5 5 −
tan θ
cosθ 3
1 2 1
3
cot θ = 1
= 1
= − 5
= − 5
cscθ = 1
= 1
= − 2
= − 2 3
tan θ − 2 5 2 5 2 sin θ
3 3 3
5
− 2
cot θ = 1
= 1
= − 3
52. cosθ 1
, tan θ > 0 4
tan θ − 3 3
Since tan θ = sin θ
cosθ > 0 and cosθ < 0 , sin θ < 0 .
54. cscθ = 3, cot θ < 0, so θ is in quadrant IISolve for sin θ : sin 2 θ + cos2 θ = 1
sin θ = ± 1 − cos2 θ
sin θ = − 1 − cos2 θ
2
Solve for sin θ : sin θ =
1 =
1
cscθ 3
Solve for cosθ : sin 2 θ + cos
2 θ = 1
2
= − 1 −
− 1
= − 1 − 1 cosθ = ± 1 − sin θ
4
16
Since θ is in quadrant II, cosθ < 0 .
15 = − = −
cosθ = − 1 − sin 2 θ
16 4 1 1 8 2 2 15
−
= − 1 −
= − 1 − = − = − 9 9 3
tan θ = sin θ 15 4
= = − ⋅ − = 15
1
cosθ
1
4
1
sin θ
3
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
4
3
−
tan θ = =
cosθ
2 2
cscθ = 1
= 1
= − 4
⋅ 15
= − 4 15
−
sin θ 15 15 15 15 1 3 2 2 − = ⋅ −
⋅ = −
4 3 2 2 2 4
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
3 2
2
4
1 1
4
3
3
4
2 = −
2
θ
−
2
secθ = 1
= 1
= − ⋅ 3 sin θ = − 1 − cos2 θ
cosθ 2 2
−
2 2 2 4
= − 1 −
− 4
16 9 3 = − 1 − = − = −
3 5 25 25 5
cot θ = 1
= 1
= − 4 ⋅
2 = − 2 2
cscθ = 1
= 1
= − 5
tan θ 2 2 2 sin θ −
3 3
−
5
4
1
55. tan θ = 3
, sin θ < 0, so θ is in quadrant III 4
57. tan θ = − , sin θ > 0, so θ is in quadrant II 3
Solve for secθ : sec2 θ = 1 + tan
2 θ Solve for secθ : sec2 θ = 1 + tan 2 θ 2
secθ = ± 1 + tan 2 θ secθ = ± 1 + tan θ
Since θ is in quadrant III, secθ < 0 . Since θ is in quadrant II, secθ < 0 .
2
secθ = − 1 + tan 2 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
− sin θ = − 1 − cos2 θ
= − 1 −
− 4
= − 1 − 16
= −
9 3
= −
sin θ =
3
1 − cos2 θ
5
25 25 5 2 = 1 −
− 3 10
=
1 − 90
cscθ = 1
sin θ = 1
− 3
= − 5
3
10 100
5 10 10
cot θ = 1
= 1
= 4
= = 100 10
1 1 tan θ 3 3 cscθ = = = 10
sin
10 10
56. cot θ = 4
, cosθ < 0, so θ is in quadrant III cot θ = 1
= 1
= −33
tan θ = 1
= 1
= 3
tan θ 1
cot θ 4 4
58. secθ = − 2, tan θ > 0, so θ is in quadrant III
Solve for secθ : sec2 θ = 1 + tan
2 θ
secθ = ± 1 + tan 2 θ
Solve for tan θ : sec2 θ = 1 + tan
2 θ
tan θ = ± sec2 θ −1
Since θ is in quadrant III, secθ < 0 . tan θ = sec2 θ −1 = (− 2)2 −1 = 4 −1 = 3
secθ = − 1 + tan 2 θ
cosθ = 1
= − 1
= − 1 + 3
= −
1 + 9
= − 25
= − 5 secθ 2
16 16 4
cot θ = 1
= 1 ⋅
3 =
3
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
cosθ = 1
= − 4
tan θ 3 3 3
secθ 5
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
12
12
2
sin θ = − 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 csccscθ =
1 =
1 = −
2 ⋅
3 = −
2 3 3 3 3
sin θ 3 3 3 3 −
77. sin 2 (40º ) + cos
2 (40º ) = 1 2
78.
sec2 (18º ) − tan
2 (18º ) = 1
59.
60.
61.
62.
sin(− 60º ) = − sin 60º = − 3
2
cos(−30º ) = cos 30º = 3
2
tan(−30º ) = − tan 30º = − 3
3
sin(−135º ) = − sin 135º = − 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 ) = sec 60º = 2
64. csc(−30º ) = − csc 30º = − 2
65. sin(−90º ) = − sin 90º = −1
66. cos(− 270º ) = cos 270º = 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
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
= − sin ⋅ = −1
73. csc
− π
= − csc π
= − 2 12
sin π
12
4
4
74.
sec (−π) = sec π = −1
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
18
18
86.
sec
− π
cos 37π
= sec π
cos 37π 92. If cot θ = − 2 , then
18
18
18
18
= sec π
cos π
+ 36π
cot θ + cot (θ − π) + cot (θ − 2π) = − 2 + (−2) + (−2) = −6
18
18 18
= sec π
cos π
+ 2π
93. sin1º + sin 2º + sin 3º +... + sin 357º
18
18
+ sin 358º + sin 359º
= sec π
cos π
= sec π
⋅ 1
= 1
= sin 1º + sin 2º + sin 3º + ⋅⋅⋅ + sin(360º −3º )
+ sin(360º − 2º ) + sin(360º −1º )
= sin 1º + sin 2º + sin 3º + ⋅ ⋅ ⋅ + sin(−3º )
+ sin(− 2º ) + sin(−1º )
87.
sin ( −20º ) + tan (200º )
cos (380º )
18 sec
π
18 = sin 1º + sin 2º + sin 3º + ⋅ ⋅ ⋅ − sin 3º − sin 2º − sin 1º
= sin (180º ) = 0
94. cos1º + cos 2º + cos 3º + ⋅ ⋅ ⋅ + cos 357º
+ cos 358º + cos 359º
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 + 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 )
+ 2 cos(180º −1º ) + cos (180º ) = 2 cos1º +2 cos 2º +2 cos 3º +... − 2 cos 2º
− 2 cos1º + cos180º
= cos180º = −1
95. The domain of the sine function is the set of all real numbers.
89. If sin θ = 0.3 , then
sin θ + 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 ) = tan θ 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 ) = cot θ
is not defined for numbers that are
91. If tan θ = 3 , then
tan θ + 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
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Section 2.3: Properties of the Trigonometric Functions Chapter 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 ) = − sin θ . 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 ) = − tan θ . Its graph is symmetric with
respect to the origin.
110. The cotangent function is odd because
cot(−θ ) = − cot θ . 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 tan θ = 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 r
2 = x2 + y 2 = 9 +1 = 10
r = 10
sin θ = 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
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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
120. a. tan θ = 1
= y for 0 < θ <
π . 123. Suppose there is a number p, 0 < p < 2π for
4 x 2 which sin(θ + p) = sin θ for all θ . If θ = 0 ,
r 2 = x2 + y
2 = 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, sin θ = 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) = sin θ
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 tan θ = 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
tan t = y
= a . Then y = ax . Now x2 + y
2 = 1 ,
for all θ must then be
1
p = 2π .
x
1 so x
2 + a2 x
2 = 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 sin θ has period 2π , so sin θ
does cscθ .
which tan t = a . In other words, 127. If P = (a, b) is the point on the unit circle−∞ < tan t < ∞ , 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
= tan θ . If there −a a
cot t = x
= a . Then x = ay . Now x2 + y
2 = 1 , exists a number p, 0 < p < π , for which
y tan(θ + p) = tan θ for all θ , then if θ = 0 ,
so a2 y 2 + y 2 = 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 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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Section 2.3: Properties of the Trigonometric Functions Chapter 2: Trigonometric Functions
a
128. cot θ = 1
: Since tan θ has period π , so tanθ
does cot θ .
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 sin θ
a cosθ
cot θ = a
= 1
= 1
b b
tan θ
130. Let P = (a, b) be the point on the unit circle
corresponding to θ . Then
tan θ = b
= sin θ
139. f (25) = 3 25 − 9 + 6
= 3 16 + 6 = 12 + 6 = 18a cosθ
So the point (25,18) is on the graph.
cot θ = a
= cosθ
b sin θ
140. y = (0)3 − 9(0)2 + 3(0) − 27
131. (sin θ cosφ )2 + (sin θ sin φ)2 + cos
2 θ
= sin 2 θ cos2 φ + sin 2 θ sin 2 φ + cos2 θ
= sin 2 θ (cos2 φ + sin 2 φ) + cos2 θ
= sin 2 θ + cos2 θ = 1
= −27
Thus the y-intercept is (0, −27) .
132 – 136. Answers will vary. Section 2.4
(− x)4 + 3
x
4 + 3 2137. 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) = −2 x2 + 12 x −13
= −2( x2 − 6 x) −13
graph by a factor of 3.
= −2
x2 − 6x +
144
4( 2)2
−13 + 2
144
4( 2)2
= −2( x2 − 6 x + 9) − 13 + 18
= −2( x2 − 6 x + 9) + 5
= −2( x − 3)2 + 5
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 2: Trigonometric Functions
, .
= − ω
2. y = 2 x 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 at2 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 = A sin(ω x) 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 = A cos(ω x) 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 = − 4 cos(2x)
b. The graph of y = sin x is increasing for This is in the form y = A cos(ω x) where
− π
< x < π
. A = − 4 and ω = 2 . Thus, the amplitude is2 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 = A sin(ω x) 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
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 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 = A sin(ω x) 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 = A cos(ω x) 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 = A cos(ω x) 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. EThis is in the form y = A cos(ω x) where
25. A1 A = −
2 and ω =
3 . Thus, the amplitude is
2
26. I
A = − 1
= 1
and the period is2 2
T = 2π
= 2π
= 4π
.
27. H
28. B3 2
29. C
20.
y = 4
sin 2
x
3
3 30. G
This is in the form
y = A sin(ω x) where
A = 4 3
31. J
and ω = 2
. Thus, the amplitude is A = 4
= 4
32. D
3 3 3 33. Comparing y = 4 cos x to y = A cos (ω x ) , 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 = 4 cos x will
lie between −4 and 4 on the y-axis. Because the
This is in the form y = A sin(ω x) 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
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 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 = A sin (ω x ) , wedomain is all real numbers, (−∞, ∞ ) and the
range is [−4, 4] . find A = −4 and ω = 1 . Therefore, the amplitude
2πis −4 = 4 and the period is = 2π . Because
134. Comparing y = 3sin x to y = A sin (ω x ) , 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 πlie 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
3πend 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 . The2 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
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 2: Trigonometric Functions
–––
From the graph we can determine that the 37. Comparing y = cos ( 4 x ) to y = A cos (ω x ) , wedomain 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 = A cos (ω x ) , we amplitude is 1, the graph of y = cos ( 4 x ) 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] .
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 2: Trigonometric Functions
, 0
38. Comparing y = sin (3x ) to y = A sin (ω x ) , we y-axis. Because the period is π , 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 ofis 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 ( 2 x ) , 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 (2 x ) .
From the graph we can determine that the Comparing y = cos ( 2 x ) to y = A cos (ω x ) , we
domain is all real numbers, (−∞, ∞ ) and the find A = 1 and ω = 2 . Therefore, the amplitude
2πrange 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 ( 2 x ) will lie
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
Comparing y = − sin ( 2 x ) to y = A sin (ω x ) , 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
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 2: Trigonometric Functions
y = − sin ( 2 x ) will lie between −1 and 1 on the 0, , , 4 2
, and π 4
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 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 ( 2 x ) , 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 = 2 cos 1
x
to
y = A cos (ω x ) ,
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 = A sin (ω x ) ,
amplitude is 2 = 2 and the period is 2π
= 8π . 1/ 4
2
Because the amplitude is 2, the graph of
y = 2 cos 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 is 8π , 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 = 2 cos 1
x
, we multiply the y-coordinatesvalues:
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
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 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 = A cos (ω x ) , range is − ,
we find
= −
2
A 1
and ω = 2 . Therefore, the
. 2 2
1
2 44. Comparing
y = −4sin
x
to
y = A sin (ω x ) , 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 ( 2 x ) will lie between − 1 and
1 on 2π
= 16π . Because the amplitude is 4, the2 2 2 1/ 8
the y-axis. Because the period is π , 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 is 16π ,
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 by4 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 ( 2 x ) , 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
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 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 = 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
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
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 2: Trigonometric Functions
2
direction to obtain the graph shown below. direction to obtain the graph shown below.
y
3 (0, 2)
(2, 2)
2
( –3– , 3)
2
2 x
(–3– , 3) 2
(–1– , 3) 2
9 (1, 8)
From the graph we can determine that the
domain is all real numbers, (−∞, ∞ ) and the
range is [−1, 5] .
From the graph we can determine that the
domain is all real numbers, (−∞, ∞ ) and the
range is [−8, 2] .
π 47. We begin by considering y = 5cos (π x ) . 48. We begin by considering y = 4sin x .
Comparing y = 5 cos (π x ) to y = A cos (ω x ) , we Comparing
y = 4sin
π x
to
y = A sin (ω x ) ,find
A = 5 and ω = π . Therefore, the amplitude
2
is 5 = 5 and the period is 2π
= 2 . Because the π
we find
A = 4 and ω = π
. Therefore, the 2
amplitude is 5, the graph of y = 5 cos (π x ) will
amplitude is 4 = 4 and the period is 2π = 4 .
lie between −5 and 5 on the y-axis. Because the
period is 2 , one cycle will begin at x = 0 and
π / 2
Because the amplitude is 4, the graph of
end at x = 2 . We divide the interval [0, 2] into y = 4sin
π x
will lie between −4 and 4 on
four subintervals, each of length 2
= 1
by 2
4 2 finding the following values:
0, 1
, 1, 3
, and 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
2 2 4These values of x determine the x-coordinates of
length = 1 by finding the following values: 4
the five key points on the graph. To obtain the y- coordinates of the five key points for
y = 5 cos (π x ) − 3 , we multiply the y-coordinates
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 y-
of the five key points for y = cos x by A = 5 coordinates of the five key points for
and then subtract 3 units. Thus, the graph of y = 4sin π
x
− 2 , we multiply the y-
y = 5cos (π x ) − 3 will lie between −8 and 2 on 2
the y-axis. The five key points are coordinates of the five key points for y = sin x
by A = 4 and then subtract 2 units. Thus, the
(0, 2) , 1
, −3
, (1, −8) , 3
, −3
, (2, 2)
π 2
2
graph of
y = 4sin
x − 2 will lie between −6
We plot these five points and fill in the graph of the curve. We then extend the graph in either
2
and 2 on the y-axis. The five key points are
(0, −2) , (1, 2) , (2, −2) , (3, −6) , (4, −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 Chapter 2: Trigonometric Functions
4
direction to obtain the graph shown below. 3
9 (0, 4) , , −2 , (3, 4) ,
2 ,10 , (6, 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 [−6, 2] .
From the graph we can determine that the
49. We begin by considering
y = −6sin
π x
. domain is all real numbers, ( −∞, ∞ ) and the
3
range is [−2,10] .
Comparing
y = −6sin π
x
to
y = A sin (ω x ) , 3
50. We begin by considering
y = −3cos
π x
. 4
we find A = −6 and ω = π
. Therefore, the 3
Comparing
y = −3cos π
x
to
y = A cos (ω x ) ,
amplitude is −6 = 6 and the period is 2π
= 6 . 4
π / 3
Because the amplitude is 6, the graph of
y = 6sin π
x
will lie between −6 and 6 on the
we find A = −3 and ω = π
. Therefore, the 4
2π
3 amplitude is −3 = 3 and the period is = 8 .
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
length 6
= 3
by finding the following values: 4 2
0, 3
, 3, 9
, and 6
π / 4
Because the amplitude is 3, the graph of
y = −3cos π
x
will lie between −3 and 3 on
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
2 2 8These values of x determine the x-coordinates of length = 2 by finding the following values:
4the five key points on the graph. To obtain the y- coordinates of the five key points for
y = −6sin π
x
+ 4 , we multiply the y-
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 y-
3
coordinates of the five key points for y = sin x
coordinates of the five key points for
y = −3cos π
x
+ 2 , we multiply the y-by A = −6 and then add 4 units. Thus, the graph
4
of y = −6sin π
x
+ 4 will lie between −2 and
coordinates of the five key points for y = cos x
3
10 on the y-axis. The five key points are
by A = −3 and then add 2 units. Thus, the graph
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 2: Trigonometric Functions
of y = −3cos π
x
+ 2 will lie between −1 and on the y-axis. The five key points are
4
π π 3π
(0, 5) , , 2 , 2
,5 , ,8 , (π ,5)
45 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.
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
10
π ,5
3π 4
,8
2
(0, 5)
π , 2
(π ,5)
4
−π π
From the graph we can determine that the
domain is all real numbers, (−∞, ∞ ) and the
range is [−1, 5] .
52.
From the graph we can determine that the
domain is all real numbers, (−∞, ∞ ) and the
range is [2, 8] .
y = 2 − 4 cos (3x ) = −4 cos (3x ) + 2
51. y = 5 − 3sin (2x ) = −3sin (2x ) + 5
We begin by considering y = −4 cos (3x ) .We begin by considering y = −3sin (2 x ) .
Comparing y = −4 cos (3x ) 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
we find A = −3 and ω = 2 . Therefore, the
amplitude is −4 = 4 and the period is 2π
.
amplitude is −3 = 3 and the period is 2π
= π . 2
3
Because the amplitude is 4, the graph of
Because the amplitude is 3, the graph of
y = −3sin (2 x ) will lie between −3 and 3 on the
y = −4 cos (3x ) will lie between −4 and 4 on
2π
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
the y-axis. Because the period is
will begin at x = 0 and end at x =
, one cycle 3
2π . We
3
length π
by finding the following values:
divide the interval 0,
2π into four
4 3
0, π
, π
, 3π
, and π
subintervals, each of length 2π / 3
= π
by4 2 4 4 6
These values of x determine the x-coordinates of finding the following values:the five key points on the graph. To obtain the y- coordinates of the five key points for
y = −3sin ( 2x ) + 5 , we multiply the y- 0,
π ,
6
π ,
π 3 2
, and 2π 3
coordinates of the five key points for
y = sin x These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-
by A = −3 and then add 5 units. Thus, the graph
of y = −3sin ( 2x ) + 5 will lie between 2 and 8
coordinates of the five key points for
y = −4 cos (3x ) + 2 , we multiply the y-
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 2: Trigonometric Functions
,0 4 3 4 3
= −
= −
= −
2
coordinates of the five key points for y = cos x 3 3 9
by A = −4 and then adding 2 units. Thus, the 0, , ,
4 2 , and 3
4
graph of y = −4 cos (3x ) + 2 will lie between −2 These values of x determine the x-coordinates of
and 6 on the y-axis. The five key points are the five key points on the graph. To obtain the y- coordinates of the five key points for
π π 2π
(0, −2) , π
, 2 , ,6 , , 2 , , −2 5 2π
6
3
2
3
y = − sin x , we multiply the y-
3
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.
coordinates of the five key points for
5
y = sin x
by A = − .The five key points are 3
(0, 0) , 3
, − 5
, 3
, 9
, 5
, (3, 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.
3 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
equivalent form y 5
sin 2π
x
.3 3
From the graph we can determine that the
domain is all real numbers, (−∞, ∞ ) and the
Comparing y 5
sin 2π
x
to
5 5 3 3 range is − , .
y = A sin (ω x ) , we find 5 A = − and ω =
2π .
3 3
3 3 54. Since cosine is an even function, we consider the
Therefore, the amplitude is
− 5
= 5
and the 9 3π
3 3 equivalent form y = cos 5 2
x . Comparing
period is 2π
= 3 . Because the amplitude is
9 3π 2π / 3 y = cos
5 2
x to y = A cos (ω x ) , we find
5 , the graph of y
5 sin
2π x
will lie
9 3π3 3 3 A = and ω =
5 . Therefore, the amplitude is
2
between − 5
and 5
on the y-axis. Because the 9 9 2π 43 3 = and the period is = . Because
5 5 3π / 2 3period is 3 , one cycle will begin at x = 0 and
end at x = 3 . We divide the interval [0, 3] into four subintervals, each of length 3
by
finding
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Section 2.4: Graphs of the Sine and Cosine Functions Chapter 2: Trigonometric Functions
the amplitude is 9
, the graph of 5
9 3π 9 9
4 y = cos 5 2
x will lie between − 5
and 5
the following values:
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