Section 2.1
1. 2 Cr =π ; 2 A r=π 2. 1 2(2)2 2 A ππ==
3. standard position
4. central angle 5. d
rθ ; 2 1 2 r θ
. b
s t ; t θ
True
Chapter 2
Trigonometric Functions
º 111 40º10'25"401025 606060 (400.16670.00694)º 40.17º
º 111 61º42'21"614221 606060 (610.70000.00583)º 61.71º
25. º 111 50º14'20"501420 606060 (500.23330.00556)º 50.24º
26. º 111 73º40'40"734040 606060 (730.66670.0111)º 73.68º
27. º 111 9º9'9"999 606060 (90.150.0025)º 9.15º
28. º 111 98º22'45"982245 606060 (980.36670.0125)º 98.38º
29. 40.32º40º0.32º
40º0.32(60')
40º19.2'
40º19'0.2'
40º19'0.2(60")
40º19'12"
40º19'12"
30. 61.24º61º0.24º
61º0.24(60')
61º14.4'
61º14'0.4'
61º14'0.4(60")
61º14'24"
61º14'24"
31. 18.255º18º0.255º
18º0.255(60')
18º15.3'
18º15'0.3'
18º15'0.3(60")
18º15'18"
18º15'18"
Section2.1: Angles and Their Measure
32. 29.411º29º0.411º
29º0.411(60')
29º24.66'
29º24'0.66'
29º0.66(60")
29º24'39.6"
29º24'40"
33. 19.99º19º0.99º
19º0.99(60')
19º59.4'
19º59'24" =+
19º59'0.4'
19º59'0.4(60")
19º59'24"
34. 44.01º44º0.01º
44º0.01(60')
44º0.6'
44º0'0.6'
44º0'0.6(60")
44º0'36"
44º0'36"
35. 3030 radian radian 1806 ππ°=⋅=
36. 2 120120 radian radians 1803 ππ°=⋅=
37. 4 240240 radian radians 1803 ππ°=⋅=
38. 11 330330 radian radians 1806 ππ°=⋅=
39. 6060 radian radian 1803 ππ−°=−⋅=−
40. 3030 radian radian 1806 ππ−°=−⋅=−
41. 180180 radian radians 180 π °=⋅=π
42. 3 270270 radian radians 1802 ππ°=⋅=
43. 3 135135 radian radians 1804 ππ−°=−⋅=−
44. 5 225225 radian radians 1804 ππ−°=−⋅=−
45. 9090 radian radians 1802 ππ−°=−⋅=−
46. 180180 radian radians 180 π −°=−⋅=−π
47. 180 degrees60 33 ππ=⋅=° π
48. 55180 degrees150 66 ππ=⋅=° π
49. 55180 degrees225 44 ππ −=−⋅=−° π
50. 22180 degrees120 33 ππ −=−⋅=−° π
51. 180 degrees90 22 ππ=⋅=° π
52. 180 44 degrees720 π=π⋅=° π
53. 180 degrees15 1212 ππ=⋅=° π
54. 55180 degrees75 1212 ππ=⋅=° π
55. 180 degrees90 22 ππ −=−⋅=−° π
56. 180 degrees180 −π=−π⋅=−° π
57. 180 degrees30 66 ππ −=−⋅=−° π
58. 33180 degrees135 44 ππ −=−⋅=−° π
59. 17 1717 radian radian0.30 radian 180180 ππ °=⋅=≈
60. 7373 radian 180 73 radians 180 1.27 radians °=⋅π π = ≈
61. 4040 radian 180 2 radian 9 0.70 radian −°=−⋅π π =− ≈−
62. 5151 radian 180 17 radian 60 0.89 radian −°=−⋅π π =− ≈−
63. 125125 radian 180 25 radians 36 2.18 radians °=⋅π π = ≈
64. 350350 radian 180 35 radians 18 6.11 radians °=⋅π π = ≈
65. 180 3.14 radians3.14 degrees179.91º=⋅≈ π
66. 180 0.75 radian0.75 degrees42.97º=⋅≈ π
67. 180 2 radians2 degrees114.59º=⋅≈ π
68. 180 3 radians3 degrees171.89º=⋅≈ π
69. 180 6.32 radians6.32 degrees362.11º=⋅≈ π
70. 180 2 radians2 degrees81.03º=⋅≈ π
71. 1 10 meters; radian; 2 r θ == 1 105 meters 2
72. 6 feet;2 radian;6212 feet rsr θθ ====⋅=
73. 1 radian; 2 feet; 3 s θ==
74. 1 radian; 6 cm; 4 s θ==
75. 5 miles;3 miles; rs== 3 0.6 radian
76. 6 meters;8 meters;
77. 2 inches; 30º30 radian; 1806 r θππ===⋅=
78. 2 3 meters; 120º120 radians 1803 r θππ
79. 1 10 meters; radian 2 r
80. 6 feet;2 radians r θ ==
ft
Section2.1: Angles and Their Measure
2 inches; 30º30 radian 1806 r θππ===⋅=
Chapter2: Trigonometric Functions
86. 2 3 meters; 120º120 radians 1803 r θππ===⋅=
87. 2 feet; radians 3 r θπ==
88. 4 meters;
89.
yards;
90. 5 9 cm; 50º50 radian 18018 r θππ===⋅= 5 97.854 cm 18 srθπ ==⋅≈
2 1154522 9=35.343 cm 22184
91. 6 inches r = In 15 minutes, 151rev360º90º radians 6042 θπ ==⋅== 639.42 inches 2 srθπ ==⋅=π≈ In 25 minutes, 2555 rev360º150º radians 60126 θπ ==⋅== 5 6515.71 inches 6 srθπ ==⋅=π≈
92. 40 inches;20º radian 9 r θπ=== 40 4013.96 inches 99 srθππ ==⋅=≈
93. 4 m; 45º45 radian 1804 r θππ===⋅= () 2 1122 426.28 m 224 Ar π θπ ===≈
94. 3 cm; 60º60 radians 1803 r θππ===⋅= () 2 11322 34.71 cm 2232 Ar θππ ===≈
95. 3 30 feet; 135º135 radians 1804 r θππ===⋅= () 2 11367522 301060.29 ft 2242 Ar θππ
96. 2 15 yards; 100 yd rA== () 2 2 1 2 1 10015 2 100112.5 1008 0.89 radian 112.59 Ar θ θ θ θ = =
8180160 50.93 9 ππ ° ⋅=≈°
97. 22 12 22 112 120 223 1212 (34)(9) 2323 Arrθθθπ ππ =−=°= =− 2 1212 (1156)(81) 2323 (1156)(81)33 115681 33 1075 1125.74 in 3 ππ ππ
98. 22 12 22 1125 125 2236 125125 (30)(6) 236236 Arrθθθπ ππ =−=°= =− 2 125125 (900)(36) 236236 2525 (450)(18)3636 11250450 3636 10800 300942.48 in 36
99. 1 5 cm; 20 seconds; radian 3 rt θ ===
() () 1/3 111 radian/sec 2032060
51/3 511 cm/sec 2032012 t sr v tt θ
100. 2 meters;20 seconds;5 meters rts ===
()() /5/2 511 radian/sec 202208 51 m/sec 204 sr tt s v t θ ω====⋅= ===
101. 25 feet;13 rev/min26 rad/min r ω ===π 2526 ft./min6502042.0 ft/min
ft.1mi60min 65023.2mi/hr min5280ft1hr vr v ω ==⋅π=π≈
102. 6.5 m;22 rev/min44 rad/min r ω ===π (6.5)44 m/min286898.5 m/min m1km60min 28653.9km/hr min1000m1hr
v
103. 4 m;8000 rev/min16000 rad/min r ω ===π (4)16000 m/min64000 cm/min cm11km60min 64000 min1001000m1hr 120.6km/hr vr m v cm
Section2.1: Angles and Their Measure
104. 5 m;5400 rev/min10800 rad/min r ω ===π (5)10800 m/min54000 cm/min cm1m1km60min 54000 min100cm1000m1hr
vr v ω ==⋅π=π =π⋅⋅⋅ ≈
101.8km/hr
105. 26 inches;13 inches;35 mi/hr drv === 35 mi5280 ft12 in.1hr hrmift60 min
36,960 in./min v =⋅⋅⋅ = 36,960 in./min 13 in.
452.5 rev/min v r ω==
2843.08 radians/min
2843.08 rad1rev min2 rad
106. 15 inches;3 rev/sec6 rad/sec r ω ===π 156 in./sec90282.7 in/sec in.1ft1mi3600sec 9016.1mi/hr sec12in.5280ft1hr
vr v ω ==⋅π=π≈ =π⋅⋅⋅≈
107. 3960 miles r = 35º9'29º57' 5º12' 5.2º 5.2 180 0.09076 radian θ=− = = π =⋅ ≈ 3960(0.09076)359 miles srθ ==≈
108. 3960 miles r = 38º21'30º20' 8º1' 8.017º 8.017 180 0.1399 radian θ=− =
3960(0.1399)554 miles srθ ==≈
109. 3429.5 miles r = 1 rev/day2 radians/day radians/hr 12 ω π ==π= 3429.5898 miles/hr 12
vrω π ==⋅≈
Chapter2: Trigonometric Functions
110. 3033.5 miles r = 1 rev/day2 radians/day radians/hr 12
miles/hr 12
111. 2.39105 miles r =×
5 1 rev/27.3 days 2 radians/27.3 days
radians/hr 1227.3 2.39102292 miles/hr 327.6
112. 9.29107 miles r =×
7 1 rev/365 days 2 radians/365 days
radians/hr 12365 9.291066,633 miles/hr 4380
113. 12 2 inches;8 inches; rr== 1 3 rev/min6 radians/min ωπ
114. 30 feet r = 1 rev2 0.09 radian/sec 70 sec70 sec35
rad6 ft 30 feet2.69 feet/sec 35 sec7 sec
115. 4 feet;10 rev/min20 radians/min r ω ===π 420 ft 80 min 80 ft1 mi60 min min5280 fthr 2.86 mi/hr
116. 26 inches;13 inches; dr== 480 rev/min960 radians/min ω==π 13960 in 12480 min 12480 in1ft1 mi60 min min12 in5280 fthr 37.13 mi/hr vr=ω =⋅π
80 mi/hr12in5280 ft1 hr1 rev 13 in1 ft1 mi60 min2 rad 1034.26 rev/min v r ω= =⋅⋅⋅⋅ π
117. 8.5 feet;4.25 feet; 9.55 mi/hr drv === 9.55 mi/hr 4.25 ft 9.55 mi15280 ft1 hr1 rev hr4.25 ftmi60 min2 31.47 rev/min v r ω== =⋅⋅⋅⋅ π
118. Let t represent the time for the earth to rotate 90 miles. 24
902(3559) 90(24) 0.0966 hours5.8 minutes 2(3559) t t = π =≈≈ π
119. 2 2 (9) 81 A rπ π π = = =
We need ¾ of this area. 3243 81 44 ππ = . Now we calculate the small area. 2 2 (3) 9 A rπ π π = = =
We need ¼ of the small area. 19 9 44 ππ =
So the total area is: 2439252 63 444 ππππ +== square feet.
120. First we find the radius of the circle. 2 82 4 Cr r r π ππ = = =
The area of the circle is 22(4)16 Arπππ=== .
The area of the sector of the circle is 4π . Now we calculate the area of the rectangle. (4)(47) 44 Alw A A = =+ =
So the area of the rectangle that is outside of the circle is 2 444 u π
121. The earth makes one full rotation in 24 hours. The distance traveled in 24 hours is the circumference of the earth. At the equator the circumference is 2(3960) π miles. Therefore, the linear velocity a person must travel to keep up with the sun is: 2(3960) 1037 miles/hr 24 s v t π ==≈
122. Find , when 3960 miles and 1'. sr θ == 1 degree radians 1'0.00029 radian 60 min180 degrees 3960(0.00029)1.15 miles sr
Thus, 1 nautical mile is approximately 1.15 statute miles.
Section2.1: Angles and Their Measure
123. We know that the distance between Alexandria and Syene to be 500 s = 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
7.27.2 radian 18025 ππ°=°⋅= ° . Therefore,
500 25 2512,5005003979 miles sr r r θ π ππ = =⋅ =⋅=≈ 12,500 2225,000Crππ π ==⋅= miles.
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 96 θ=° with 200 r = feet.
20096335.10 feet 180 sr θπ =⋅=⋅°⋅≈ ° 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 96 θ=° . One sector with 200 r = feet and the other with 190 r = feet.
2222 22 11 222 96 200190 2180 4 39003267.26 15
()
The area of the warning track is about 3267.26 square feet.
Chapter2: Trigonometric Functions
125. 11 rotates at rev/min r ω , so 111vr=ω 22 rotates at rev/min r ω , so 222vr=ω
Since the linear speed of the belt connecting the pulleys is the same, we have:
126. Answers will vary.
127. If the radius of a circle is r and the length of the arc subtended by the central angle is also r, then the measure of the angle is 1 radian. Also, 180 1 radian degrees π = . 1 1 360 °= revolution
128. Note that radians 110.017 radian 180 π °=°⋅≈
and 180 1 radian57.296 radians °
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 radians =π , the arc length formula can be rewritten as follows: 180 s rr π θθ==
131 – 133. Answers will vary.
134. ()37 037 7 37 3 fxx x xx =+ =+ =−→=−
135. 2 9 x cannot be zero so the domain is: {} |3xx ≠±
136. Shift to the left 3 units would give 3 yx=+ Reflecting about the x-axis would give 3 yx=−+ . Shifting down 4 units would result in 34 yx=−+−
137. 1212 , 22 3(4)62 , 22 181 222,,4 xxyy ++ +−+ = ==−
Section 2.2
1. 222 cab =+ 2. ()() 53571578 f =−=−=
3. True 4. equal; proportional 5. 13 , 22 6. 1 2 7. b
8. () 0,1 9. 22 , 22
10. a
11. ; yx rr
12. False
Section2.2: Trigonometric Functions: Unit Circle Approach
Section2.2: Trigonometric Functions: Unit Circle Approach
=−
26. ()() ()() sec8sec08 sec042sec01 π π π=+ =+⋅==
27. 33 coscos 22 4 cos 22 cos(1)2 2 cos 2 0 π π
=
=
28. ()() ()() sin3sin3 sin2sin0 π −π=−π =−π+=−π=
29. ()() secsec1 −π=π=−
30. () ()() tan3tan(3) tan03tan00 π −π=−π =−+=−=
31. 2112 sin45ºcos60º 222 + +=+=
32. 1212 sin30ºcos45º 222 −=−=
33. sin90ºtan45º112 +=+=
34. cos180ºsin180º101 −=−−=−
35. 2221 sin45ºcos45º 2242 =⋅==
36. 33 tan45ºcos30º1 22 =⋅=
37. csc45ºtan60º236 =⋅=
38. 2323 sec30ºcot45º1 33 =⋅=
39. 4sin90º3tan180º41304 −=⋅−⋅=
40. 5cos90º8sin270º508(1)8 −=⋅−−=
41. 33 2sin3tan23330 3623
42. 2 2sin3tan23123 442 ππ +=⋅+⋅=+
43. 343 2sec4cot22422 4333 ππ +=⋅+⋅=+
44. 23 3csccot31231 343 ππ +=⋅+=+
45. csccot101 22 ππ +=+=
46.
47. The point on the unit circle that corresponds to 2 120º 3 θπ== is 13 , 22
48. The point on the unit circle that corresponds to
49. The point on the unit circle that corresponds to
Section2.2: Trigonometric Functions: Unit Circle Approach
50.
52. The point on the unit circle that corresponds to
51.
53. The point on the unit circle that corresponds to 8
55. The point on the unit circle that corresponds to 9 405º 4 θπ == is 22 , 22
csc405º12 22 2 2 ==⋅⋅=
122 sec405º12 22 2 2 ==⋅⋅=
56. The point on the unit circle that corresponds to 13 390º 6 θπ == is 31 , 22
57. The point on the unit circle that corresponds to
Trigonometric Functions: Unit Circle Approach
59. The point on the unit circle that corresponds to
58.
60. The point on the unit circle that corresponds to 4 240º 3 θπ =−=− is 13 , 22
Chapter2: Trigonometric Functions
61. The point on the unit circle that corresponds to 5 450º 2 θπ== is () 0,1
5 sin1 2 π = 51 csc1 21 π == 5 cos0 2 π = 51 secundefined 20 π == 51 tanundefined 20 π == 50 cot0 21 π ==
62. The point on the unit circle that corresponds to 5900ºθπ== is () 1,0 .
63. The point on the unit circle that corresponds to 14
65. Set the calculator to degree mode: sin28º0.47 ≈
66. Set the calculator to degree mode: cos14º0.97 ≈ .
67. Set the calculator to degree mode: 1 sec21º1.07 cos21
68. Set the calculator to degree mode: 1 cot70º0.36
69. Set the calculator to radian mode: tan0.32 10
Section2.2: Trigonometric Functions: Unit Circle Approach
70. Set the calculator to radian mode: sin0.38 8
(3,
71.
72.
73. Set the calculator to radian mode: sin10.84 ≈ 74. Set the calculator to radian mode: tan11.56 ≈ 75. Set the calculator to degree mode: sin1º0.02
78. For the point (5,12) , 5 x = , 12 y =− , 22 2514416913 rxy=+=+== 1213 sin csc 1312 513 cos sec 135 125
79. For the point (2,3) , 2 x = , 3 y =− , 22 4913 rxy=+=+= 31331313 sin csc 133 1313 21321313 cossec 132 1313
Chapter2: Trigonometric Functions
80. For the point (1,2) , 1 x =− , 2 y =− , 22
sin45ºsin135ºsin225ºsin315º
81. For the point (2,2) , 2 x =− ,
82.
tan60ºtan150º3
87. () () sin40ºsin130ºsin220ºsin310º sin40ºsin130ºsin40º180º sin130º180º sin40ºsin130ºsin40ºsin130º 0 +++ =++++ + =+−− =
88. () tan40ºtan140ºtan40ºtan180º40º tan40ºtan40º 0 +=+− =− =
89. If () sin0.1 f θθ== , then () sin()0.1 f θπθ+=+π=− .
90. If () cos0.3 f θθ== , then () cos()0.3 f θπθ+=+π=−
91. If () tan3 f θθ== , then () tan()3 f θπθ+=+π=
92. If () cot2 f θθ==− , then () cot()2 f θπθ+=+π=−
Section2.2: Trigonometric Functions: Unit Circle Approach
93.
94.
112. a. 3 cos 662 g ππ
The point 3 , 62 π
b. The point 3 , 26 π
is on the graph of g
is on the graph of 1 g
c. 22(0) 66 2cos(0) 21 2 gg
Thus, the point ,2 6 π
is on the graph of 2 6 ygx π
113. Answers will vary. One set of possible answers is 115713 ,,,, 33333
114. Answers will vary. One ser of possible answers is 13531119 ,,,, 44444 πππππ
115. sin sin
0.50.47940.9589
0.40.38940.9735
0.20.19870.9933
0.10.09980.9983
0.010.01001.0000
0.0010.00101.0000
0.00010.00011.0000
0.000010.000011.0000
() sin f θθ θ = approaches 1 as θ approaches 0.
116. cos1 cos1
0.50.12240.2448
0.40.07890.1973
0.20.01990.0997
0.10.00500.0050
0.010.000050.0050
0.0010.00000.0005
0.00010.00000.00005
0.000010.00000.000005
θ () cos1 g θ θ θ = approaches 0 as θ approaches 0.
117. Use the formula ()() 2 0 sin2 v R g θ θ= with
2 32.2ft/sec g = ; 45º θ= ; 0 100 ft/sec v = :
2 (100)sin(245º) 45º310.56 feet 32.2 R =≈
()
Use the formula ()()
2 2 0 sin 2 v H g θ θ= with
2 32.2ft/sec g = ; 45º θ= ; 0 100 ft/sec v = : ()
22 100(sin45º) 45º77.64 feet 2(32.2) H =≈
118. Use the formula ()() 2 0 sin2 v R g θ θ= with
2 9.8 m/sec g = ; 30º θ= ; 0 150 m/sec v = : ()
2 150sin(230º) 30º1988.32 m 9.8 R =≈
Use the formula ()()
2 2 0 sin 2 v H g θ θ= with
2 9.8 m/sec g = ; 30º θ= ; 0 150 m/sec v = : ()
22 150(sin30º) 30º286.99 m 2(9.8) H =≈
119. Use the formula ()() 2 0 sin2 v R g θ θ= with 2 9.8 m/sec g = ; 25º θ= ; 0 500 m/sec v = :
() 2 19,541.95 500sin(225º) 25º m 9.8 R =≈
Use the formula ()() 2 2 0 sin 2 v H g θ θ= with 2 9.8 m/sec g = ; 25º θ= ; 0 500 m/sec v = : () 22 500(sin25º) 25º2278.14 m 2(9.8) H =≈
120. Use the formula ()() 2 0 sin2 v R g θ θ= with 2 32.2ft/sec g = ; 50º θ= ; 0 200 ft/sec v = : () 2 200sin(250º) 50º1223.36 ft 32.2 R =≈
Use the formula ()() 2 2 0 sin 2 v H g θ θ= with 2 32.2ft/sec g = ; 50º θ= ; 0 200 ft/sec v = : () 22 200(sin50º) 50º364.49 ft 2(32.2) H =≈
121. Use the formula () 2 sincos a t g θ θθ = with 2 32 ft/sec g = and 10 feet a = :
a. ()() 210 301.20 seconds 32sin30ºcos30º t =≈
b. ()() 210 451.12 seconds 32sin45ºcos45º t =≈
c. ()() 210 601.20 seconds 32sin60ºcos60º t =≈
122. Use the formula () cos160.5cos(2) x θθθ =++ ()() ()() 30cos30º160.5cos(230º) cos30º160.5cos60º 4.90 cm x =++⋅ =++ ≈ ()() ()() 45cos45º160.5cos(245º) cos45º160.5cos90º 4.71 cm x =++⋅ =++ ≈
Section2.2: Trigonometric Functions: Unit Circle Approach
123. Note: distance on road time on road rate on road = 82
1
4 1 1 4tan x x θ θ
1
=− =− a. 21 (30º)1 3sin30º4tan30º 21 1 11 34 2 3 43 11.9 hr 34 T =+− =+− =+−≈
Sally is on the paved road for 1 10.57 hr 4tan30º −≈ .
b. 21 (45º)1 3sin45º4tan45º 21 1 1 41 3 2 221 11.69 hr 34 T =+− =+− =+−≈
Sally is on the paved road for o 1 10.75 hr 4tan45 −=
c. oo 21 (60º)1 3sin604tan60 21 1 343 3 2 41 1 3343 1.63 hr T =+− =+− ⋅ ⋅ =+− ≈
Sally is on the paved road for o 1 10.86 hr 4tan60 −≈ .
d. 21 (90º)1 3sin90º4tan90º T =+− .
But tan90º is undefined, so we cannot use the function formula for this path. However, the distance would be 2 miles in the sand and 8 miles on the road. The total time would be: 25 11.67 33 +=≈ hours. The path would be to leave the first house walking 1 mile in the sand straight to the road. Then turn and walk 8 miles on the road. Finally, turn and walk 1 mile in the sand to the second house.
124. When 30º θ= :
1sec30º 1 30º(2)251.42 cm 3 tan30º V
When 45º θ= :
125. tan
Arletha is 15.4 feet from the car.
126. tan 22 8555 tan 22 2tan4555 555 3968 2tan4 H D D
The tourist is 3968 feet from the monument.
127. tan 22 20 tan 22(200) 400tan10 71 H D H H H θ = ° = =° ≈
The tree is approximately 71 feet tall.
128. tan 22 0.52 tan 22(384400) 768800tan0.26 3488 H D H H H θ = ° = =° =
The moon has a radius of 1744 km.
129. 222 41 cossin;cossin1 49 θθθθ +=+=
Substitute cos;sinxyθθ== and solve these simultaneous equations for y. 222 22 2 2 41 ;1 49 1 41 (1) 49 8 0 49 xyxy yx xx xx +=+= =− +−= −−=
Using the quadratic formula: 2 8 49 3281 9 4949 7 8 1,1, 49 (1)(1)4(1)() 2 111 1 81 22277 abc x or ==−=− −−±−−− = ±+±± ====−
Since the point is in quadrant III then 1 7 x =− and 2 2 1148 11 74949 4843 497 y y =−−=−= =−=−
130. 222 1 cossin;cossin1 9 θθθθ −=−+=
Substitute cos;sinxyθθ== and solve these simultaneous equations for y.
22 2 2 2 1 ;1 9 1 1 (1) 9 10 0 9 99100 xyxy xy yy yy yy
Using the quadratic formula: 2 9,9,10 (9)(9)4(9)(10) 18 9813609441921 181818 abc y
Since the point is in quadrant II then 321122 18183 y
and 2 2 245 11 399 55
Section2.2: Trigonometric Functions: Unit Circle Approach
c. Using the MAXIMUM feature, we find: 0 20 90°
when 67.5º R θ=
Slope of
Since L is parallel to M, the slope of L is equal to the slope of M. Thus, the slope of tan L =θ
133. a. When 1 t = , the coordinate on the unit circle is approximately (0.5, 0.8) . Thus,
Set the calculator on RADIAN mode:
b. When 5.1 t = , the coordinate on the unit circle is approximately (0.4,0.9) . Thus,
Set the calculator on RADIAN mode:
Chapter2: Trigonometric Functions
134. a. When 2 t = , the coordinate on the unit circle is approximately (0.4, 0.9) . Thus,
Set the calculator on RADIAN mode:
b. When 4 t = , the coordinate on the unit circle is approximately (0.7,0.8) . Thus, sin40.8
Answers will vary.
Set the calculator on RADIAN mode:
135 – 137. Answers will vary.
142. 22 22 (13)(0(6)) (2)(6) 43640210 d =−+−− =−+ =+==
Section 2.3
1. All real numbers except 1 2 ; 1 | 2 xx ≠−
2. even
3. False
4. True
5. 2π , π
6. All real number, except odd multiples of 2 π
7. b.
8. a
9. 1
10. False; 1 sec cos θ θ =
11. 2 sin405ºsin(360º45º)sin45º 2 =+==
12. 1 cos420ºcos(360º60º)cos60º 2 =+==
13. tan405ºtan(180º180º45º)tan45º1 =++==
14. 1 sin390ºsin(360º30º)sin30º 2 =+==
15. csc450ºcsc(360º90º)csc90º1 =+==
16. sec540ºsec(360º180º)sec180º1 =+==−
17. cot390ºcot(180º180º30º)cot30º3 =++==
Section2.3: Properties of the Trigonometric Functions
21. ()() tan21tan(021)tan00 π=+π==
22. 9 csccsc4 22 ππ =+π
23. 17 secsec4 44 ππ
=
csc22 2 csc 2 1
24. 17 cotcot4 44
25. 193 tantan3tan 6663
26. 25 secsec4 66 ππ
18. sec420ºsec(360º60º)sec60º2 =+== 19. 33 coscos8 44
27. Since sin0 θ> for points in quadrants I and II, and cos0 θ< for points in quadrants II and III, the angle θ lies in quadrant II.
28. Since sin0 θ< for points in quadrants III and IV, and cos0 θ> for points in quadrants I and IV, the angle θ lies in quadrant IV.
29. Since sin0 θ< for points in quadrants III and IV, and tan0 θ< for points in quadrants II and IV, the angle θ lies in quadrant IV.
30. Since cos0 θ> for points in quadrants I and IV, and tan0 θ> for points in quadrants I and III, the angle θ lies in quadrant I.
31. Since cos0 θ> for points in quadrants I and IV, and tan0 θ< for points in quadrants II and IV, the angle θ lies in quadrant IV.
32. Since cos0 θ< for points in quadrants II and III, and tan0 θ> for points in quadrants I and III, the angle θ lies in quadrant III.
33. Since sec0 θ< for points in quadrants II and III, and sin0 θ> for points in quadrants I and II, the angle θ lies in quadrant II.
34. Since csc0 θ> for points in quadrants I and II, and cos0 θ< for points in quadrants II and III, the angle θ lies in quadrant II.
35. sin,cos34
quadrant
47. 5 sin,90º180º, in quadrant II 13
Solve
48. 4 cos,270º360º; in quadrant IV
Section2.3: Properties of the Trigonometric Functions
51. 2 sin,tan0, sois in quadrant II 3 θθθ =<
Since
53. sec2,sin0,so is in quadrant IV θθθ =<
Solve
Solve
113
tan3
54. csc3,cot0,so is in quadrant II θθθ =<
Solve for sin θ : 11 sin csc3 θ θ ==
Solve for cos θ : 22 2 sincos1 cos1sin
in quadrant II,
55.
Solve
Since
58. sec2,tan0,so is in quadrant III θθθ =−>
Solve for tan θ : 22 2 sec1tan tansec1 θθ θθ =+ =±−
Chapter2: Trigonometric Functions
59. 3 sin(60º)sin60º 2 −=−=−
60. 3 cos(30º)cos30º 2 −==
61. 3 tan(30º)tan30º 3 −=−=−
62. 2 sin(135º)sin135º 2 −=−=−
63. sec(60º)sec60º2 −==
64. csc(30º)csc30º2 −=−=−
65. sin(90º)sin90º1 −=−=−
66. cos(270º)cos270º0 −==
67. tantan1 44
68. sin()sin0 −π=−π=
69. 2 coscos 442
70. 3 sinsin 332
71. tan()tan0 −π=−π=
72. 33 sinsin(1)1 22
73. csccsc2 44
74. () secsec1 −π=π=−
75. 23 secsec 663 ππ
76. 23 csccsc 333 ππ
77. ()() 22 sin40ºcos40º1 +=
78. ()()22 sec18ºtan18º1 −=
79. ()()()() 1 sin80ºcsc80ºsin80º1 sin80º =⋅=
80. ()()()() 1 tan10ºcot10ºtan10º1 tan10º =⋅=
81. ()()()()()sin40º tan40ºtan40ºtan40º0 cos40º −=−=
82. ()()()()()cos20º cot20ºcot20ºcot20º0 sin20º −=−=
83. ()()()() ()() ()() cos400ºsec40ºcos40º360ºsec40º cos40ºsec40º 1 cos40º1 cos40º ⋅=+⋅ =⋅ =⋅=
84. ()()()() ()() ()() tan200ºcot20ºtan20º180ºcot20º tan20ºcot20º 1 tan20º1 tan20º ⋅=+⋅ =⋅ =⋅= 85. sincscsincsc2525 12121212
24 sincsc 121212 sincsc2 1212 sincsc 1212 1 sin1 12 sin 12
86.
92. If cot2 θ=− , then ()() ()() cotcotcot2 2226 θθθ +−π+−π =−+−+−=−
93. sin1ºsin2ºsin3º...sin357º sin358ºsin359º sin1ºsin2ºsin3ºsin(360º3º) sin(360º2º)sin(360º1º) sin1ºsin2ºsin3ºsin(3 ++++ ++ =+++⋅⋅⋅+− +−+− =+++⋅⋅⋅+− () º) sin(2º)sin(1º) sin1ºsin2ºsin3ºsin3ºsin2ºsin1º sin180º0 +−+− =+++⋅⋅⋅−−− ==
87. () ()() () ()() () ()() ()() sin20º tan200º cos380º sin20º tan20º180º cos20º360º sin20º tan20º cos20º tan20ºtan20º0 + =++ + =+ =−+=
sin70º tan70º cos430º
88. () ()() () ()()
sin70º tan70º cos70º360º
sin70º tan70º cos70º tan70ºtan70º0 =− + =− =−=
sin70º tan70º cos430º +− =− () ()() () ()() ()()
89. If sin0.3 θ= , then ()()sinsin2sin4 0.30.30.30.9 θθθ ++π++π =++=
90. If cos0.2 θ= , then ()()coscos2cos4
0.20.20.20.6 θθθ ++π++π =−++=
91. If tan3 θ= , then ()()tantantan2 3339 θθθ ++π++π =++=
94. cos1ºcos2ºcos3ºcos357º cos358ºcos359º +++⋅⋅⋅+ ++ cos1ºcos2ºcos3º...cos(360º3º) cos(360º2º)cos(360º1º) cos1ºcos2ºcos3º...cos(3º) cos(2º)cos(1º) cos1ºcos2ºcos3º...cos3º =++++− +−+− =++++− +−+− =++++ cos2ºcos1º 2cos1º2cos2º2cos3º...2cos178º 2cos179ºcos180º 2cos1º2cos2º2cos3º...2cos(180º2º) ++ =++++ ++ =++++− ()2cos(180º1º)cos180º 2cos1º2cos2º2cos3º...2cos2º 2cos1ºcos180º cos180º1 +−+ =+++− −+ ==−
95. The domain of the sine function is the set of all real numbers.
96. The domain of the cosine function is the set of all real numbers.
97. ()tan f θθ = is not defined for numbers that are odd multiples of 2 π .
98. ()cot f θθ = is not defined for numbers that are multiples of π .
99. ()sec f θθ = is not defined for numbers that are odd multiples of 2 π
Chapter2: Trigonometric Functions
100. ()csc f θθ = is not defined for numbers that are 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.
103. The range of the tangent function is the set of all 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()sinθθ−=− . Its graph is symmetric with respect to the origin.
108. The cosine function is even because cos()cosθθ−= . Its graph is symmetric with respect to the y-axis.
109. The tangent function is odd because tan()tanθθ−=− . Its graph is symmetric with respect to the origin.
110. The cotangent function is odd because cot()cotθθ−=− . Its graph is symmetric with respect to the origin.
111. The secant function is even because sec()secθθ−= . Its graph is symmetric with respect to the y-axis.
112. The cosecant function is odd because csc()cscθθ−=− . Its graph is symmetric with respect to the origin.
113. a. 1 ()() 3 fafa−=−=−
b. ()(2)(4) ()()() 111 1 333 fafafa fafafa ++π++π =++ =++=
114. a. 1 ()() 4 fafa−==
b. ()(2)(2) ()()() 111 444 3 4 fafafa fafafa ++π+−π =++ =++ =
115. a. ()()2 fafa−=−=−
b. ()()(2) ()()() 2226 fafafa fafafa ++π++π =++ =++=
116. a. ()()(3)3 fafa−=−=−−= b. ()()(4) ()()() 3(3)(3) 9 fafafa fafafa ++π++π =++ =−+−+− =−
117. a. ()()4fafa−==−
b. ()(2)(4) ()()() 4(4)(4) 12 fafafa fafafa ++π++π =++ =−+−+− =−
118. a. ()()2 fafa−=−=−
b. ()(2)(4) ()()() 222 6 fafafa fafafa ++π++π =++ =++ =
119. Since 5001 tan 15003 y x θ=== , then 222 9110 10 rxy r =+=+= = 11 sin 1910 θ== + 55 5 1 1 3 3 10 55510 51015.8 minutes T =−+
=−+ =≈
120. a. 1 tan 4 y x θ== for 0 2 θπ<< . 222 16117 17 rxy r =+=+= = Thus, 1 sin 17 θ= 21 ()1 1 1 4 3 4 17 217217 112.75 hours
b. Since 1 tan 4 θ= , 4 x = . Sally heads directly across the sand to the bridge, 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.
121. Let (,)Pxy = be the point on the unit circle that corresponds to an angle t. Consider the equation
tan y ta x == . Then yax = . Now 22 1 xy+= ,
so 222 1 xax+= . Thus, 2 1 1 x a =± + and
2 1 a y a =± + . That is, for any real number a ,
there is a point (,)Pxy = on the unit circle for which tan ta = . In other words, tan t −∞<<∞ , and the range of the tangent function is the set of all real numbers.
122. Let (,)Pxy = be the point on the unit circle that corresponds to an angle t. Consider the equation cot x ta y == . Then x ay = . Now 22 1 xy+= ,
so 222 1 ayy+= . Thus, 2 1 1 y a =± + and
2 1 a x a =± + . That is, for any real number a ,
there is a point (,)Pxy = on the unit circle for which cot ta = . In other words, cot t −∞<<∞ , and the range of the tangent function is the set of all real numbers.
Section2.3: Properties of the Trigonometric Functions
123. Suppose there is a number p, 02 p <<π for which sin()sin p θθ += for all θ . If 0 θ= , then () sin0sinsin00 pp +=== ; so that p =π . If 2 θπ = then sinsin 22 p ππ
But p =π . Thus, 3 sin1sin1 22 ππ =−==
, or 11−= . This is impossible. The smallest positive number p for which sin()sin p θθ += for all θ must then be 2 p =π
124. Suppose there is a number p, 02 p <<π , for which cos()cos p θθ += for all θ . If 2 θπ = , then coscos0 22 p ππ +==
; so that p =π .
If 0 θ= , then ()() cos0cos0 p += . But p =π . Thus ()() cos1cos01 π=−== , or 11−= . This is impossible. The smallest positive number p for which cos()cos p θθ += for all θ must then be 2 p =π
125. 1 sec cos θ θ = : Since cos θ has period 2π , so does sec. θ
126. 1 csc sin θ θ = : Since sin θ has period 2π , so does csc. θ
127. If (,)Pab = is the point on the unit circle corresponding to θ , then (,)Qab =−− is the point on the unit circle corresponding to θ +π . Thus, tan()tan bb aa θθ +π=== . If there exists a number , 0,pp<<π for which tan()tan p θθ += for all θ , then if 0 θ= , ()() tantan00. p == But this means that p is a multiple of π . Since no multiple of π exists in the interval () 0, π , this is impossible. Therefore, the fundamental period of () tan f θθ = is π
Chapter2: Trigonometric Functions
128. 1 cot = tan θ θ : Since tan θ has period π , so does cot θ
129. Let (,)Pab = be the point on the unit circle corresponding to θ . Then 11 csc sin b θ θ == 11 sec cos a θ θ == 11 cot tan a b b a θ θ ===
130. Let (,)Pab = be the point on the unit circle corresponding to θ . Then sin tan cos b a θ θ θ == cos cot sin a b θθ θ ==
131. 222 22222 2222 22 (sincos)(sinsin)cos sincossinsincos sin(cossin)cos sincos 1 θφθφθ
132 – 136. Answers will vary.
137. 44 22 ()33 ()() ()55 xx f xfx xx −++ −=== . Thus, () f x is even.
138. We need to use completing the square to put the function in the form
()() f xaxhk =−+
()21213
2(69)1318 2(69)5 2(3)5 fxxx xx xx xx xx x =−+− =−−− =−−+−+ =−−+−+ =−−++ =−−+
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:
139. (25)32596 316612618 f =−+ =+=+=
So the point (25,18) is on the graph.
140. 32 (0)9(0)3(0)27 27 y =−+− =−
Thus the y-intercept is (0,27) .
Section 2.4 1. 2 3 y x =
Using the graph of 2 y x = , vertically stretch the graph by a factor of 3.
2. 2 y x =
Using the graph of y x = , compress horizontally by a factor of 1 2 .
3. 1; 2 π
4. 3; π
5. 3; 2 63 ππ =
6. True
7. False; The period is 2 2 π π = .
8. True
9. d
10. d
11. a. The graph of sin yx = crosses the y-axis at the point (0, 0), so the y-intercept is 0.
b. The graph of sin yx = is increasing for 22 x ππ −<<
c. The largest value of sin yx = is 1.
d. sin0 when 0,,2 xx==ππ .
e. 3 sin1 when ,; 22 3 sin1 when ,. 22 xx xx ππ ==− ππ =−=−
f. 15711 sin when ,,, 26666 xx ππππ =−=−−
Section2.4: Graphs of the Sine and Cosine Functions
g. The x-intercepts of sin x are {} |, k an integer xxk=π
12. a. The graph of cos yx = crosses the y-axis at the point (0, 1), so the y-intercept is 1.
b. The graph of cos yx = is decreasing for 0 x <<π
c. The smallest value of cos yx = is 1 .
d. 3 cos0 when , 22 xx ππ ==
e. cos1 when 2,0,2; cos1 when ,. xx xx ==−ππ =−=−ππ
f. 31111 cos when ,,, 26666 xx ππππ ==−−
g. The x-intercepts of cos x are () 21 |, k an integer 2 k xx +π
13. 2sin yx =
This is in the form sin()yAx =ω where 2 A = and 1 ω= . Thus, the amplitude is 22 A == and the period is 22 2 1 T ω ππ ===π
14. 3cos y x =
This is in the form cos()yAx =ω where 3 A = and 1 ω= . Thus, the amplitude is 33 A == and the period is 22 2 1 T ω ππ ===π .
15. 4cos(2) yx =−
This is in the form cos()yAx =ω where 4 A =− and 2 ω= . Thus, the amplitude is 44 A =−= and the period is 22 2 T ω ππ ===π
16. 1 sin 2 y x =−
This is in the form sin()yAx =ω where 1 A =− and 1 2 ω= . Thus, the amplitude is 11 A =−= and the period is 1 2
22 4 T ω ππ ===π
Chapter2: Trigonometric Functions
17. 6sin() yx =π
This is in the form sin()yAx =ω where 6 A = and ωπ = . Thus, the amplitude is 66 A == and the period is 22 2 T ω ππ === π
18. 3cos(3)yx =−
This is in the form cos()yAx =ω where 3 A =− and 3 ω= . Thus, the amplitude is 33 A =−= and the period is 22 3 T ω ππ == .
19. 13 cos 22 y x
This is in the form cos()yAx =ω where 1 2 A =− and 3 2 ω= . Thus, the amplitude is 11 22 A =−= and the period is 3 2 224 3 T ω πππ
20. 42 sin 33 yx
This is in the form sin()yAx =ω where 4 3 A = and 2 3 ω= . Thus, the amplitude is 44 33 A == and the period is 2 3 22 3 T ω ππ ===π
21. 5252 sinsin 3333 yxx
This is in the form sin()yAx =ω where 5 3 A =− and 2 3
. Thus, the amplitude
and the
22. 9393 coscos 5252 yxx
This is in the form cos()yAx =ω where 9 5 A = and 3 2 π ω= . Thus, the amplitude is 99 55 A == and the period is 3 2 224 3 T ωπ ππ === 23. F
E
A
I
H
B
C
G
J
D
33. Comparing 4cos y x = to () cos y Ax=ω , we find 4 A = and 1 ω= . Therefore, the amplitude is 44 = and the period is 2 2 1 π =π . Because the amplitude is 4, the graph of 4cos y x = will lie between 4 and 4 on the y-axis. Because the period is 2π , one cycle will begin at 0 x = and end at 2 x =π . We divide the interval [] 0,2π into four subintervals, each of length 2 42 ππ = by finding the following values: 0, 2 π , π , 3 2 π , and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 4cos y x = , we multiply the y-coordinates of the five key points for cos yx = by 4 A = . The five key points are
() 0,4 , ,0 2 π
, () ,4π , 3 ,0 2 π , () 2,4 π
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 [] 4,4 .
34. Comparing 3sin yx = to () sin y Ax=ω , we find 3 A = and 1 ω= . Therefore, the amplitude is 33 = and the period is 2 2 1 π =π . Because the amplitude is 3, the graph of 3sin yx = will lie between 3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at 0 x = and end at 2 x =π . We divide the interval [] 0,2π
into four subintervals, each of length 2 42 ππ = by finding the following values: 0, 2 π , π , 3 2 π , and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 3sin yx = , we multiply the y-coordinates of the five key points for sin yx = by 3 A = . The five key points are () 0,0 , ,3 2 π
, () ,0π , 3 ,3 2 π
, () 2,0 π
We plot these five points and fill in the graph of the curve. We then extend the graph in either
Section2.4: Graphs of the Sine and Cosine Functions
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 .
35. Comparing 4sin yx =− to () sin y Ax=ω , we find 4 A =− and 1 ω= . Therefore, the amplitude is 44−= and the period is 2 2 1 π =π . Because the amplitude is 4, the graph of 4sin yx =− will lie between 4 and 4 on the y-axis. Because the period is 2π , one cycle will begin at 0 x = and end at 2 x =π . We divide the interval [] 0,2π into four subintervals, each of length 2 42 ππ = by
finding the following values: 0, 2 π , π , 3 2 π , 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 4sin yx =− , we multiply the y-coordinates of the five key points for sin yx = by 4 A =− . The five key points are () 0,0 , ,4 2 π , () ,0π , 3 ,4 2 π , () 2,0 π
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.
Chapter2: Trigonometric Functions
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 4,4
36. Comparing 3cos y x =− to () cos y Ax=ω , we find 3 A =− and 1 ω= . Therefore, the amplitude is 33−= and the period is 2 2 1 π =π . Because the amplitude is 3, the graph of 3cos y x =− will lie between 3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at 0 x = and end at 2 x =π . We divide the interval [] 0,2π
into four subintervals, each of length 2 42 ππ = by finding the following values:
0, 2 π , π , 3 2 π , and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 3cos y x =− , we multiply the y-coordinates of the five key points for cos yx = by 3 A =− . The five key points are () 0,3 , ,0 2 π
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 2 5 ( , 3) ( , 3) (0, 3) 5 y x (2 , 3) 3 2 ( , 0) , 0) 2 (
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 3,3 .
37. Comparing ()cos4 y x = to () cos y Ax=ω , we find 1 A = and 4 ω= . Therefore, the amplitude is 11 = and the period is 2 42 ππ = . Because the amplitude is 1, the graph of ()cos4 y x = will lie between 1 and 1 on the y-axis. Because the period is 2 π , one cycle will begin at 0 x = and end at 2 x π = . We divide the interval 0, 2 π
into four subintervals, each of length /2
= by finding the following values: 0, 8 π , 4 π , 3 8 π , and 2 π
These values of x determine the x-coordinates of the five key points on the graph. The five key points are () 0,1 , ,0 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.
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 1,1 .
38. Comparing ()sin3 y x = to () sin y Ax=ω , we find 1 A = and 3 ω= . Therefore, the amplitude is 11 = and the period is 2 3 π . Because the amplitude is 1, the graph of ()sin3 y x = will lie between 1 and 1 on the y-axis. Because the period is 2 3 π , one cycle will begin at 0 x = and end at 2 3 x π = . We divide the interval
0, 3
into four subintervals, each of length 2/3 46 ππ = by finding the following values: 0, 6 π , 3 π , 2 π , and 2 3 π
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.
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 1,1
39. Since sine is an odd function, we can plot the equivalent form ()sin2 yx =−
Comparing ()sin2 y x =− to () sin y Ax=ω , we find 1 A =− and 2 ω= . Therefore, the amplitude is 11−= and the period is 2 2 π =π
Because the amplitude is 1, the graph of ()sin2 y x =− will lie between 1 and 1 on the
y-axis. Because the period is π , one cycle will begin at 0 x = and end at x =π . We divide the interval [] 0, π into four subintervals, each of length 4 π by finding the following values: 0, 4 π , 2 π , 3 4 π , and π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for ()sin2 y x =− , we multiply the y-coordinates of the five key points for sin yx = by 1 A =− .The five key points are () 0,0 , ,1 4 π
, ,0 2 π
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
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 ()cos2 yx = .
Comparing ()cos2 y x = to () cos y Ax=ω , we find 1 A = and 2 ω= . Therefore, the amplitude is 11 = and the period is 2 2 π =π . Because the amplitude is 1, the graph of ()cos2 y x = will lie between 1 and 1 on the y-axis. Because the period is π , one cycle will begin at 0 x = and end at x =π . We divide the interval [] 0, π into four subintervals, each of length 4 π by finding the following values:
0, 4 π , 2 π , 3 4 π , and π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for ()cos2 y x = , we multiply the y-coordinates of the five key points for cos yx = by 1 A = .The five key points are
() 0,1 , ,0 4 π
, ,1 2 π
, 3 ,0 4 π
, () ,1π
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 [] 1,1
41. Comparing 1 2sin 2 y x = to () sin y Ax=ω , we find 2 A = and 1 2 ω= . Therefore, the amplitude is 22 = and the period is 2 4 1/2 π =π
Because the amplitude is 2, the graph of 1 2sin 2 yx = will lie between 2 and 2 on the y-axis. Because the period is 4π , one cycle will begin at 0 x = and end at 4 x =π . We divide the interval [] 0,4π into four subintervals, each of length 4 4 π =π by finding the following values:
0, π , 2π , 3π , and 4π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 2sin 2 yx = , we multiply the y-coordinates of
the five key points for sin yx = by 2 A = . The five key points are () 0,0 , () ,2π , () 2,0 π , () 3,2 π , () 4,0 π
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 1 2cos 4 y x = to () cos y Ax=ω , we find 2 A = and 1 4 ω= . Therefore, the amplitude is 22 = and the period is 2 8 1/4 π =π
Because the amplitude is 2, the graph of 1 2cos 4 yx = will lie between 2 and 2 on the y-axis. Because the period is 8π , one cycle will begin at 0 x = and end at 8 x =π . We divide the interval [] 0,8π into four subintervals, each of length 8 2 4 π =π by finding the following values: 0, 2π , 4π , 6π , and 8π These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 2cos 4 yx = , we multiply the y-coordinates of the five key points for cos yx = by 2 A = .The five key points are () 0,2 , () 2,0 π , () 4,2 π , () 6,0 π , () 8,2 π
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 2 8 4 8 4 y x (0, 2) (2 , 0) (6 , 0) (8 , 2) ( 8 , 2) ( 2 , 0) ( 4 , 2)(4 , 2)
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 2,2
43. Comparing () 1 cos2 2 yx =− to () cos y Ax=ω , we find 1 2 A =− and 2 ω= . Therefore, the amplitude is 11 22 −= and the period is 2 2 π =π
Because the amplitude is 1 2 , the graph of () 1 cos2 2 yx =− will lie between 1 2 and 1 2 on the y-axis. Because the period is π , one cycle will begin at 0 x = and end at x =π . We divide the interval [] 0, π into four subintervals, each of length 4 π by finding the following values:
0, 4 π , 2 π , 3 4 π , and π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for () 1 cos2 2 yx =− , we multiply the y-coordinates of the five key points for cos yx = by 1 2 A =− .The five key points are 1 0, 2
We plot these five points and fill in the graph of the curve. We then extend the graph in either
Section2.4: Graphs of the Sine and Cosine Functions
direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is 11 , 22
44. Comparing 1 4sin 8 y x =−
to () sin y Ax=ω , we find 4 A =− and 1 8 ω= . Therefore, the amplitude is 44−= and the period is 2 16 1/8 π =π . Because the amplitude is 4, the graph of 1 4sin 8 y x =− will lie between 4 and 4 on the y-axis. Because the period is 16π , one cycle will begin at 0 x = and end at 16 x =π . We divide the interval [] 0,16π into four subintervals, each of length 16 4 4 π =π by finding the following values: 0, 4π , 8π , 12π , and 16π These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 4sin 8 y x =− , we multiply the y-coordinates of the five key points for sin yx = by 4 A =− . The five key points are () 0,0 , () 4,4 π , () 8,0 π , () 12,4 π , () 16,0 π We plot these five points and fill in the graph of the curve. We then extend the graph in either
Chapter2: Trigonometric Functions
direction to obtain the graph shown below.
8 16 5 (12 , 4)
(4 , 4) (16 , 0) (8 , 0) (0, 0) 5 y x
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 4,4
45. We begin by considering 2sin yx = . Comparing 2sin yx = to () sin y Ax=ω , we find 2 A = and 1 ω= . Therefore, the amplitude is 22 = and the period is 2 2 1 π =π . Because the amplitude is 2, the graph of 2sin yx = will lie between 2 and 2 on the y-axis. Because the period is 2π , one cycle will begin at 0 x = and end at 2 x =π . We divide the interval [] 0,2π into four subintervals, each of length 2 42 ππ = by finding the following values: 0, 2 π , π , 3 2 π , and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 2sin3yx=+ , we multiply the y-coordinates of the five key points for sin yx = by 2 A = and then add 3 units. Thus, the graph of 2sin3yx=+ will lie between 1 and 5 on the yaxis. The five key points are () 0,3 , ,5 2 π
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 [] 1,5 .
46. We begin by considering 3cos y x = . Comparing 3cos y x = to () cos y Ax=ω , we find 3 A = and 1 ω= . Therefore, the amplitude is 33 = and the period is 2 2 1 π =π . Because the amplitude is 3, the graph of 3cos y x = will lie between 3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at 0 x = and end at 2 x =π . We divide the interval [] 0,2π
into four subintervals, each of length 2 42 ππ = by finding the following values:
0, 2 π , π , 3 2 π , and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 3cos2yx=+ , we multiply the y-coordinates of the five key points for cos yx = by 3 A = and then add 2 units. Thus, the graph of 3cos2yx=+ will lie between 1 and 5 on the y-axis. The five key points are
() 0,5 , ,2 2 π
, () ,1π , 3 ,2 2 π
, () 2,5 π
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 [] 1,5
47. We begin by considering () 5cos yx =π
Comparing () 5cos y x =π to () cos y Ax=ω , we find 5 A = and ωπ = . Therefore, the amplitude is 55 = and the period is 2 2 π π = . Because the amplitude is 5, the graph of () 5cos y x =π will lie between 5 and 5 on the y-axis. Because the period is 2 , one cycle will begin at 0 x = and end at 2 x = . We divide the interval [] 0,2 into four subintervals, each of length 21 42 = by finding the following values: 0, 1 2 , 1, 3 2 , and 2
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for () 5cos3yx π =− , we multiply the y-coordinates of the five key points for cos yx = by 5 A = and then subtract 3 units. Thus, the graph of () 5cos3yx π =− will lie between 8 and 2 on the y-axis. The five key points are () 0,2 , 1 ,3 2
, () 1,8 , 3 ,3 2
, () 2,2
We plot these five points and fill in the graph of the curve. We then extend the graph in either
Section2.4: Graphs of the Sine and Cosine Functions
direction to obtain the graph shown below. 2
3 9 y x 3 2 ( , 3) 1
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 8,2 .
48. We begin by considering 4sin 2 yx π =
.
Comparing 4sin 2 yx π = to () sin y Ax=ω , we find 4 A = and 2 π ω= . Therefore, the amplitude is 44 = and the period is 2 4 /2 π π = . Because the amplitude is 4, the graph of 4sin 2 yx π
= will lie between 4 and 4 on the y-axis. Because the period is 4 , one cycle will begin at 0 x = and end at 4 x = . We divide the interval [] 0,4 into four subintervals, each of length 4 1 4 = by finding the following values: 0, 1, 2, 3, and 4
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 4sin2 2 yx π =− , we multiply the ycoordinates of the five key points for sin yx = by 4 A = and then subtract 2 units. Thus, the graph of 4sin2 2 yx π =− will lie between 6 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
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
49. We begin by considering 6sin 3 yx π =−
Comparing 6sin 3 yx π =−
to () sin y Ax=ω ,
we find 6 A =− and 3 π ω= . Therefore, the amplitude is 66−= and the period is 2 6 /3 π π = .
Because the amplitude is 6, the graph of 6sin 3 y x π = will lie between 6 and 6 on the y-axis. Because the period is 6, one cycle will begin at 0 x = and end at 6 x = . We divide the interval [] 0,6 into four subintervals, each of length 63 42 = by finding the following values: 0, 3 2 , 3, 9 2 , and 6
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 6sin4 3 yx
=−+
, we multiply the ycoordinates of the five key points for sin yx = by 6 A =− and then add 4 units. Thus, the graph of 6sin4 3 yx
=−+
will lie between 2 and 10 on the y-axis. The five key points are
() 0,4 , 3 ,2 2
, () 3,4 , 9 ,10 2
, () 6,4
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 [] 2,10
50. We begin by considering 3cos 4 yx π
=−
Comparing 3cos 4 yx π
=− to () cos y Ax=ω ,
we find 3 A =− and 4 π ω= . Therefore, the amplitude is 33−= and the period is 2 8 /4 π π =
Because the amplitude is 3, the graph of 3cos 4 yx π =− will lie between 3 and 3 on the y-axis. Because the period is 8, one cycle will begin at 0 x = and end at 8 x = . We divide the interval [] 0,8 into four subintervals, each of length 8 2 4 = by finding the following values: 0, 2, 4, 6, and 8
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 3cos2 4 yx π =−+ , we multiply the ycoordinates of the five key points for cos yx = by 3 A =− and then add 2 units. Thus, the graph
of 3cos2 4 yx π =−+ will lie between 1 and 5 on the y-axis. The five key points are () 0,1 , () 2,2 , () 4,5 , () 6,2 , () 8,1
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 1,5
51. ()() 53sin23sin25yxx =−=−+
We begin by considering ()3sin2 yx =−
Comparing ()3sin2 yx =− to () sin y Ax=ω , we find 3 A =− and 2 ω= . Therefore, the amplitude is 33−= and the period is 2 2 π =π
Because the amplitude is 3, the graph of ()3sin2 yx =− will lie between 3 and 3 on the y-axis. Because the period is π , one cycle will begin at 0 x = and end at x =π . We divide the interval [] 0, π into four subintervals, each of length 4 π by finding the following values:
0, 4 π , 2 π , 3 4 π , and π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for () 3sin25yx=−+ , we multiply the ycoordinates of the five key points for sin yx = by 3 A =− and then add 5 units. Thus, the graph of () 3sin25yx=−+ will lie between 2 and 8
Section2.4: Graphs of the Sine and Cosine Functions
on the y-axis. 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.
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 2,8
52. ()() 24cos34cos32yxx =−=−+
We begin by considering ()4cos3 yx =− . Comparing ()4cos3 yx =− to () cos y Ax=ω , we find 4 A =− and 3 ω= . Therefore, the amplitude is 44−= and the period is 2 3 π
Because the amplitude is 4, the graph of ()4cos3 yx =− will lie between 4 and 4 on the y-axis. Because the period is 2 3 π , one cycle will begin at 0 x = and end at 2 3 x π = . We
divide the interval 2 0, 3 π into four subintervals, each of length 2/3 46 ππ = by finding the following values: 0, 6 π , 3 π , 2 π , and 2 3 π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for () 4cos32yx=−+ , we multiply the y-
Chapter2: Trigonometric Functions
coordinates of the five key points for cos yx = by 4 A =− and then adding 2 units. Thus, the graph of () 4cos32yx=−+ will lie between 2 and 6 on the y-axis. The five key points are () 0,2 , ,2 6 π
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. 33
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 52 sin 33 yx π =−
. Comparing 52 sin 33 yx π
to () sin y Ax=ω , we find 5 3 A =− and 2 3 π ω= .
Therefore, the amplitude is 55 33 −= and the period is 2 3 2/3 π π = . Because the amplitude is 5 3 , the graph of 52 sin 33 yx π =−
will lie between 5 3 and 5 3 on the y-axis. Because the period is 3 , one cycle will begin at 0 x = and end at 3 x = . We divide the interval [] 0,3 into four subintervals, each of length 3 4 by finding the following values:
0, 3 4 , 3 2 , 9 4 , and 3
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 52 sin 33 yx π =− , we multiply the ycoordinates of the five key points for sin yx = by 5 3 A =− .The five key points are () 0,0 , 35 , 43
, 3 ,0 2
, 95 , 43 , () 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.
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is 55 , 33
54. Since cosine is an even function, we consider the equivalent form 93 cos 52 yx π =
. Comparing
=
cos 52 yx π
to () cos y Ax=ω , we find 9 5 A = and 3 2 π ω= . Therefore, the amplitude is 99 55 = and the period is 24 3/23 π π = . Because the amplitude is 9 5 , the graph of
will lie between 9 5 and 9 5
on the y-axis. Because the period is 4 3 , one cycle will begin at 0 x = and end at 4 3 x = . We divide the interval 4 0, 3 into four subintervals, each of length 4/31 43 = by finding the following values:
0, 1 3 , 2 3 , 1 , and 4 3
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 93 cos 52 yx π =
, we multiply the y-coordinates of the five key points for cos yx = by 9 5 A =
Thus, the graph of 93 cos 52 yx π
=−
will lie between 9 5 and 9 5 on the y-axis. The five key points are 9 0, 5
, 1 ,0 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.
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is 99 , 55
.
Section2.4: Graphs of the Sine and Cosine Functions
55. We begin by considering 3 cos 24 yx π
=−
Comparing 3 cos 24 yx π =− to () cos y Ax=ω , we find 3 2 A =− and 4 π ω=
Therefore, the amplitude is 33 22 −= and the period is 2 8 /4 π π = . Because the amplitude is 3 2 , the graph of 3 cos 24 yx π =−
will lie between 3 2 and 3 2 on the y-axis. Because the period is 8, one cycle will begin at 0 x = and end at 8 x = . We divide the interval [] 0,8 into four
subintervals, each of length 8 2 4 = by finding the following values: 0, 2, 4, 6, and 8
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 31 cos 242yx π =−+ , we multiply the ycoordinates of the five key points for cos yx = by 3 2 A =− and then add 1 2 unit. Thus, the graph of 31 cos 242yx π =−+ will lie between 1 and 2 on the y-axis. The five key points are () 0,1 , 1 2, 2 , () 4,2 , 1 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. 2 2 8 48 4 y x (0, 1) (8, 1) ( 8, 1) (6, ) (4, 2) ( 4, 2) 1 2 (2, ) 1 2
Chapter2: Trigonometric Functions
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 1,2
56. We begin by considering 1 sin 28 yx π =−
Comparing 1 sin 28 yx π =− to () sin y Ax=ω ,
we find 1 2 A =− and 8 π ω= . Therefore, the amplitude is 11 22 −= and the period is 2 16 /8 π π = . Because the amplitude is 1 2 , the graph of 1 sin 28 yx π =− will lie between 1 2 and 1 2 on the y-axis. Because the period is 16, one cycle will begin at 0 x = and end at 16 x = We divide the interval [] 0,16 into four
subintervals, each of length 16 4 4 = by finding the following values: 0, 4, 8, 12, and 16
These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 13 sin 282yx π =−+ , we multiply the ycoordinates of the five key points for sin yx = by 1 2 A =− and then add 3 2 units. Thus, the graph of 13 sin 282yx π =−+ will lie between 1 and 2 on the y-axis. The five key points are 3 0, 2
, () 4,1 , 3 8, 2 , () 12,2 , 3 16, 2
2.5 y x (16, ) (12, 2) ( 4, 2) (4, 1) 3 2 (8, ) 3 2 (0, ) 3 2
From the graph we can determine that the domain is all real numbers, () , −∞∞ and the range is [] 1,2
57. 22 3;;2AT T ω ππ ==π=== π 3sin(2)yx =±
58. 221 2;4; 42 AT T ω ππ ==π=== π 1 2sin 2 yx =±
59. 22 3;2; 2 AT T ω ππ =====π 3sin()yx =±π
60. 22 4;1;2 1 AT T ω ππ =====π 4sin(2) yx =±π
61. The graph is a cosine graph with amplitude 5 and period 8.
Find ω : 2 8 82 2 84 ω ω ω π = =π ππ ==
The equation is: 5cos 4 yx π =
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 8 12 16481216 0.5
62. The graph is a sine graph with amplitude 4 and period 8π.
Find
63. The graph is a reflected cosine graph with amplitude
67. The graph is a reflected sine graph with amplitude 1 and period 4 3 π
Find
64. The graph is a reflected sine graph with amplitude 2 and period 4.
The equation is: 3 sin 2 yx
68. The graph is a reflected cosine graph with amplitude π and period 2π.
Find
:
The equation is: cos yx =−π
69. The graph is a reflected cosine graph, shifted up 1 unit, with amplitude 1 and period 3 2
65. The graph is a sine graph with amplitude 3
and period 1.
66. The graph is a reflected cosine graph with amplitude 5 2 and period 2. Find
The
The equation is: 4 cos1 3 yx
70. The graph is a reflected sine graph, shifted down 1 unit, with amplitude 1 2 and period 4 3 π
Find ω :
The equation is: 13sin1 22 yx
Chapter2: Trigonometric Functions
71. The graph is a sine graph with amplitude 3 and period 4.
Find
72. The graph is a reflected cosine graph with amplitude 2 and period 2.
Find
The equation is: 2cos() yx =−π
73. The graph is a reflected cosine graph with amplitude 4 and period 2 3 π
Find
74. The graph is a sine graph with amplitude 4 and period π.
Find
The average rate of change is 2 π
76. ()()()() /20cos/2cos0 /20/2 012 /2 ffππ ππ ππ = ==− The average rate of change is 2 π 77. ()() ()() 11 sinsin0 /20 222 /20/2 sin/4sin0 /2 2 222 2 /22 ff π π ππ π π πππ
The average rate of change is 2 π
78. () cos2cos(20) /2(0) 2 /20/2 cos()cos(0)11 /2/2 24 2 ff π π ππ π ππ ππ
The average rate of change is 4 π .
79. () ()() sin4 f gxx = () (())4sin4sin g fxxx ==
80. 1 (())cos 2 f gxx =
() 11 (())coscos 22 g fxxx ==
81. () (())2cos2cos f gxxx =−=−
()(())cos2 g fxx =−
82. () (())3sin3sin f gxxx =−=−
Section2.4: Graphs of the Sine and Cosine Functions
()(())sin3 g fxx =−
85. () 220sin(60),0Ittt=π≥ 221
Period: second 6030
Amplitude:220220 amperes T A ω ππ === π ==
Chapter2: Trigonometric Functions
86. ()120sin(30),0 Ittt=π≥
221
Period: second 3015
Amplitude:120120 amperes T A ω ππ === π ==
87. ()220sin(120) Vtt =π
c. VIR = 120sin(120)20 6sin(120) ()6sin(120) tI tI I tt π= π= =π
d. Amplitude:66 amperes A ==
221
Period: second 12060 T ω ππ === π
a. Amplitude: 220220 volts
221
Period: second 12060 A T ω == ππ === π
b, e.
c. VIR = 220sin(120)10 22sin(120) ()22sin(120) tI tI I tt π π π = = =
d. Amplitude:2222 amperes A ==
221
Period: second 12060 T ω ππ === π
88. ()120sin(120) Vtt =π
a. Amplitude:120120 volts
221 Period: second 12060 A T ω == ππ === π
b, e.
89. a. [] () () () 2 2 0 22 0 2 2 0 () () sin2 sin2 sin2 Vt Pt R Vft R Vft R V ft R = π
= π = =π
b. The graph is the reflected cosine graph translated up a distance equivalent to the amplitude. The period is 1 2 f , so 4 f ω=π .
The amplitude is 22 00 1 22 VV RR ⋅= .
The equation is: ()() () 22 00 2 0 cos4 22 1cos4 2 VV PtftRR V ft R =−π+ =−π
c. Comparing the formulas: ()() () 2 1 sin21cos4 2 ftft π=−π
90. a. Since the tunnel is in the shape of one-half a sine cycle, the width of the tunnel at its base is one-half the period. Thus, 2 2(28)56 T π ω === or 28 π ω=
The tunnel has a maximum height of 15 feet so we have 15 A = . Using the form sin()yAx =ω , the equation for the sine curve that fits the opening is 15sin 28 x y π = .
b. Since the shoulders are 7 feet wide and the road is 14 feet wide, the edges of the road correspond to 7 x = and 21 x =
The tunnel is approximately 10.6 feet high at the edge of the road.
91. a. Physical potential:
b.
c. No.
d.
Physical potential peaks at 15 days after the 20th birthday, with minimums at the 3rd and 26th days. Emotional potential is 50% at the 17th day, with a maximum at the 10th day and a minimum at the 24th day. Intellectual potential starts fairly high, drops to a minimum at the 13th day, and rises to a maximum at the 29th day. 92. cos, 22yxxππ=−≤≤
94. Answers may vary.
95. Answers may vary.
511151 ,,,,,,, 32323232
96. 2sin2 sin1 x x =− =− Answers may vary.
3711 ,2,,2,,2,,2 2222
97. Answers may vary.
359 ,1,,1,,1,,1 4444 ππππ
98 – 102. Answers will vary.
93.
Chapter2: Trigonometric Functions
103. 22 222 222 2 ()() ()5()1(51) (2)(55)151 255151 25(25) 25 fxhfx h xhxhxx h x xhhxhxx h xxhhxhxx h xhhhhxh xh hh +− +−++−−+ = ++−++−+−
104. 115 44 52 sin 42 ππ
105. The y-intercept is: 3021 615 y y =+− =−= () 0,5
The x-interecpts are: 0321 1 1322 3 11 2 or 2 33 57 or 33 x xx xx xx =+− =+→=+ =+−=+ =−=− 57 33,0,,0
106. () 2 2 2 40 1809 1 2 12 5 29 25 sq. units 9 Ar ππ
Section 2.5
1. 4 x =
2. True
3. origin; x = odd multiples of 2 π
4. y-axis; x = odd multiples of 2 π
5. b
6. True
7. The y-intercept of tan y x = is 0.
8. cot y x = has no y-intercept.
9. The y-intercept of sec yx = is 1.
10. csc yx = has no y-intercept.
11. sec1 when 2,0,2; sec1 when , xx xx ==−ππ =−=−ππ
12. 3 csc1 when ,; 22 3 csc1 when , 22 xx xx ππ ==− ππ =−=−
13. sec yx = has vertical asymptotes when 33 ,,, 2222 x ππππ =−−
14. csc yx = has vertical asymptotes when 2,,0,,2 x =−π−πππ
15. tan y x = has vertical asymptotes when 33 ,,, 2222 x ππππ =−− .
16. cot y x = has vertical asymptotes when 2,,0,,2 x =−π−πππ
17. 3tan y x = ; The graph of tan y x = is stretched vertically by a factor of 3.
Section2.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
The domain is , is an odd integer 2
The range is the set of all real number or (,) −∞∞ .
18. 2tan y x =− ; The graph of tan y x = is stretched vertically by a factor of 2 and reflected about the x-axis.
The domain is , is an odd integer 2
The range is the set of all real number or (,) −∞∞
19. 4cot y x = ; The graph of cot y x = is stretched vertically by a factor of 4.
The domain is {} , is an integer xxkk ≠π . The range is the set of all real number or (,) −∞∞
20. 3cot y x =− ; The graph of cot y x = is stretched vertically by a factor of 3 and reflected about the x-axis.
The domain is {} , is an integer xxkk ≠π . The range is the set of all real number or (,) −∞∞ .
21. tan 2 yx π
=
; The graph of tan y x = is horizontally compressed by a factor of 2 π
The domain is {} does not equal an odd integer xx The range is the set of all real number or (,) −∞∞
22. 1 tan 2 yx =
; The graph of tan y x = is horizontally stretched by a factor of 2.
The domain is {} , is an integer xxkk ≠π . The range is the set of all real number or (,) −∞∞ .
23. 1 cot 4 yx = ; The graph of cot y x = is horizontally stretched by a factor of 4.
Chapter2: Trigonometric Functions
The domain is {} 4, is an integer xxkk ≠π . The range is the set of all real number or (,) −∞∞ .
24. cot 4 yx π =
; The graph of cot y x = is horizontally stretched by a factor of 4 π
The domain is {} 4, is an integer xxkk ≠ . The range is the set of all real number or (,) −∞∞
25. 2sec y x = ; The graph of sec yx = is stretched vertically by a factor of 2.
The domain is , is an odd integer 2 k xxk π
The range is {} 2 or 2 yyy≤−≥
26. 1 csc 2 yx = ; The graph of csc yx = is vertically compressed by a factor of 1 2
The domain is {} , is an integer xxkk ≠π . The range is 11 or 22yyy
27. 3csc y x =− ; The graph of csc yx = is vertically stretched by a factor of 3 and reflected about the x-axis.
The domain is {} , is an integer xxkk ≠π . The range is {} 3 or 3 yyy≤−≥
28. 4sec y x =− ; The graph of sec yx = is vertically stretched by a factor of 4 and reflected about the x-axis.
The domain is , is an odd integer 2 k xxk π
The range is {} 4 or 4 yyy≤−≥
29. 1 4sec 2 yx =
; The graph of sec yx = is horizontally stretched by a factor of 2 and
Section2.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
vertically stretched by a factor of 4.
The domain is {} , is an odd integer xxkk ≠π
The range is {} 4 or 4 yyy≤−≥ .
30. () 1 csc2 2 yx = ; The graph of csc yx = is horizontally compressed by a factor of 1 2 and vertically compressed by a factor of 1 2 .
The domain is , is an integer 2 k xxk
. The range is 11 or 22yyy
31. () 2csc yx =−π ; The graph of csc yx = is horizontally compressed by a factor of 1 π , vertically stretched by a factor of 2, and reflected about the x-axis.
The domain is {} does not equal an integer xx
The range is {} 2 or 2 yyy≤−≥
32. 3sec 2 yx π
=−
; The graph of sec yx = is horizontally compressed by a factor of 2 π , vertically stretched by a factor of 3, and reflected about the x-axis.
The domain is {} does not equal an odd integer xx The range is {} 3 or 3 yyy≤−≥
Copyright © 2016 Pearson Education, Inc.
Chapter2: Trigonometric Functions
33. 1 tan1 4 yx =+
; The graph of tan y x = is horizontally stretched by a factor of 4 and shifted up 1 unit.
The domain is {} 2, is an odd integer xxkk ≠π . The range is the set of all real number or (,) −∞∞
34. 2cot1 y x =− ; The graph of cot y x = is vertically stretched by a factor of 2 and shifted down 1 unit.
The domain is {} , is an integer xxkk ≠π . The range is the set of all real number or (,) −∞∞ .
35. 2 sec2 3 yx π
=+
; The graph of sec yx = is horizontally compressed by a factor of 3 2π and shifted up 2 units.
The domain is 3 , is an odd integer 4 xxkk
The range is {} 1 or 3 yyy≤≥
36. 3 csc 2 yx π
=
; The graph of csc yx = is horizontally compressed by a factor of 2 3π .
The domain is 2 , is an integer 3 xxkk
. The range is {} 1 or 1 yyy≤−≥ .
Section2.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
37. 11 tan2 24 yx
=−
; The graph of tan y x = is horizontally stretched by a factor of 4, vertically compressed by a factor of 1 2 , and shifted down 2 units.
The domain is {} 2, is an odd integer xxkk ≠π . The range is the set of all real number or (,) −∞∞ .
38. 1 3cot2 2 yx
=−
; The graph of cot y x = is horizontally stretched by a factor of 2, vertically stretched by a factor of 3, and shifted down 2 units.
The domain is {} 2, is an integer xxkk ≠π . The range is the set of all real number or (,) −∞∞ .
39. 1 2csc1 3 yx
=−
; The graph of csc yx = is horizontally stretched by a factor of 3, vertically stretched by a factor of 2, and shifted down 1 unit.
The domain is {} 3, is an integer xxkk ≠π
The range is {} 3 or 1 yyy≤−≥
40. 1 3sec1 4 yx
=+
; The graph of sec yx = is horizontally stretched by a factor of 4, vertically stretched by a factor of 3, and shifted up 1 unit.
The domain is {} 2, is an odd integer xxkk ≠π
The range is {} 2 or 4 yyy≤−≥ .
41. ()()() 3 0 0 tan/6tan0 6 3 /6/6 0 6 3623 3 ff π π πππ ππ
The average rate of change is 23 π
Chapter2: Trigonometric Functions
45. ()(())tan4 f gxx = () (())4tan4tan g fxxx ==
The average rate of change is 6
1 (())2sec 2 f gxx = () 1 (())2secsec 2 g fxxx == 47. () (())2cot2cot f gxxx =−=−
Section2.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
48. () 1 (())2csccsc 2 f gxxx == 1 (())2csc 2 g fxx = 49.
© 2016 Pearson Education, Inc. ()(())cot2 g fxx =−
51. a. Consider the length of the line segment in two sections, x, the portion across the hall that is 3 feet wide and y, the portion across that hall that is 4 feet wide. Then, 3 cos 3 cos x x θ θ = = and 4 sin 4 sin y y θ θ = = Thus, 34 3sec4csc cossin Lxy
b. Let 1 34 cossin Y x x =+ 0 0 25 2 π
c. Use MINIMUM to find the least value: 0 0 25 2 π L is least when 0.83 θ≈ d. ()() 34 9.86 feet cos0.83sin0.83 L ≈+≈ .
Note that rounding up will result in a ladder that won’t fit around the corner. Answers will vary.
52. a. () 10tan() dtt =π
b. () 10tan() dtt =π is undefined at 1 2 t = and 3 2 t = , or in general at is an odd integer 2 k tk =
. At these instances, the length of the beam of light approaches infinity. It is at these instances in the rotation of the beacon when the beam of light being cast on the wall changes from one side of the beacon to the other.
c. () 10tan() 00
0.13.2492
0.27.2654
0.313.764
0.430.777 t dtt =π
53. Yes, the two functions are equivalent.
54. 2 4 yx=−
Test x-axis symmetry: Let yy =− ()2 2 4 4 same yx yx −=− =−
d. (0.1)(0)3.24920 32.492
0.100.10 dd =≈
0.20.10.20.1 dd =≈
(0.2)(0.1)7.26543.2492 40.162
0.30.20.30.2 dd =≈
(0.3)(0.2)13.7647.2654 64.986
0.40.30.40.3 dd =≈
(0.4)(0.3)30.77713.764 170.13
e. The first differences represent the average rate of change of the beam of light against the wall, measured in feet per second. For example, between 0 t = seconds and 0.1 t = seconds, the average rate of change of the beam of light against the wall is 32.492 feet per second.
Test y-axis symmetry: Let x x =− 2 4 yx=−− different
Test origin symmetry: Let x x =− and yy =− ()2 2 4 4 different yx yx −=−− =−−
Therefore, the graph will have x-axis symmetry.
55. 222 222 22 ()() ((3))(1)(3) (3)(1)9 x hykr xy xy −+−= −−+−= ++−=
56. 2 46 (4) (4)10
57. The relation is a circle with center (0,4) and radius 4 so it is not a function.
Section 2.6
1. phase shift
2. False
3. 4sin(2)yx=−π Amplitude:44
Section2.6: Phase Shift; Sinusoidal Curve Fitting
4. 3sin(3)yx=−π Amplitude:33
9. 4sin(2)5yx=π+−
Amplitude:44
Subinterval
Section2.6: Phase Shift; Sinusoidal Curve Fitting
11. 3cos(2)5yx=π−+
Amplitude:33
Period:2
10. 2cos(24)4yx=π++
Amplitude:22
Interval
Subinterval
Key
15.
Assuming
Assuming
Assuming
19. ()2tan4 yx π =−
Begin with the graph of tan y x = and apply the following transformations:
1) Shift right π units () tan yx π
2) Horizontally compress by a factor of 1 4
3) Vertically stretch by a factor of 2
20. () 1 cot2 2 yx π =−
Begin with the graph of cot y x = and apply the following transformations:
1) Shift right π units () cot yx π =−
2) Horizontally compress by a factor of 1 2
cot2 yx
3) Vertically compress by a factor of 1 2
Chapter2: Trigonometric Functions
21. 3csc2 4 yx
Begin with the graph of csc yx = and apply the following transformations:
1) Shift right 4
2)
3)
Begin with the graph of sec yx = and apply the following transformations:
2) Horizontally compress by a factor of 1 3
sec3 yx
3) Vertically compress by a factor of 1 2
Begin with the graph of cot y x = and apply the following transformations:
1) Shift left 2
units cot 2
2) Horizontally compress by a factor of 1 2
3) Reflect about the x-axis cot2 2 yx
Begin with the graph of tan y
and apply the following transformations:
1) Shift
2) Horizontally compress
3) Reflect about
25. ()sec2 yxππ=−+
Begin with the graph of sec yx = and apply the following transformations:
1) Shift left
2) Horizontally
3)
Begin with the graph of csc yx = and apply the following transformations:
1) Shift left 4
units csc
2) Reflect about the y-axis csc
3) Horizontally compress by a factor of
3) Reflect about the
Period: second
Amplitude:120120 amperes
28. () 220sin60,0 6 Ittt
29.
Section2.6: Phase Shift; Sinusoidal Curve Fitting
d. () 21.68sin0.5162.12457.81yx =−+ / / e. 20 0 90 13 31. a. b.
75.428.147.3 Amplitude:23.65 22 A ===
75.4+28.1103.5
24.25sin(0.4931.927)51.61yx =−+
Chapter2: Trigonometric Functions
c.
d. 21.73sin(0.5182.139)54.82yx =−+
e. 20 0 80 13
33. a. 6.5 + 12.4167 = 18.9167 hours which is at 6:55 PM.
b. ()5.860.38 6.24 Ampl:3.12 22 A ===
5.86(0.38)5.48
34. a. 0.10 + 12.4167 = 12.5167 hours which is at 12:31 PM.
b. 9.97(0.59)9.38 Ampl:4.69 22 A === 9.97(0.59)10.56 Vertical Shift:5.28 22 224 12.41676.20835149
Phase Shift (use 9.97,0.10):yx
πππ ω + == === == 24
9.974.69sin0.105.28 149 24 4.694.69sin0.10 149 2.4 1sin 149 2.4 2149 1.5202 π φ π
Thus, 24 4.69sin1.52025.28 149 yx π =++ or () 24 4.69sin3.00425.28 149 yx π =++
Vertical Shift:2.74 22
224
ω +− ==
12.41676.20835149
Phase Shift (use 5.86,6.5):yx
== 24 5.863.12sin6.52.74 149 24 3.123.12sin6.5 149 156 1sin 149 156 2149 1.7184 π φ π φ π
Thus, 24 3.12sin1.71842.74 149 yx π =−+ or () 24 3.12sin3.39592.74 149 yx π =−+
c. () 24 3.12sin151.71842.74 149 1.49 feet y π =−+ ≈
c. () 24 4.69sin181.52025.28 149 0.90 feet y π =++ ≈
35. a. 13.7510.52 Amplitude:1.615 2 A == 13.7510.52 Vertical Shift:12.135 2 2 365
Phase Shift (use 13.75,172):yx ω + = π = == 2 13.751.615sin17212.135 365 2 1.6151.615sin172 365 344 1sin 365 344 2365 1.3900 φ
d.
actual
Chapter2: Trigonometric Functions
d. The actual hours of sunlight on April 1, 2014 was 13.37 hours. This is close to the predicted amount of 13.63 hours.
38. a. 13.4210.83 Amplitude:1.295 2
d. The actual hours of sunlight on April 1, 2014 were 12.38 hours. This is very close to the predicted amount of 12.35 hours.
39 – 40. Answers will vary.
41. () 32 23814fxxxx=−−+ on the interval () 4,5 Use MAXIMUM and MINIMUM on the graph of 32 1 23814yxxx=−−+ .
(0.76)17.47
(1.76)1.53
16(2)5(2)16(1)5(1) 2(1) 16(4)10165
c.
44. 2 2 95 4 2 x x x −=− = =±
But only the 2 is in the domain of the part of the piecewise function.
375 312 4 x x x −+=− −=− =
Thus the values are {} 2,4
Chapter 2 Review Exercises
1. 3 135135 radian radians 1804 ππ°=⋅=
2. 1818 radian radian 18010 ππ°=⋅=
3. 33180 degrees135 44 ππ=⋅=° π
4. 55180 degrees450 22 ππ −=−⋅=−° π
5. 11 tansin1 4622 ππ −=−=
6. 3sin45º4tan34233243 62323 π −=⋅−⋅=−
7. () 32 6cos2tan623 432 3223
8. 55 seccotseccot213 3434 ππππ
9. tansin000 π+π=+=
10. cos540ºtan(405º)1(1) 110 −−=−−−
11. 222 2 1 sin20ºsin20ºcos20º1 sec20º +=+=
12. 1 sec50ºcos50ºcos50º1 cos50º ⋅=⋅=
13. cos(40º)cos40º 1 cos40ºcos40º ==
14. sin(40º)sin40º 1 sin40ºsin40º ==−
15. () () sin400ºsec50ºsin400ºsec50º 1 sin40º360º cos50º sin40ºsin40º cos50ºsin(90º50º) sin40º 1 sin40º ⋅−=⋅ =+⋅ == ==
16. 4 sin 5 θ= and 0 2 θπ<< , so θ lies in quadrant I.
Using the Pythagorean Identities: 22 2 2 cos1sin 4169 cos11 52525 93 cos 255 θθ θ θ =−
=−=−=
=±=±
Note that cos θ must be positive since θ lies in quadrant I. Thus, 3 cos 5 θ=
4 5 3 5 sin454 tan cos533 θ θ θ = ==⋅=
4 5 1155 csc1 sin44 θ θ = ==⋅=
3 5 1155 sec1 cos33 θ θ = ==⋅=
4 3 1 1133 cot tan44 θ θ = ==⋅=
17. 12 tan 5 θ= and sin0 θ< , so θ lies in quadrant III.
Using the Pythagorean Identities:
Chapter2: Trigonometric Functions
19. 12 sin 13 θ= and θ lies in quadrant II.
Using the Pythagorean Identities: 22
Note that sec θ must be negative since θ lies in quadrant III. Thus, 13
18. 5 sec 4 θ=− and tan0 θ< , so θ lies in quadrant II.
Using the Pythagorean Identities:
Note that cos θ must be negative because θ lies in quadrant II. Thus, 5 cos 13 θ=−
20. 5 sin 13 θ=− and 3 2 2 πθ<<π (quadrant IV)
Using the Pythagorean Identities: 22 2 2 cos1sin 525144 cos11 13169169 14412 cos 16913 θθ
Note that cos θ must be positive because θ lies in quadrant IV. Thus, 12 cos 13 θ= .
21. 1 tan 3 θ= and 180º270º θ << (quadrant III)
Using
Pythagorean Identities:
Note that sec θ must be negative since θ lies in quadrant III. Thus, 10 sec 3 θ=−
22. sec3 θ= and 3 2 2 πθ<<π (quadrant IV)
Using the Pythagorean Identities: 22
Note that tan θ must be negative since θ lies in quadrant IV. Thus,. tan22 θ=− 11 cos sec3 θ θ == sin tan cos θ θ θ = , so ()() 122 sintancos22 33 θθθ
23. cot2 θ=− and 2 πθ<<π (quadrant II)
Using the Pythagorean Identities: () 22 2 2 csc1cot csc12145 csc5 θθ θ θ =+ =+−=+= =±
Note that csc θ must be positive because θ lies in quadrant II. Thus, csc5 θ= 1155 sin csc5 55 θ θ ==⋅= cos cot sin θ θ θ = , so ()() 525 coscotsin2 55 θθθ
24. 2sin(4)yx =
The graph of sin yx = is stretched vertically by a factor of 2 and compressed horizontally by a factor of 1 4
Domain: () , −∞∞
Range: [] 2,2
Chapter2: Trigonometric Functions
25. 3cos(2) yx =−
The graph of cos yx = is stretched vertically by a factor of 3, reflected across the x-axis, and compressed horizontally by a factor of 1 2
Domain: () , −∞∞
Range: [] 3,3
26. tan()yx=+π
The graph of tan y x = is shifted π units to the left.
Domain:
Range: () , −∞∞
27. 2tan(3)yx =−
The graph of tan y x = is stretched vertically by a factor of 2, reflected across the x-axis, and compressed horizontally by a factor of 1 3
Domain: |,k is an integer
Range: () ,
The graph of cot y x = is shifted 4 π units to the left.
Domain: |,k is an integer
Range: ()
29. ()4sec2 y x =
The graph of sec yx = is stretched vertically by a factor of 4 and compressed horizontally by a factor of 1 2 . 4 3 2 1 2 3 4 5 x y ( , 4) 2 ( , 4) 2 (0, 4) ( , 4) ( , 4) 4 3
Domain: |,k is an odd integer 4
Range: {} |4 or 4 yyy≤−≥
30. csc 4 yx π
=+
The graph of csc yx = is shifted 4 π units to the left.
Domain: |,k is an integer 4
Range: {} |1 or 1 yyy≤−≥
31. () 4sin242yx=+−
The graph of sin yx = is shifted left 4 units, compressed horizontally by a factor of 1 2 , stretched vertically by a factor of 4, and shifted down 2 units.
Domain: () , −∞∞
Range: [] 6,2
32. 5cot 34 x y π
=−
The graph of cot y x = is shifted right 4 π units, stretched horizontally by a factor of 3, and stretched vertically by a factor of 5.
Domain: 3 |3,k is an integer 4 xxk π
Range: () , −∞∞
33. sin(2)yx =
Amplitude = 1 1 = ; Period = 2 2 π =π
34. 2cos(3) yx =−π
Amplitude = 22−= ; Period = 22 33 π π =
Chapter2: Trigonometric Functions
35. 4sin(3)yx = Amplitude:44
36. 1 cos 22yx
39. The graph is a cosine graph with amplitude 5 and period 8π.
40. The graph is a reflected sine graph with amplitude 7 and period 8.
41. Set the calculator to radian mode: sin0.38 8 π ≈ .
42. Set the calculator to degree mode: o o 1 sec101.02 cos10 =≈
43. Terminal side of θ in quadrant III implies sin0csc0 cos0sec0
44. cos0, tan0θθ>< ; θ lies in quadrant IV.
45.
46. The point (2,5) P =− is on a circle of radius 22 (2)542529 r =−+=+= with the center at the origin. So, we have
47. The domain of sec yx = is odd multiple of 2
The range of sec yx = is {} 1 or 1. yyy≤−≥
The period is 2π .
48. a. oo 2035 3220'35"3232.34 603600 =++≈
b. 63.18o o 0.18(0.18)(60')10.8' 0.8'(0.8)(60")48" == == Thus, oo 63.186310'48" =
49. 2 feet,30º or 6 r θθπ=== () 2 2 21.047 feet 63 11 21.047 square feet 2263 sr Ar θ θ ππ ==⋅=≈
50. In 30 minutes: 8 inches,180º or r θθ ===π 8825.13 inches srθ ==⋅π=π≈ In 20 minutes: 2 8 inches,120º or 3 r θθπ=== 216 816.76 inches 33 srθππ ==⋅=≈
51. 180 mi/hr v = ; 1 mile 2 1 0.25 mile 4 d r = == 180 mi/hr 0.25 mi 720 rad/hr
720 rad1 rev hr2 rad 360 rev hr 114.6 rev/hr v r ω== = =⋅ π = π ≈
52. Since there are two lights on opposite sides and the light is seen every 5 seconds, the beacon makes 1 revolution every 10 seconds: 1 rev2 radians radians/second 10 sec1 rev5 ω ππ =⋅=
Chapter2: Trigonometric Functions
53. ()220sin30,0 6 Ittt π =π+≥
a. 21 Period 3015 π == π
b. The amplitude is 220.
c. The phase shift is: 11 6 30630180 φ ω π π
54. a.
b. 955540 Amplitude:20 22 A === 95+55150 Vertical Shift:75 22 2 126
==
== Phase shift (use 55,1):yx== 5520sin175 6 2020sin 6 1sin 6 26 2 3
Thus, 2 20sin75 63yx ππ =−+ , or () 20sin475 6 yx π =−+
d. () 19.81sin0.5432.29675.66yx =−+
Chapter 2 Test
1.
2.
3.
4.
15. Set the calculator to degree mode:
5.
6.
16. Set the calculator to radian mode:
7.
17. To remember the sign of each trig function, we primarily need to remember that sin θ is positive in quadrants I and II, while cos θ is positive in quadrants I and IV. The sign of the other four trig functions can be determined directly from sine and cosine by knowing sin
Chapter2: Trigonometric Functions
sincostanseccsccot in QI in QII+ in QIII+ in QIV+
18. Because ()sin f xx = is an odd function and since 3 ()sin 5 faa== , then 3 ()sin()sin 5 faaa −=−=−=− .
19. 5 sin 7 θ= and θ in quadrant II.
Using the Pythagorean Identities: 2 22
Note that cos θ must be negative because θ lies in quadrant II. Thus, 26 cos 7 θ=−
20. 2 cos 3 θ= and 3 2 2 πθπ << (in quadrant IV).
21. 12 tan 5 θ=− and 2 πθπ << (in quadrant II)
Using the Pythagorean Identities:
Note
Note that sec θ must be negative since θ lies in quadrant II. Thus, 13 sec 5 θ=− 13 5 115 cos sec13 θ θ ===− sin tan cos θθ θ = , so ()() 12512 sintancos 51313 θθθ
22. The point () 2,7 lies in quadrant I with 2 x = and 7 y = . Since 222 x yr+= , we have 22 2753 r =+= . So, 7753753 sin 53 535353 y r θ===⋅=
23. The point () 5,11 lies in quadrant II with 5 x =− and 11 y = . Since 222 x yr+= , we have () 2 2 511146 r =−+= . So, 551465146 cos 146 146146146 x r θ===⋅=− .
24. The point () 6,3 lies in quadrant IV with 6 x = and 3 y =− . Since 222 x yr+= , we have () 2 2 634535 r =+−== . So, 31 tan 62 y x θ===−
25. Comparing 2sin 36 x y π =− to () sin yAxωφ=− , we see that
2 A = , 1 3 ω= , and 6 φπ = . The graph is a sine curve with amplitude 2 A = , period
22 6 1/3 T ππ π ω === , and phase shift 6 1/32 φππ ω === . The graph of 2sin 36 x y π =−
will lie between 2 and 2 on the y-axis. One period will begin at 2 x φπ ω == and end at
213 6 22 x πφππ π ωω =+=+= . We divide the
interval 13 , 22 ππ into four subintervals, each of length 63 42 ππ = . 7713 ,2,2,,,5,5, 2222
The five key points on the graph are ()()713 ,0,2,2,,0,5,2,,0 222
We plot these five points and fill in the graph of the sine function. The graph can then be extended in both directions.
26. tan2 4 yx π
=−++
Begin with the graph of tan y x = , and shift it 4 π units to the left to obtain the graph of
. Next, reflect this graph about the y-axis to obtain the graph of tan
=−+
. Finally, shift the graph up 2 units to obtain the graph of tan2 4 yx π
=−++
.
27. For a sinusoidal graph of the form () sin yAxωφ=− , the amplitude is given by A , the period is given by 2π ω , and the phase shift is given by φ ω . Therefore, we have 3 A =− , 3 ω= , and 3 3 44 φππ
. The equation for the graph is
yx
28. The area of the walk is the difference between the area of the larger sector and the area of the smaller shaded sector.
a l k 50 ° 3 ft The area of the walk is given by () 22 22 11 22 2 A Rr Rr θθ θ =− =− , where R is the radius of the larger sector and r is
Chapter2: Trigonometric Functions
the radius of the smaller sector. The larger radius is 3 feet longer than the smaller radius because the walk is to be 3 feet wide. Therefore, 3 Rr=+ , and
Chapter 2 Cumulative Review
1. ()() 2 210 2110 xx xx +−= −+= 1 or 1 2 xx==−
The solution set is 1 1, 2
The shaded sector has an arc length of 25 feet and a central angle of 5 50 radians 18 °=π . The radius of this sector is 5 18 2590 feet s r
Thus, the area of the walk is given by 5 18 22 905540 699
29. To throw the hammer 83.19 meters, we need 2
m/s
m/
m/s
Linear speed and angular speed are related according to the formula vr ω =⋅ . The radius is 190 cm1.9 m r == . Thus, we have () 28.553 28.5531.9
15.028 radians per second radians60 sec1 revolution 15.028 sec1 min2 radians 143.5 revolutions per minute (rpm) r ω
To throw the hammer 83.19 meters, Adrian must have been swinging it at a rate of 143.5 rpm upon release.
2. Slope3 =− , containing (–2,5) 11 Using ()yymxx −=− () 53(2) 53(2) 536 31 yx yx yx yx −=−−− −=−+ −=−− =−−
3. radius = 4, center (0,–2) ()() 22 2 Using x hykr −+−= ()() () () 2 2 2 2 2 024 216 xy xy −+−−= ++=
4. 2312 xy−=
This equation yields a line. 2312 3212 2 4 3 xy yx yx −= −=−+ =−
The slope is 2 3 m = and the y-intercept is 4 Let 0 y = : 23(0)12 212 6 x x x −= = =
The x-intercept is 6.
5. 22 2440xyxy+−+−= ()() ()() 22 22 22 2 2144414 129 123 xxyy xy xy −++++=++ −++= −++=
This equation yields a circle with radius 3 and center (1,–2).
6. 2 (3)2yx=−+
Using the graph of 2 y x = , horizontally shift to the right 3 units, and vertically shift up 2 units.
7. a. 2 y x =
b. 3 y x =
c. sin yx =
d. tan y x =
8. ()32fxx=− () 1 32 32Inverse 23 2 3 21 ()2 33 yx xy xy x y x fxx =− =− += + = + ==+
9. Since ()() 22 sincos1 θθ+= , then ()() 22 oo sin14cos143132 +−=−=−
Chapter2: Trigonometric Functions
Chapter 2 Projects
Project I – Internet Based Project Project II
1. November 15: High tide: 11:18 am and 11:15 pm November 19: low tide: 7:17 am and 8:38 pm
12. The graph is a cosine graph with amplitude 3 and period 12.
2. The low tide was below sea level. It is measured against calm water at sea level. 3. Nov
4. The data seems to take on a sinusoidal shape (oscillates). The period is approximately 12 hours. The amplitude varies each day:
Nov 14: 0.1, 0.7
Nov 15: 0.4, 0.4
Nov 16: 0.7, 0.3
Nov 17: 1.0, 0.1
Nov 18: 1.3, 0.1
Nov 19: 1.6, 0.1
Nov 20: 1.8
5. Average of the amplitudes: 0.66 Period : 12
Average of vertical shifts: 2.15 (approximately) There is no phase shift. However, keeping in mind the vertical shift, the amplitude () sin y ABxD=+
A = 2 12
Thus, () 0.66sin0.522.15yx =+ (Answers may vary)
6. () 0.848sin0.521.252.23yx =++
The two functions are not the same, but they are similar.
7. Find the high and low tides on November 21 which are the min and max that lie between 168 t = and 192 t = . Looking at the graph of the equation for part (5) and using MAX/MIN for values between 168 t = and 192 t = :
Low tides of 1.49 feet when t = 178.2 and t = 190.3.
High tides of 2.81 feet occur when t = 172.2 and t = 184.3.
Looking at the graph for the equation in part (6) and using MAX/MIN for values between t = 168 and t = 192:
A low tide of 1.38 feet occurs when t = 175.7 and 187.8 t =
A high tide of 3.08 feet occurs when t = 169.8 and 181.9 t = .
8. The low and high tides vary because of the moon phase. The moon has a gravitational pull on the water on Earth.
Project III
() 0 ()1sin2 s tft =π
4. Let 0 1 f = =1. Let 012 x ≤≤ , with 0.5 x Δ=
Label the graph as 0 012 x T ≤≤ , and each tick mark is at
7. 00 ()1sin(20)Stft π =+ , 10 ()1sin(2)Stftππ=+
8.
4
T
S
IV
miles
Project
1. Lanai:
Maui: Oahu
Molokai:
3960 cos(0.0101) 3960
39600.9999(3960)
0.346 miles
0.34652802090 feet h h h hx
5. Kamakou, Haleakala, and Lanaihale are all visible from Oahu.
Project V
Answers will vary.
Project at Motorola
Digital Transmission over the Air
Digital communications is a revolutionary technology of the century.For many years,Motorola has been one of the leading companies to employ digital communication in wireless devices,such as cell phones.
Figure1shows a simplified overview of a digital communication transmission over the air.The information source to be transmitted can be audio,video, or data.The information source may be formatted into a digital sequence of symbols from a finite set E an F ={0,1}.So 0110100 is an example of a digital sequence.The period of the symbols is denoted by T.
The principle of digital communication systems is that,during the finite interval of time T,the information symbol is represented by one digital waveform from a finite set of digital waveforms before it is sent.This technique is called modulation.
Modulation techniques use a carrier that is modulated by the information to be transmitted.The modulated carrier is transformed into an electromagnetic field and propagated in the air through an antenna.The unmodulated carrier can be represented in its general form by a sinusoidal function s(t)=A sin A v0 t+f B , where A is the amplitude, v0 is the radian frequency, and f is the phase.
Let’s assume that A=1, f=0,and v0=2pf0 radian,where f0 is the frequency of the unmodulated carrier.
1. Write s(t) using these assumptions.
2. What is the period, T0 ,of the unmodulated carrier?
3. Evaluate s(t) for t=0, 1/
and 1/f0 .
4. Graph s(t) for 0 t 12T0 .That is,graph 12 cycles of the function.
5. For what values of t does the function reach its maximum value?
[Hint: Express t in terms of f0].
Three modulation techniques are used for transmission over the air:amplitude modulation,frequency modulation and phase modulation.In this project,we are interested in phase modulation. Figure2illustrates this process.An information symbol is mapped onto a phase that modulates the carrier.The modulated carrier is expressed by Si(t)=sin A 2pf0 t+ci B .
Let’s assume the following mapping scheme:
c0 = 0
c1 = p
6. Map the binary sequence M=010 into a phase sequence P.
7. What is the expression of the modulated carrier S0(t) for ci=c0 and S1(t) for ci=c1 ?
8. Let’s assume that in the sequence M the period of each symbol is T=4T0 .For each of the three intervals C 0,4T0 D , C 4T0 ,8T0 D ,and C 8T0 ,12T0 D ,indicate which of S0(t) or S1(t) is the modulated carrier.On the same graph,illustrate M, P,and the modulated carrier for 0 t 12T0
Figure 1 Simplified Overview of a Digital Communication Transmission
Figure 2 Principle of Phase Modulation
2. Identifying Mountain Peaks in Hawaii
Suppose that you are standing on the southeastern shore of Oahu and you see three mountain peaks on the horizon.You want to determine which mountains are visible from Oahu.The possible mountain peaks that can be seen from Oahu and the height (above sea level) of their peaks are given in thetable.
Lanai65Lanaihale3,370
Maui110Haleakala10,023
Hawaii190MaunaKea13,796 Molokai40Kamakou4,961
(a)To determine which of these mountain peaks would be visible from Oahu,consider that you are standing on the shore and looking “straight out”so that your line of sight is tangent to the surface of Earth at the point where you are standing.Make a sketch of the right triangle formed by your sight line,the radius from the center of Earth to the point where you are standing,and the line from the center of Earth through Lanai.
(b)Assuming that the radius of Earth is 3960 miles, determine the angle formed at the center of Earth.
(c)Determine the length of the hypotenuse of the triangle.Is Lanaihale visible from Oahu?
(d)Repeat parts (a)–(c) for the other three islands.
(e)Which three mountains are visible from Oahu?
4. CBL Experiment Using a CBL,the microphone probe,and a tuning fork,record the amplitude,frequency,and period of the sound from the graph of the sound created by the tuning fork over time.Repeat the experiment for different tuning forks.
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