Cha p - Test Bank 123 · 10 ) = )) = = = = = ) = = = ) = ) = ) = ) = ) = = ) = =

127127

Copyright © 2016 Pearson Education, Inc.

127127


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

Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-A-Unit-Circle-Approach-10th-Edition-Sullivan


https://testbank123.eu/Solutions-Manual-for-Trigonometry-A-Unit-Circle-Approach-10th-Edition-Sullivan

128128


128128


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





129129


129129



180 2





130130


130130



= −

= −

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


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


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





131131


131131



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





132132


132132



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





133133


133133



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





134134


134134



( )

( )

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





135135


135135



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.





136136


136136



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





137137


137137



12. False





138138


138138


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





139139


139139



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





140140


140140



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





141141


141141



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






2

138138


138138


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


θ = 2π

= 120º is

− 1

, 3

. 3 2 2

6 1 2 1


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






2

139139


139139


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






2

140140


140140


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


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






2

141141


141141


= −

2

2


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






2

142142


142142


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


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






2

143143


143143


cot 390º = 2 = 3

⋅ 2

= 3

1 2 1






2

144144


144144


−

⋅ 3


θ π

= − 30º is 3

, − 1

.


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


2


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

⋅ =






2

145145


145145


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






−

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


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

−


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

−


sec 21º = 1

≈ 1.07 . −

1

cot −

= − 1

⋅

− 2

3 3

⋅ =

cos 21°

3

3

2

3

3 3 −






= −

= −


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






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 .






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






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 :






2

R (25º ) = 500 sin(2 ⋅ 25º )

≈ 19, 541.95 m 9.8






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 +

3 ⋅ 1

2

− 4 ⋅1

c. t (60) = 2(10)


= 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






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






[ ]

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






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





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.






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






5

5

tan θ = sin θ

= 5 4 5 4

= ⋅ − = −

cosθ

3

5

3

3 −

cscθ = 1

= 1

= 5

sin θ 4 4






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






3

sin θ

1

1 13

−






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


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


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






cosθ

4

5

4

4 −

= − 1 −

25 = −

144 12 = −

5 169 169 13






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


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







2 θ 5 5 5

−

3


1 1 5 5 5Since θ is in quadrant II, sin θ > 0 . cot θ = = = ⋅ =

tan θ 2 5 2 5 5 2

5






= −

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


2 θ = 1


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

2

Solve for sin θ : sin θ =

1 =

1

cscθ 3


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






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






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






cosθ = 1

= − 4

tan θ 3 3 3

secθ 5






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






= − sin ⋅ = −1

73. csc

− π

= − csc π

= − 2 12

sin π

12

4

4

74.

sec (−π) = sec π = −1






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






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


110. The cotangent function is odd because

cot(−θ ) = − cot θ . Its graph is symmetric with


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


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






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 .






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





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






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 .


= 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


= 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






, 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


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


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


= π

by 4 2

finding the following values:




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



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






–––

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



end at x = 2π . We divide the interval [0, 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


0, π

, π , 3π

, and 2π

8 4 8 2


2 2 the five key points on the graph. The five key



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


five key points are

y = cos x by

3π

A = −3 . The the curve. We then extend the graph in either


(0, −3) , π

, 0

, (π ,3) ,

,0 , (2π , −3) 2

2




y

5( , 3) ( , 3)

( 3

, 0) 2

x

2 2

(–

– , 0) 2


(0, 3)

5 (2 , 3)


range is [−1,1] .



range is [−3, 3] .






, 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


the five key points on the graph. The five key

points are




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)



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






y = − sin ( 2 x ) will lie between −1 and 1 on the 0, , , 4 2

, and π 4






,0 ,0


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


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


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





domain is all real numbers, ( −∞, ∞ ) and the

range is [−2, 2] .

42. Comparing

y = 2 cos 1

x

to

y = A cos (ω x ) ,


4


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


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


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






= −

= −

8

= −

= −


y


( 2 , 0) 2

(0, 2)

(8 , 2)

( 8 , 2) (2 , 0) (6 , 0)

x

8 4 4 8

2

( 4 , 2) (4 , 2)



range is [−2, 2] .



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-


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



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








,5


y

5 (12 , 4)


16

8

(0, 0)

x

(16 , 0)

(8 , 0)

5 (4 , 4)



range is [−4, 4] .



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



= π

by 4 2


between −3 and 3 on the y-axis. Because the



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


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








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)



range is [−1, 5] .



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 .


period is 2 , one cycle will begin at x = 0 and

π / 2


end at x = 2 . We divide the interval [0, 2] into y = 4sin

π x


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



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


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







4

direction to obtain the graph shown below. 3

9 (0, 4) , , −2 , (3, 4) ,

2 ,10 , (6, 4)

2





domain is all real numbers, ( −∞, ∞ ) and the

range is [−6, 2] .


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 ) ,


= 6 . 4

π / 3


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


y = −3cos π

x


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


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






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



4 We plot these five points and fill in the graph of



y

10

π ,5

3π 4

,8

2

(0, 5)

π , 2

(π ,5)

4

−π π



range is [−1, 5] .

52.



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


.


= π . 2

3



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


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


y = −4 cos (3x ) + 2 , we multiply the y-






,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




5

y = sin x

by A = − .The five key points are 3

(0, 0) , 3

, − 5

, 3

, 9

, 5

, (3, 0)




3 3



range is [−2, 6] .

53. Since sine is an odd function, we can plot the

equivalent form y 5

sin 2π

x

.3 3



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






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:





Cha p - Test Bank 123 · 10 ) = )) = = = = = ) = = = ) = ) = ) = ) = ) = = ) = =

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Transcript of Cha p - Test Bank 123 · 10 ) = )) = = = = = ) = = = ) = ) = ) = ) = ) = = ) = =