Chapter 1 Angles and the Trigonometric Functions
1. The radian measure of a central angle is the length of the intercepted arc, s, divided by the circle’s radius, r. The length of the intercepted arc is 42 feet: s = 42 feet. The circle’s radius is 12 feet: r = 12 feet. Now use the formula for radian measure to find the radian measure of .θ 42 feet 3.5 12 feet s r θ ===
Thus, the radian measure of θ is 3.5
2. a. radians60 6060 radians 180180 radians 3 ππ π °=°⋅= ° =
b. radians270 270270 radians 180180 3 radians 2 ππ
c. radians300 300300 radians 180180 5 radians 3 ππ
3. a. radians180o radians 4 radians ππ
4 180o 45o 4
b. o 44 radians180 radians ππ
c. 180o 6 radians 6 radians radians π =⋅ 6180o
d. 180o 4.7 radians 4.7 radians radians
5. a. For a 400º angle, subtract 360º to find a positive coterminal angle. 400o – 360o = 40o b. For a –135º angle, add 360º to find a positive coterminal angle. –135o + 360o = 225o 6. a. 1313103 2 5555 ππππ π −=−= b. 3029 2 15151515 ππππ
b.
c. d.
7. a. 8553602855720135 °−°⋅=°−°=°
b. 1717 224 33 ππ −⋅=−ππ 17125 333 πππ =−=
c. 2525 236 66 ππ −+⋅=−+ππ 253611 666 πππ =−+=
d. () 17.417.457991.8 ≈=
For a 991.8 angle, we need to subtract two multiples of 360 to find a positive coterminal angle less than 360. 17.42217.444.8 ππ −⋅=−≈
8. The formula s rθ = can only be used when θ is expressed in radians. Thus, we begin by converting 45° to radians. Multiply by radians 180 π ° radians45 4545 radians 180180 radians 4 π π π °=°⋅= ° =
Now we can use the formula s rθ = to find the length of the arc. The circle’s radius is 6 inches : r = 6 inches. The measure of the central angle in radians is : 44 ππ θ = . The length of the arc intercepted by this central angle is (6 inches) 4 6 inches 4 3 inches 2 4.71 inches. srθ π π π = = = = ≈
9. The formula 2 1 2 A r θ = can only be used when θ is expressed in radians. Thus, we begin by converting 150° to radians. Multiply by radians 180 π ° . radians 150150 180 150 radians 180 5 radians 6 π π π °=°⋅ ° = =
The circle’s radius is 6 feet. Now use the formula 2 1 2 A r θ = to find the area of the sector.: 2 1 2 2 1 2 5 (6 feet) 6 15 square feet 47.12 square feet Ar θ π π = = = ≈
10. We are given ω , the angular speed.
45 ω = revolutions per minute We use the formula r ν ω = to find ,v the linear speed. Before applying the formula, we must express ω in radians per minute.
45 revolutions2 radians 1 minute1 revolution 90 radians 1 minute π ω π =⋅
The angular speed of the propeller is 90π radians per minute. The linear speed is 90 1.5 inches 1 minute 135 inches minute inches 135 minute inches 424 minute r ν ω π π π = =⋅
The linear speed is 135π inches per minute, which is approximately 424 inches per minute.
Concept and Vocabulary Check 1.1
C1. origin; x-axis
C2. counterclockwise; clockwise
C3. acute; right; obtuse; straight
C4. s r
C5. 180 π °
C6. 180 π °
C7. coterminal; 360° ; 2π
C8. rθ
C9. 2 1 2 r θ
C10. false
C11. rω ; angular
Exercise Set 1.1
1. obtuse
2. obtuse
3. acute
4. acute
5. straight
6. right
7. 40 inches 4 radians 10 inches s r θ ===
8. 30 feet 6 radians 5 feet s r θ ===
9. 8 yards4 radians 6 yards3 s r θ ===
10. 18 yards 2.25 radians 8 yards s r θ ===
4545
centimeters 4
1818
135135
radians 150150
300300
radians 330330
radians 225225
20. radians 270270 180 270 radians 180 3 radians 2 π π π −°=−°⋅ ° =− =−
21. radians180o radians 22 radians ππ π =⋅ 180o 2 90o = =
22. radians180o radians 99 radians ππ π =⋅ 180o 20o 9 ==
23. o 22 radians180 radians 33 radians ππ π =⋅ 2180o 3 120o ⋅ = =
24. 0o 3 radians1803180 135o 4 radians4 π π ⋅ ⋅==
25. o 77 radians180 radians 66 radians ππ π =⋅ 7180o 6 210o = =
26. oo 11 radians18011180 330o 6 radians6 π π ⋅ ⋅==
27. 180o 3 radians3 radians radians ππ π −=−⋅ 3180o 540o =−⋅ =−
28. 180o oo 4 radians 4180720 radians π π −⋅=−⋅=−
29. radians 1818 180 18 radians 180 0.31radians π π °=°⋅ ° = ≈
30. radians 7676 180 76 radians 180 1.33radians π π °=°⋅ ° = ≈
31. radians 4040 180 40 radians 180 0.70radians π π −°=−°⋅ ° =− ≈−
32. radians 5050 180 50 radians 180 0.87radians π π −°=−°⋅ ° =− ≈−
33. radians 200200 180 200 radians 180 3.49radians π π °=°⋅ ° = ≈
34. radians 250250 180 250 radians 180 4.36radians π π °=°⋅ ° = ≈
35. 180o 2 radians2 radians radians π =⋅ 2180o 114.59o π = ≈
36. oo 1803180 o 3 radians171.89 radians ππ ⋅=≈
37. radians180o radians 1313 radians ππ π =⋅ 180o 13 13.85o = ≈
38. 180180oo radians10.59o 17radians17
39. 180o 4.8 radians4.8 radians radians π
40. oo 1805.2180 5.2 radians radians ππ
57. 39536035 °−°=°
58. 41536055 °−°=°
59. 150360210 −°+°=°
60. 160360200 −°+°=°
61. 76536037651080315 −°+°⋅=−°+°=°
62. 76036037601080320 −°+°⋅=−°+°=°
63. 1919127 2 6666 ππππ
64. 1717107 2 5555 ππππ π −=−=
65. 232323203 224 55555 πππππ ππ −⋅=−=−=
66. 25252524 224 66666 πππππ ππ −⋅=−=−=
67. 10099 2 50505050 ππππ π −+=−+=
68. 8079 2 40404040 ππππ π −+=−+=
69. 3131 236 77 ππ −+⋅=−+ππ 314211 777 πππ =−+=
70. 3838 236 99 ππ −+⋅=−+ππ 385416 999 πππ =−+=
71. 12inches,45 r θ ==°
Begin by converting 45° to radians, in order to use the formula s rθ = radians 4545radians 1804 °=°⋅=ππ °
Now use the formula s rθ = 123inches9.42inches 4 sr π θπ ==⋅=≈
72. 16inches,60 r θ ==°
Begin by converting 60° to radians, in order to use the formula s rθ = . radians 6060radians 1803 °=°⋅=ππ °
Now use the formula s rθ = . 16 16inches16.76inches 33 sr ππ θ ==⋅=≈
73. 8feet,225 r θ ==°
Begin by converting 225° to radians, in order to use the formula s rθ = . radians5 225225radians 1804 °=°⋅=ππ °
Now use the formula s rθ = . 5 810feet31.42feet 4 sr π θπ ==⋅=≈
74. 9yards,315 r θ ==°
Begin by converting 315° to radians, in order to use the formula s rθ = radians7 315315radians 1804 °=°⋅=ππ °
Now use the formula s rθ = 763 9yards49.48yards 44 sr ππ θ ==⋅=≈
75. Begin by converting the angle to radians. radians 1818 radians 18010 °=°⋅=ππ °
Now use the formula 2 1 2 A r θ = to find the area of the sector.: 2 1 2 2 1 2 (10 meters) 10 5 square meters 15.71 square meters
76. Begin by converting the angle to radians. radians 1515 180 radians 12 π
°
Now use the formula 2 1 2 A r θ = to find the area of the sector.: 2 1 2 2 1 2 (6 yards) 12 3 square yards 2 4.71 square yards
77. Begin by converting the angle to radians. radians 240240 180 4 radians 3
78. Begin by converting the angle to radians. radians 330330 180 11 radians 6 π π °=°⋅ ° = Now use the formula 2 1 2 A r θ = to find the area of the sector.: 2 1 2 2 1 2 11 (3 inches) 6 33 square inches 4 25.92 square inches Ar θ π π =
79. 6 revolutions per second 6 revolutions2 radians12 radians 1 second1 revolutions1 seconds 12 radians per second ππ
80. 20 revolutions per second 20 revolutions2 radians40 radians 1 second1 revolution1 second 40 radians per second ππ π =⋅=
81. 42 and 33 ππ
82. 75 and 66 ππ 83. 35 and 44 ππ 84. 7 and 44 ππ 85. 3 and 22 ππ 86. π and π 87. 5511 2 606 π π ⋅= 88. 357 2 606 π π ⋅=
89. 3 minutes and 40 seconds equals 220 seconds. 22022 2 603
90. 4 minutes and 25 seconds equals 265 seconds. 26553 2 606
93. First, convert to degrees. 11360 revolutionrevolution 661revolution 1 36060 6
Now, convert 60° to radians. radians60 6060radians 180180 radians 3 ππ
Therefore, 1 6 revolution is equivalent to 60° or 3
radians.
94. First, convert to degrees. 11360o revolutions revolutions 331 revolution 1 360120oo
Now, convert 120° to radians. radians1202 120120radiansradians 1801803 °=°⋅==πππ °
Therefore, 1 3 revolution is equivalent to 120° or 2 3 π radians.
95. The distance that the tip of the minute hand moves is given by its arc length, s. Since s rθ = , we begin by finding r and θ . We are given that r = 8 inches. The minute hand moves from 12 to 2 o'clock, or 1 6 of a complete revolution. The formula s rθ = can only be used when θ is expressed in radians. We must convert 1 6 revolution to radians.
112 radians revolution revolution 661 revolution radians 3 π π =⋅ =
The distance the tip of the minute hand moves is 8 (8inches)inches8.38inches.
96. The distance that the tip of the minute hand moves is given by its arc length, s. Since s rθ = , we begin by finding r and θ . We are given that r = 6 inches. The minute hand moves from 12 to 4 o’clock, or 1 3 of a complete revolution. The formula s rθ = can only be used when θ is expressed in radians. We must convert 1 3 revolution to radians.
112 radians revolutionrevolution 331 revolution 2 radians 3 π π =⋅ =
The distance the tip of the minute hand moves is 212 (6inches)inches 33 4inches12.57inches.
97. The length of each arc is given by s rθ = . We are given that r = 24 inches and 90. θ =° The formula s rθ = can only be used when θ is expressed in radians. radians90 9090radians 180180 =radians 2 ππ
The length of each arc is (24inches)12inches 2 37.70inches.
98. The distance that the wheel moves is given by s rθ = . We are given that r = 80 centimeters and θ = 60°. The formula s rθ = can only be used when θ is expressed in radians. radians60 6060radians 180180 =radians 3
The length that the wheel moves is 80 (80centimeters)centimeters 33 83.78centimeters.
99. Begin by converting the angle to radians. radians 135135 180 3 radians 4
Now use the formula 2 1 2 A r θ = to find the area of the sector.:
100. Begin by converting the angle to radians. radians 120120 180 2 radians 3
Now use the formula 2 1 2 A r θ = to find the area of the sector.: 2 1 2 2 1 2 2 (30 feet) 3 300 square feet 942.48 square feet Ar θ π π =
101. Recall that s r θ = . We are given that s = 8000 miles and r = 4000 miles. 8000 miles 2 radians 4000 miles s r θ === Now, convert 2 radians to degrees. 180o o 2 radians2 radians114.59 radians π =⋅≈
102. Recall that s r θ = . We are given that s = 10,000 miles and r = 4000 miles. 10,000 miles 2.5 radians 4000 miles s r θ === Now, convert 2.5 radians to degrees. 180o o 2.5 radians143.24 2 radians π ⋅≈
103. Recall that s rθ = . We are given that r = 4000 miles and 30 θ =° . The formula s rθ = can only be used when θ is expressed in radians. radians30 3030radians 180180
radians 6 ==(4000miles)2094miles 6 sr
To the nearest mile, the distance from A to B is 2094 miles.
104. Recall that s rθ = . We are given that r = 4000 miles and 10 θ =° . We can only use the formula s rθ = when θ is expressed in radians. radians10
1010radians 180180
radians 18
==(4000miles)698miles 18 sr
To the nearest mile, the distance from A to B is 698 miles.
105. Linear speed is given by r ν ω = . We are given that 12 π ω = radians per hour and r = 4000 miles. Therefore, (4000miles) 12 4000 milesperhour 12 1047milesperhour r π
The linear speed is about 1047 miles per hour.
106. Linear speed is given by r ν ω = . We are given that r = 25 feet and the wheel rotates at 2 revolutions per minute. We need to convert 2 revolutions per minute to radians per minute.
2 revolutions per minute
2radians
2revolutionsperminute 1revolution
4radiansperminute
= (25feet)(4)314feetperminute r
The linear speed of the Ferris wheel is about 314 feet per minute.
107. Linear speed is given by r ν ω = . We are given that r = 12 feet and the wheel rotates at 20 revolutions per minute.
20 revolutions per minute
2radians
20revolutionsperminute 1revolution
40radiansperminute
(12feet)(40) 1508feetperminute
The linear speed of the wheel is about 1508 feet per minute.
108. Begin by converting 2.5 revolutions per minute to radians per minute.
2.5 revolutions per minute
2radians
2.5revolutionsperminute 1revolution
5radiansperminute
The linear speed of the animals in the outer rows is (20feet)(5)100feetperminute
The linear speed of the animals in the inner rows is (10feet)(5)50feetperminute
The difference is 1005050πππ −= feet per minute or about 157 feet per minute.
109. – 121. Answers may vary.
122. 3015'10''30.25 =
123. 6545'20''65.76 =
124. 30.423025'12'' =
125. 50.425025'12'' =
126. does not make sense; Explanations will vary. Sample explanation: Angles greater than π will exceed a straight angle.
127. does not make sense; Explanations will vary. Sample explanation: It is possible for π to be used in an angle measured using degrees.
128. makes sense
129. does not make sense; Explanations will vary. Sample explanation: That will not be possible if the angle is a multiple of 2. π
130. A right angle measures 90° and 90 2 π °= radians 1.57 radians.
If 3 radians1.5radians, 2 θθ == is smaller than a right angle.
131. s rθ = Begin by changing 20 θ =° to radians.
2020radians 1809
100= 9
To the nearest mile, a radius of 286 miles should be used. 132.
square feet(30 feet)
square feet (30 feet)
radians to degrees: 1414180radians89
The sprinkler should rotate about 89. °
133. s rθ = Begin by changing 26 θ =° to radians.
To the nearest mile, Miami, Florida is 1815 miles north of the equator.
134. First find the hypotenuse.
2 2 512 25144 169 13 cab c c c c =+ =+ =+ =
Next write the ratio. 5 13 a c =
135. First find the hypotenuse.
222 2 2 11 11 2 2 cab c c c c =+ =+ =+ = = Next write the ratio and simplify. 1 2 12 22 2 2 a c = =⋅ = 136. 22 22 22 22 2 abab cc cc ab c +=+ + = Since 222 , cab =+ continue simplifying by substituting 2 c for 22ab + which gives 1.
Section1.2 Right Triangle Trigonometry
Section 1.2
Check Point Exercises
1. Use the Pythagorean Theorem, 222 cab =+ , to find c.
Referring to these lengths as opposite, adjacent, and hypotenuse, we have opposite3 sin hypotenuse5 adjacent4 cos hypotenuse5 opposite3 tan adjacent4
hypotenuse5 csc opposite3 hypotenuse5 sec adjacent4 adjacent4 cot opposite3
2. Use the Pythagorean Theorem, 222 cab =+ , to find b.
Note that side a is opposite θ and side b is adjacent to θ . opposite1 sin hypotenuse5 adjacent26 cos hypotenuse5 opposite16 tan adjacent12 26 hypotenuse5 csc5 opposite1 hypotenuse556 sec adjacent12 26 adjacent26 cot26 opposite1
3. Apply the definitions of these three trigonometric functions. length of hypotenuse csc45 length of side opposite 45 2 2 1 length of hypotenuse sec45 length of side adjacent to 45 2 2 1 length of side adjacent to 45 cot45 length of side opposite 45 1 1 1
4. length of side opposite 60 tan60 length of side adjacent to 60 3 3
opposite
6. We can find the value of cos θ by using the Pythagorean identity.
7. a. oooo sin46cos(9046)cos44 =−=
cottan
8.
Function
Many Scientific Calculators
Many Graphing Calculators
Function Mode
9. Because we have a known angle, an unknown opposite side, and a known adjacent side, we select the tangent function.
tan24o 750 750tan24o 750(0.4452)333.9
The distance across the lake is approximately 333.9 yards.
10. side opposite14 tan side adjacent10 θ ==
Use a calculator in degree mode to find θ
Many Scientific Calculators
Many Graphing Calculators 1 TAN(1410)ENTER ÷
( 14 ÷ 10 ) ENTER
The display should show approximately 54. Thus, the angle of elevation of the sun is approximately 54°.
Concept and Vocabulary Check 1.2
C1. sin
C2. opposite; adjacent to; hypotenuse
C3. true
C4. sin θ ; cos θ ; tan θ ;
C5. tan θ ; cot θ
C6. 1; 2 sec θ ; 2 csc θ
C7. sin θ ; tan θ ; sec θ
Exercise Set 1.2
1. 222912225 22515 c c =+= == opposite93 sin hypotenuse155 adjacent124 cos hypotenuse155 opposite93 tan adjacent124 hypotenuse155 csc opposite93 hypotenuse155 sec adjacent124 adjacent124 cot opposite93
2. 22268100 10010 c c =+= == opposite63 sin hypotenuse105 adjacent84 cos hypotenuse105 opposite63 tan adjacent84 hypotenuse105 csc opposite63 hypotenuse105 sec adjacent84 adjacent84 cot opposite63
3. 222 2 2129 841441400 40020 a a a += =−= == opposite20 sin hypotenuse29 adjacent21 cos hypotenuse29 opposite20 tan adjacent21 hypotenuse29 csc opposite20 hypotenuse29 sec adjacent21 adjacent21 cot opposite20
4. 222 2 1517 28922564 648 a a a += =−= == opposite8 sin hypotenuse17 adjacent15 cos hypotenuse17 opposite8 tan adjacent15 hypotenuse17 csc opposite8 hypotenuse17 sec adjacent15 adjacent15 cot opposite8
5.
222 2 1026 676100576 57624 b b b += =−= == opposite105 sin hypotenuse2613 adjacent2412 cos hypotenuse2613 opposite105 tan adjacent2412 hypotenuse2613 csc opposite105 hypotenuse2613 sec adjacent2412 adjacent2412 cot opposite105
6. 222 2 4041 1681160081 819 a a a += =−= == opposite9 sin hypotenuse41 adjacent40 cos hypotenuse41 opposite9 tan adjacent40 hypotenuse41 csc opposite9 hypotenuse41 sec adjacent40 adjacent40 cot opposite9
7.
222 2 2135 1225441784 78428 b b b += =−= == opposite284 sin hypotenuse355 adjacent213 cos hypotenuse355 opposite284 tan adjacent213 hypotenuse355 csc opposite284 hypotenuse355 sec adjacent213 adjacent213 cot opposite284
8. 222 2 2425 62557649 497 a a a
opposite24 sin hypotenuse25 adjacent7 cos hypotenuse25 opposite24 tan adjacent7 hypotenuse25 csc opposite24 hypotenuse25 sec adjacent7 adjacent7 cot opposite24
9. lengthofsideadjacentto30 cos30 lengthofhypotenuse
10. length of side opposite 30 tan30 length of side adjacent to 30 1133
11. lengthofhypotenuse sec45 lengthofsideadjacentto45 2 2
12. length of hypotenuse csc45 length of side opposite 45 2 2
13. tantan60 3 lengthofsideopposite60 lengthofsideadjacentto60 3
14.
of side adjacent to 60 cotcot60 3length of side opposite 60 1133
19.
25. 1 000 sin37csc37sin371 sin370 =⋅= 26. 000 0 1 cos53sec53cos531 cos52 == 27. 22 sincos1 θθ+= 22 sincos1 99 ππ +=
28. We can find the value of the expression by using the Pythagorean identity 22 sincos1 θθ+= 22 sincos1 1010 ππ +=
29. 1tansec22 +=θθ 2020 1tan23sec23 2020 1sec23tan23 += =−
30. We can find the value of the expression by using the Pythagorean identity 1cotcsc22 +=θθ 2020 csc63cot631 −=
31. sin7cos(907)cos83 °=°−°=°
32. () sin19cos9019cos71 °=°−°=°
33. csc 25° = sec(90° – 25o) = sec 65°
34. csc35sec(9035)sec55 °=°−°=°
35. tancot
37. 22 cossin 525
38. 3343 cossinsinsin
39. Many Scientific Calculators
Function Mode Keystrokes
Many Graphing Calculators
Function Mode Keystrokes
40. Many Scientific Calculators
Function Mode Keystrokes
Many Graphing Calculators
Function Mode Keystrokes
41.
Function
Function
44. Many
46.
Many Scientific Calculators
Function Mode Keystrokes Display (rounded to four places)
3 sin 10
Many Graphing Calculators
Function Mode Keystrokes Display (rounded to four places)
3 sin 10
47.
Many Scientific Calculators
Function Mode Keystrokes
Many Graphing Calculators
Function Mode Keystrokes
48. Many Scientific Calculators
Function Mode Keystrokes
Display (rounded to four places)
Function
49. tan37 250 250tan37 250(0.7536)188cm
50. tan61 10 10tan61 10(1.8040)18cm
51. cos34 220 220cos34 220(0.8290)182in.
52. sin34 13 13sin34 13(0.5592)7m
Many Graphing Calculators
53. 16 sin23 1616 41m sin230.3907 c c °= =≈≈ °
54. 23 tan44 2323 24yd tan440.9657 b b °= =≈≈ °
55. Scientific Calculator
Display (rounded to the nearest degree) 0.2974 1 SIN 1 SIN0.2974ENTER 17
Graphing Calculator
If sin0.2974,then17. θθ=≈°
56. Scientific Calculator Graphing Calculator
Display (rounded to the nearest degree)
0.877 COS-1 COS-1 0.877 ENTER 29
If cos0.877, then 29 θθ=≈°
57. Scientific Calculator Graphing Calculator
Display (rounded to the nearest degree) 1 4.6252TAN 1 TAN4.6252ENTER 78
If tan4.6252,then78. θθ=≈°
58. Scientific Calculator Graphing Calculator
Display (rounded to the nearest degree) 26.0307 TAN-1 TAN-1 26.0307 ENTER 88
If tan26.0307, then 88.θθ=≈°
59. Scientific Calculator Graphing Calculator
Display (rounded to three places) 0 1 .4112COS 1 COS0.4112ENTER 1.147 If cos0.4112,then1.147radians. θθ =≈
60. Scientific Calculator Graphing Calculator
Display (rounded to three places)
0.9499 SIN-1 SIN-1 0.9499 ENTER 1.253
If sin0.9499, then 1.253 radians. θθ ==
61. Scientific Calculator Graphing Calculator
Display (rounded to three places) 0 1 .4169TAN 1 TAN0.4169ENTER 0.395
If tan0.4169,then0.395radians. θθ =≈
62. Scientific Calculator Graphing Calculator
63.
65. 22 1sin40sin50 +°+° 22 22 1sin(9050)sin50 1cos50sin50 11 2 =+°−°+° =+°+° =+ =
66. 22 1tan10csc80 −°+° 22 22 1cot80csc80 1csc80cot80 11 2 =−°+° =+°−° =+ =
67. csc37sec53tan53cot37 °°−°° 22 sec53sec53tan53tan53 sec53tan53 1 =°°−°° =°−°
68. cos12sin78cos78sin12 °°+°° 22 sin78sin78cos78cos78 sin78cos78 1 =°°+°° =°+° =
69.
73. tan40 630 630tan40 630(0.8391)529 a a a °= =°
The distance across the lake is approximately 529 yards.
74. tan40 35 35tan40 35(0.8391)29 h h h °= =°
The tree’s height is approximately 29 feet.
75. 125 tan 172 θ =
Use a calculator in degree mode to find θ . Many Scientific Calculators Many Graphing
The display should show approximately 36. Thus, the angle of elevation of the sun is approximately 36°.
76. 555 tan 1320
Use a calculator in degree mode to find θ .
Many Scientific Calculators
Many Graphing Calculators 555 ÷ 1320 = TAN-1 TAN-1 ( 555 ÷ 1320 ) ENTER
The display should show approximately 23. Thus, the angle of elevation is approximately 23°.
77. 500 sin10 500500 2880 sin100.1736 c c °= =≈≈ °
The plane has flown approximately 2880 feet.
78. sin5 5000 5000sin55000(0.0872)436 a a °= =°≈=
The driver’s increase in altitude was approximately 436 feet.
79. 60 cos 75 θ =
Use a calculator in degree mode to find θ
Many Scientific Calculators Many Graphing Calculators 1 6075COS ÷= 1 COS(6075)ENTER ÷
The display should show approximately 37. Thus, the angle between the wire and the pole is approximately 37°.
80. 55 cos 80 θ =
Use a calculator in degree mode to find θ
Many Scientific Calculators
Many Graphing Calculators 55 ÷ 80 = COS-1 COS-1 ( 55 ÷ 80 ) ENTER
The display should show approximately 47. Thus, the angle between the wire and the pole is approximately 47°.
81. – 91. Answers may vary. 92.
θ
sin θ θ approaches 1 as θ approaches 0.
cos1 θ θ approaches 0 as θ approaches 0.
94. does not make sense; Explanations will vary. Sample explanation: An increase in the size of a triangle does not affect the ratios of the sides.
95. does not make sense; Explanations will vary. Sample explanation: This value is irrational. Irrational numbers are rounded on calculators.
96. makes sense
97. makes sense
98. false; Changes to make the statement true will vary. A sample change is: tan4545 tan tan1515 °°
99. true
100. false; Changes to make the statement true will vary. A sample change is:
sin45cos451
101. true
102. In a right triangle, the hypotenuse is greater than either other side. Therefore both oppositeadjacent and hypotenusehypotenuse must be less than 1 for an acute angle in a right triangle.
103. Use a calculator in degree mode to generate the following table. Then use the table to describe what happens to the tangent of an acute angle as the angle gets close to 90°.
As θ approaches 90°, tan increases without bound. At 90°, tan is undefined.
104. a. Let a = distance of the ship from the lighthouse. 250 tan35 250250 357 tan350.7002 a a °=
The ship is approximately 357 feet from the lighthouse.
b. Let b = the plane’s height above the lighthouse. tan22 357 357tan22357(0.4040)144 144250394 b b °= =°≈≈ +=
The plane is approximately 394 feet above the water.
105. a. y r
b. First find r: 22 (3)422 5 rxy r r =+ =−+ = 4 , 5 y r = which is positive.
106. a. x r
b. First find r: 22 (3)522 34 rxy r r =+ =−+ = 3334334 , 34 343434 x r ==⋅= which is negative.
107. a. ooo 36034515 θ′ =−=
b. 565 6666 ππππ θπ ′ =−=−=
Section 1.3
Check Point Exercises 1. 22 rxy =+ 22 1(3)1910 r =+−=+=
Now that we know x, y, and r, we can find the six trigonometric functions of θ 3310 sin 10 10
2. a. 00 θ =°= radians
The terminal side of the angle is on the positive x-axis. Select the point P = (1,0): 1,0,1xyr===
Apply the definitions of the cosine and cosecant functions. 1 cos0cos01 1 1
csc0csc0, undefined 0 x r r y °==== °===
b. 90 2 π θ =°= radians
The terminal side of the angle is on the positive y-axis. Select the point
P = (0,1): 0,1,1xyr===
Apply the definitions of the cosine and cosecant functions. 0 cos90cos0 21 1 csc90csc1 21 x r r y π π °==== °====
c. 180 θπ =°= radians
The terminal side of the angle is on the negative x-axis. Select the point
P = (–1,0): 1,0,1 x yr =−==
Apply the definitions of the cosine and cosecant functions. 1
cos180cos1 1 1
csc180csc, undefined 0 x r r y π
d. 3 270 2 π θ =°= radians
The terminal side of the angle is on the negative y-axis. Select the point
P = (0,–1): 0,1,1 x yr ==−=
Apply the definitions of the cosine and cosecant functions.
30 cos270cos0 21 31 csc270csc1 21 x r r y π π °====
3. Because sin0,θθ < cannot lie in quadrant I; all the functions are positive in quadrant I. Furthermore, θ cannot lie in quadrant II; sin θ is positive in quadrant II. Thus, with sin0,θθ < lies in quadrant III or quadrant IV. We are also given that cos0 θ < . Because quadrant III is the only quadrant in which cosine is negative and the sine is negative, we conclude that θ lies in quadrant III.
4. Because the tangent is negative and the cosine is negative, θ lies in quadrant II. In quadrant II, x is negative and y is positive. Thus, 11 tan 33 3,1 y x xy θ =−== =−= Furthermore, 2222(3)19110rxy=+=−+=+=
Now that we know x, y, and r, we can find sinandsecθθ 111010 sin 10 101010 1010 sec 33 y r r x θ θ ===⋅= ===−
5. a. Because 210° lies between 180° and 270°, it is in quadrant III. The reference angle is 21018030 θ′ =°−°=°
b. Because 7 4 π lies between 36 24 ππ = and 8 2 4 π π = , it is in quadrant IV. The reference angle is 787 2 4444 ππππ θπ ′ =−=−= .
c. Because –240° lies between –180° and –270°, it is in quadrant II. The reference angle is 24018060 θ =−=°
d. Because 3.6 lies between 3.14 π ≈ and 3 4.71 2 π ≈ , it is in quadrant III. The reference angle is 3.60.46θπ ′ =−≈
6. a. 665360305 °−°=°
This angle is in quadrant IV, thus the reference angle is 36030555 θ′ =°−°=°
b. 151587 2 4444 ππππ π −=−=
This angle is in quadrant IV, thus the reference angle is 787 2 4444 ππππ θπ ′ =−=−= .
c. 111112 22 3333 ππππ π −+⋅=−+=
This angle is in quadrant I, thus the reference angle is 3 π θ′ =
7. a. 300° lies in quadrant IV. The reference angle is 36030060 θ′ =°−°=° . 3 sin60 2 °=
Because the sine is negative in quadrant IV, 3 sin300sin60 2 °=−°=− .
b. 5 4 π lies in quadrant III. The reference angle is 554 4444 θπππππ ′ =−=−= . tan1 4 π =
Because the tangent is positive in quadrant III, 5 tantan1 44 ππ =+=
c. 6 π lies in quadrant IV. The reference angle is 6 π θ′ = . 23 sec 63 π =
Because the secant is positive in quadrant IV, 23 secsec 663 ππ −=+= .
8. a. 1717125 2 6666 ππππ π −=−= lies in quadrant II.
The reference angle is 5 66 ππ θπ ′ =−=
The function value for the reference angle is 3 cos 62 π = .
Because the cosine is negative in quadrant II, 1753 coscoscos 6662 πππ ==−=−
b. 2222242 8 3333 ππππ π +=+= lies in quadrant II. The reference angle is 2 33 ππ θπ ′ =−=
The function value for the reference angle is 3 sin 32 π = .
Because the sine is positive in quadrant II, 2223 sinsinsin 3332 πππ ===
Concept and Vocabulary Check 1.3
C1. sin y r θ = ; csc r y θ = ; cos x r θ = ; sec r x θ = ' tan y x θ = ; cot x y θ =
C2. tan θ ; sec θ ; cot θ ; csc θ ; tan θ ; cot θ
C3. sin θ ; csc θ ;
C4. tan θ ; cot θ ;
C5. cos θ ; sec θ
C6. terminal; x
C7. (a) 180 θ °− ; (b) 180 θ −° ; (c) 360 θ °−
Exercise Set 1.3
1. We need values for x, y, and r. Because P = (–4, 3) is a point on the terminal side of ,4and3 xy θ =−= . Furthermore, 2222(4)3169255rxy=+=−+=+== No w that we know x, y, and r, we can find the six trigonometric functions of θ . 3 sin 5 44 cos 55 33 tan 44 5 csc 3 y r x r y x r y θ θ
2. We need values for x, y, and r, Because P = (–12, 5) is a point on the terminal side of ,12and5 xy θ =−= . Furthermore, 2222(12)514425 16913 rxy=+=−+=+
Now that we know x, y, and r, we can find the six trigonometric functions of θ 5 sin 13 1212 cos 1313 55
3. We need values for x, y, and r. Because P = (2, 3) is a point on the terminal side of ,2and3 xy θ == Furthermore, 2222234913rxy=+=+=+=
Now that we know x, y, and r, we can find the six trigonometric functions of θ 3313313 sin 13 131313 2213213 cos 13 131313 3 tan 2 13 csc 3 13
4. We need values for x, y, and r, Because P = (3, 7) is a point on the terminal side of ,3and7 xy θ == . Furthermore, 22223794958rxy=+=+=+=
Now that we know x, y, and r, we can find the six trigonometric functions of θ . 7758758 sin 58 585858 3358358 cos 58 585858 7 tan 3 58 csc 7 58 sec 3 3
5. We need values for x, y, and r. Because P = (3, –3) is a point on the terminal side of ,3and3 xy θ ==− . Furthermore, 22223(3)99 1832 rxy=+=+−=+
Now that we know x, y, and r, we can find the six trigonometric functions of θ .
6. We need values for x, y, and r, Because P = (5, –5) is a point on the terminal side of ,5and–5 xy θ == . Furthermore, 2225(5)252550
Now that we know x, y, and r, we can find the six trigonometric functions of θ .
7. We need values for x, y, and r. Because P = (–2, –5) is a point on the terminal side of ,2and5 xy θ =−=− . Furthermore, 2222(2)(5)42529rxy=+=−+−=+= No w that we know x, y, and r, we can find the six trigonometric functions of θ 5529529
sin 29 292929 2229229
cos 29 292929 55
tan 22 2929
csc 55 2929
sec 22 22 cot 55 y r x r y x r y r x x y θ θ θ θ
8. We need values for x, y, and r, Because P = (–1, –3) is a point on the terminal side of ,–1and–3 xy θ == . Furthermore, 2222(1)(3)1910rxy=+=−+−=+=
Now that we know x, y, and r, we can find the six trigonometric functions of θ 3310310 sin 10 101010 111010 cos 10 101010 3 tan3 1 1010 csc 33 10 sec10 1 11 cot 33 y r x r y x r y r x x y θ θ θ θ θ
9. θπ = radians
The terminal side of the angle is on the negative x-axis. Select the point P = (–1, 0): 1,0,1 x yr =−== Apply the definition of the cosine function. 1 cos1 1 x r π ===−
10. θπ = radians
The terminal side of the angle is on the negative x-axis. Select the point P = (–1, 0): 1,0,1 x yr =−== Apply the definition of the tangent function. 0 tan0 1 y x π ===
11. θπ = radians
The terminal side of the angle is on the negative x-axis. Select the point P = (–1, 0): 1,0,1 x yr =−== Apply the definition of the secant function. 1 sec1 1 r x π ===−
12. θπ = radians
The terminal side of the angle is on the negative x-axis. Select the point P = (–1, 0): 1,0,1 x yr =−== Apply the definition of the cosecant function. 1 csc,undefined 0 r y π ==
13. 3 2 π θ = radians
The terminal side of the angle is on the negative y-axis. Select the point P = (0, –1): 0,–1,1xyr=== Apply the definition of the tangent function. 31 tan,undefined 20 y x π ==
14. 3 2 π θ = radians
The terminal side of the angle is on the negative y-axis. Select the point P = (0, –1): 0,1,1 x yr ==−= Apply the definition of the cosine function.
30 cos0 21 x r π ===
15. 2 π θ = radians
The terminal side of the angle is on the positive y-axis. Select the point P = (0, 1): 0,1,1 x yr === Apply the definition of the cotangent function. 0 cot0 21 x y π ===
16. 2 π θ = radians
The terminal side of the angle is on the positive y-axis. Select the point P = (0, 1): 0,1,1xyr=== Apply the definition of the tangent function.
1 tan,undefined 20 y x π ==
17. Because sin0,θθ > cannot lie in quadrant III or quadrant IV; the sine function is negative in those quadrants. Thus, with sin0,θθ > lies in quadrant I or quadrant II. We are also given that cos0 θ > . Because quadrant I is the only quadrant in which the cosine is positive and sine is positive, we conclude that θ lies in quadrant I.
18. Because sin0,θθ < cannot lie in quadrant I or quadrant II; the sine function is positive in those two quadrants. Thus, with sin0,θθ < lies in quadrant III or quadrant IV. We are also given that cos0. θ > Because quadrant IV is the only quadrant in which the cosine is positive and the sine is negative, we conclude that θ lies in quadrant IV.
19. Because sin0,θθ < cannot lie in quadrant I or quadrant II; the sine function is positive in those two quadrants. Thus, with sin0, θ < θ lies in quadrant III or quadrant IV. We are also given that cos0 θ < Because quadrant III is the only quadrant in which the cosine is positive and the sine is negative, we conclude that θ lies in quadrant III.
20. Because tan0,θθ < cannot lie in quadrant I or quadrant III; the tangent function is positive in those two quadrants. Thus, with tan0,θθ < lies in quadrant II or quadrant IV. We are also given that sin0 θ < . Because quadrant IV is the only quadrant in which the sine is negative and the tangent is negative, we conclude that θ lies in quadrant IV.
21. Because tan0,θθ < cannot lie in quadrant I or quadrant III; the tangent function is positive in those quadrants. Thus, with tan0,θθ < lies in quadrant II or quadrant IV. We are also given that cos0 θ < . Because quadrant II is the only quadrant in which the cosine is negative and the tangent is negative, we conclude that θ lies in quadrant II.
22. Because cot0,θθ > cannot lie in quadrant II or quadrant IV; the cotangent function is negative in those two quadrants. Thus, with cot0,θθ > lies in quadrant I or quadrant III. We are also given that sec0 θ < . Because quadrant III is the only quadrant in which the secant is negative and the cotangent is positive, we conclude that θ lies in quadrant III.
23. In quadrant III x is negative and y is negative. Thus, 33 cos= 55 x r θ −== , x = –3, r = 5. Furthermore,
222 222 2 5(3) 25916 164 rxy y y y =+ =−+ =−= =−=−
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ 44 sin 55 44 tan 33 55 csc 44 55 sec 33 33 cot 44 y r y x r y r x x y
24. In quadrant III, x is negative and y is negative. Thus, 1212 sin=,12,13 1313 y yr r θ =−==−=
Furthermore, 222 222 2 (12)13 16914425 255 xyr x x x += +−= =−= =−=−
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ 55 cos 1313 1212 tan 55 1313 csc 1212 1313 sec 55 55 cot 1212 x r y x r y r x x y
25. In quadrant II x is negative and y is positive. Thus, 5 sin=,5,13 13 y yr r θ === . Furthermore, 222 222 2 513 16925144 14412 xyr x x x += += =−= =−=−
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . 1212 cos 1313 55 tan 1212 13 csc 5 1313 sec 1212 1212 cot 55 x r y x r y r x x y θ
26. In quadrant IV, x is positive and y is negative. Thus, 4 cos=,4,5 5 x xr r θ ===
Furthermore, 222 222 2 45 25169 93 xyr y y y += += =−= =−=−
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ 33 sin 55 33 tan 44 55 csc 33 5 sec 4 44 cot 33 y r y x r y r x x y θ θ θ
27. Because 270360,°<<°θθ is in quadrant IV. In quadrant IV x is positive and y is negative. Thus, 8 cos= 17 x r θ = , x = 8, r = 17. Furthermore 222 222 2 817 28964225 22515 xyr y y y += += =−= =−=−
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . 1515 sin 1717 1515 tan 88 1717 csc 1515 17 sec 8 88 cot 1515 y r y x r y r x x y θ
28. Because 270360,°<<°θθ is in quadrant IV. In quadrant IV, x is positive and y is negative. Thus, 1 cos=,1,3 3 x xr r θ === . Furthermore, 222 222 2 13 918 822 xyr y y y += +=
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . 2222
29. Because the tangent is negative and the sine is positive, θ lies in quadrant II. In quadrant II, x is negative and y is positive. Thus, 22
tan=,3,2. 33 y xy x θ −===−= Furthermore, 2222(3)29413rxy=+=−+=+=
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . 2213213 sin 13 131313 3313313 cos 13 131313 13 csc 2 1313 sec 33 33 cot 22 y r x
30. Because the tangent is negative and the sine is positive, θ lies in quadrant II. In quadrant II, x is negative and y is positive. Thus, 11 tan=,1,3 33 y yx x θ =−===− . Furthermore, 2222(3)19110rxy=+=−+=+=
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ 111010 sin 10 101010 3310310 cos 10 101010 y r x r θ θ ===⋅= ===⋅=− 10 csc10 1 1010 sec 33 3 cot3 1 r y r x x y θ θ θ === ===− ===−
31. Because the tangent is positive and the cosine is negative, θ lies in quadrant III. In quadrant III, x is negative and y is negative. Thus, 44 tan= 33 y x θ == , x = –3, y = –4. Furthermore,
2222(3)(4)916 255 rxy=+=−+−=+ ==
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . 44 sin 55 33 cos 55 y r x r θ θ ===− ===− 55 csc 44 55 sec 33 33 cot 44 r y r x x y θ θ θ ===− ===− ===
32. Because the tangent is positive and the cosine is negative, θ lies in quadrant III. In quadrant III, x is negative and y is negative. Thus, 55 tan=,12,5 1212 y xy x θ ===−=− . Furthermore,
2222(12)(5)14425 16913 rxy=+=−+−=+ ==
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ 55 sin 1313 1212 cos 1313 1313 csc 55 1313 sec 1212 1212 cot 55 y r x r r y r x x y θ θ θ θ θ ===− ===−
33. Because the secant is negative and the tangent is positive, θ lies in quadrant III. In quadrant III, x is negative and y is negative. Thus, 3
sec3,1,3 1 r xr x θ =−===−= . Furthermore,
222
222 2 (1)3 918 822 xyr y y y += −+= =−= =−=−
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ . 2222
sin 33 11 cos 33 y r x r θ θ ===− ===− 22 tan22 1 33232
csc 4 22222 1122 cot 4 22222 y x r y x y θ θ θ ===
34. Because the cosecant is negative and the tangent is positive, θ lies in quadrant III. In quadrant III, x is negative and y is negative. Thus, 4
csc4,1,4 1 r yr y θ =−===−= . Furthermore,
222
222 2 (1)4 16115 15 xyr x x x += +−= =−= =−
Now that we know x, y, and r, we can find the remaining trigonometric functions of θ 11 sin 44 1515
cos 44 y r x r θ θ ===− ===− 111515
36. Because 170° lies between 90° and 180°, it is in quadrant II. The reference angle is 18017010 θ′ =°−°=°
37. Because 205° lies between 180° and 270°, it is in quadrant III. The reference angle is 20518025 θ′ =°−°=°
38. Because 210° lies between 180° and 270°, it is in quadrant III. The reference angle is 21018030 θ′ =°−°=°
39. Because 355° lies between 270° and 360°, it is in quadrant IV. The reference angle is 3603555 θ′ =°−°=°
40. Because 351° lies between 270° and 360°, it is in quadrant IV. The reference angle is 3603519 θ′ =°−°=° .
41. Because 7 4 π lies between 36 24 ππ = and 8 2 4 π π = , it is in quadrant IV. The reference angle is 787 2 4444 ππππ θπ ′ =−=−=
42. Because 5 4 π lies between 4 4 π π = and 36 24 ππ = , it is in quadrant III. The reference angle is 554 4444 θπππππ ′ =−=−=
43. Because 5 6 π lies between 3 26 ππ = and 6 6 π π = , it is in quadrant II. The reference angle is 565 6666 ππππ θπ ′ =−=−=
tan 15 151515 4415415
sec 15 151515 15 cot15 1 y x r x x y θ θ θ ===⋅= ===−⋅=− ===
35. Because 160° lies between 90° and 180°, it is in quadrant II. The reference angle is 18016020 θ′ =°−°=° .
44. Because 510 714 ππ = lies between 7 214 ππ = and 14 14 π π = , it is in quadrant II. The reference angle is 5752 7777 ππππ θπ ′ =−=−=
45. 150360210 −°+°=°
Because the angle is in quadrant III, the reference angle is 21018030 θ′ =°−°=° .
46. 250360110 −°+°=°
Because the angle is in quadrant II, the reference angle is 18011070 θ′ =°−°=° .
47. 33536025 −°+°=°
Because the angle is in quadrant I, the reference angle is 25 θ′ =°
48. 3593601 −°+°=°
Because the angle is in quadrant I, the reference angle is 1 θ′ =°
49. Because 4.7 lies between 3.14 π ≈ and 3 4.71 2 π ≈ , it is in quadrant III. The reference angle is 4.71.56θπ ′ =−≈
50. Because 5.5 lies between 3 4.71 2 π ≈ and 26.28 π ≈ , it is in quadrant IV. The reference angle is 25.50.78θπ ′ =−≈
51. 565360205 °−°=°
Because the angle is in quadrant III, the reference angle is 20518025 θ′ =°−°=°
52. 553360193 °−°=°
Because the angle is in quadrant III, the reference angle is 19318013 θ′ =°−°=°
53. 1717125 2 6666 ππππ π −=−=
Because the angle is in quadrant II, the reference angle is 5 66 ππ θπ ′ =−=
54. 111183 2 4444 ππππ π −=−=
Because the angle is in quadrant II, the reference angle is 3 44 ππ θπ ′ =−=
55. 2323167 4 4444 ππππ π −=−=
Because the angle is in quadrant IV, the reference angle is 7 2 44 ππ θπ ′ =−=
56. 1717125 4 3333 ππππ π −=−=
Because the angle is in quadrant IV, the reference angle is 5 2 33 ππ θπ ′ =−=
57. 1111165 4 4444 ππππ π −+=−+=
Because the angle is in quadrant III, the reference angle is 5 44 θπππ ′ =−= .
58. 1717247 4 6666 ππππ π −+=−+=
Because the angle is in quadrant III, the reference angle is 7 66 θπππ ′ =−=
59. 25253611 6 6666 ππππ π −+=−+=
Because the angle is in quadrant IV, the reference angle is 11 2 66 ππ θπ ′ =−=
60. 1313185 6 3333 ππππ π −+=−+=
Because the angle is in quadrant IV, the reference angle is 5 2 33 ππ θπ ′ =−=
61. 225° lies in quadrant III. The reference angle is 22518045 θ′ =°−°=° . 2 cos45 2 °=
Because the cosine is negative in quadrant III, 2 cos225=cos45= 2 °−°− .
62. 300° lies in quadrant IV. The reference angle is 36030060 3 sin60 2 θ′ =°−°=° °=
Because the sine is negative in quadrant IV, 3 sin300sin60 2 °=−°=−
63. 210° lies in quadrant III. The reference angle is 21018030 θ′ =°−°=° . 3 tan30 3 °=
Because the tangent is positive in quadrant III, 3 tan210=tan30 3 °°=
64. 240° lies in quadrant III. The reference angle is 24018060 sec602 θ′ =°−°=° °= .
Because the secant is negative in quadrant III, sec240sec602 °=−°−
65. 420° lies in quadrant I. The reference angle is 42036060 θ′ =°−°=° tan603 °=
Because the tangent is positive in quadrant I, tan420=tan60=3 °°
66. 405° lies in quadrant I. The reference angle is 40536045 tan451 θ′ =°−°=° °= .
Because the tangent is positive in quadrant I, tan405tan451 °=°=
67. 2 3 π lies in quadrant II. The reference angle is 232 3333 ππππ θπ ′ =−=−= 3 sin 32 π =
Because the sine is positive in quadrant II, 23 sin=sin 332 ππ = .
68. 3 4 π lies in quadrant II. The reference angle is 343 . 4444 2 cos 42 ππππ θπ π ′ =−=−= =
Because the cosine is negative in quadrant II, 32 cos=–cos 442 ππ =−
69. 7 6 π lies in quadrant III. The reference angle is 776 6666 θπππππ ′ =−=−= csc2 6 π =
Because the cosecant is negative in quadrant III, 7 csc=csc2 66 ππ −=− .
70. 7 4 π lies in quadrant IV. The reference angle is 787 2. 4444 cot1 4 ππππ θπ π ′ =−=−= =
Because the cotangent is negative in quadrant IV, 7 cot=cot1 44 ππ −=−
71. 9 4 π lies in quadrant I. The reference angle is 998 2 4444 θπππππ ′ =−=−= tan1 4 π =
Because the tangent is positive in quadrant I, 9 tan=tan1 44 ππ =
72. 9 2 π lies on the positive y-axis. The reference angle is 998 4 2222 θπππππ ′ =−=−= .
Because tan 2 π is undefined, 9 tan 2 π is also undefined.
73. –240° lies in quadrant II. The reference angle is 24018060 θ′ =°−°=° . 3 sin60 2 °=
Because the sine is positive in quadrant II, 3 sin(240)=sin60= 2 −°° .
74. –225° lies in quadrant II. The reference angle is 22518045. 2 sin45 2 θ′ =°−°=° °=
Because the sine is positive in quadrant II, 2 sin(225)sin45 2 −°=°=
75. 4 π lies in quadrant IV. The reference angle is 4 π θ′ = tan1 4 π =
Because the tangent is negative in quadrant IV, tan=tan1 44 ππ −−=−
76. 6 π lies in quadrant IV. The reference angle is 3 . tan 663 ππ θ ==
Because the tangent is negative in quadrant IV, 3 tan. 63 π −=−
77. sec495sec1352 °=°=−
78. 23 sec510sec150 3 °=°=−
79. 197 cotcot3 66 ππ ==
80. 133 cotcot 333 ππ ==
81. 2372 coscos 442 ππ ==
82. 35113 coscos 662 ππ ==
83. 1773 tantan 663 ππ −==
84. 11 tantan1 44 ππ −==
85. 173 sinsin 332 ππ −==
86. sinsin351 662 ππ −==
87. 3 sincoscossin 332 πππ π () () 31 11 22 31 22 13 2 =−−− =−+ =
88. sincos0sincos 46 ππ π () () 21 11 22 21 22 21 2 =−− =+ + =
89. 115115 sincoscossin 4646 ππππ + 2321 2222 62 44 62 4 =−+− =−− + =−
90. 175175 sincoscossin 3434 ππππ + 3212 2222 62 44 62 4 =−−+−
91. sintancos3155 243 πππ ()() 1 11 2 1 1 2 21 22 3 2 =−− =−− =−− =−
