Section 2.2 Trigonometric Functions of Non-Acute Angles

6. B; 480 360 120 and 180
(120° is in quadrant II)

1 bh, we have
2
1
1 2 s2 . s s s or A
2
2
2

85. Since A

A

Section 2.2 Trigonometric Functions of
Non-Acute Angles

8. Answers may vary. in QIII is
The reference angle for an angle given by
180 . The trigonometric functions of are as follows: sin sin csc csc cos cos sec sec tan tan cot cot

1. C; 180 98 82
(98° is in quadrant II)
2. F; 212 180 = 32°
(212° is in quadrant III)
3. A; 135 360 225 and
225 180 = 45°
(225° is in quadrant III)

9. Answers may vary.
Two coterminal angle have the same values for their trigonometric functions because the two angles have the same reference angle.

4. B; 60 360 300 and
360 300 60
(300° is in quadrant IV)

10. Answers may vary.
In quadrant II, the sine function is positive while the cosine and tangent functions are negative. 5. D; 750 2 360 30
(30° is in quadrant I)

cos

tan

11.

30

1
2

3
2

12.

45

2
2

13.

60

3
2

2
2
1
2
cos 120 cos 60
1
2

14. 120

3
2

15. 135

2
2

16. 150

sin150 sin 30
1
2

2
2
3
2

60

7. 2 is a good choice for r because in a 30 60 right triangle, the hypotenuse is twice the length of the shorter side (the side opposite to the 30 angle). By choosing 2, one avoids introducing a fraction (or decimal) when determining the length of the shorter side.
Choosing any even positive integer for r would have this result; however, 2 is the most convenient value.

86. Yes, the third angle can be found by subtracting the given acute angle from 90°, and the remaining two sides can be found using a trigonometric function involving the known angle and side.

sin

120

55

cot

sec

csc

3
3

3

2 3
3

2

1

1

2

2

3

3

tan135 tan 45
1
3
3

3
3
cot 120 cot 60
3
3 cot135 cot 45
1

2

2 3
3

sec120 sec 60
2

2 3
3

cot150 cot 30
3

sec150 sec 30
2 3
3

2

Copyright © 2013 Pearson Education, Inc.

2

2

56

Chapter 2 Acute Angles and Right Triangles

sin
17.

18.

210

240

1
2

3
2

tan

cos cos 210 cos 30
3
2
1
2

3
3
tan 240 tan 60
3

19. To find the reference angle for 300 , sketch this angle in standard position.

cot

3 cot 240 cot 60
3
3

sec

csc

sec 210 sec 30
2 3
3

2

2 3
3

2

sin 315 cos 315

20. To find the reference angle for 315 , sketch this angle in standard position.

tan 45 cot 45 sec 45 csc 45

2

21. To find the reference angle for 405 , sketch this angle in standard position.

The reference angle for 405 is
405 360 45 . Because 405 lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 45 . See the
Function Values of Special Angles table on page 50.) sin 405

cos 45

tan 405 cot 405 sec 405 csc 405

tan 45 cot 45 sec 45 csc 45

2
2
2
2

sin 45

cos 405

The reference angle is 360 315 45 .
Since 315° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.

2
2
1
1
2

cos 45

tan 315 cot 315 sec 315 csc 315

The reference angle is 360 300 60 .
Since 300° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.
3
sin 300 sin 60
2
1 cos 300 cos 60
2
tan 300 tan 60
3
3 cot 300 cot 60
3
sec 300 sec 60 2
2 3 csc 300 csc 60
3

2
2

sin 45

Copyright © 2013 Pearson Education, Inc.

1
1
2
2

Section 2.2 Trigonometric Functions of Non-Acute Angles

22. To find the reference angle for 420º, sketch this angle in standard position.

sec 80 csc 480

sec 60 csc 60

57

2
2 3
3

24. To find the reference angle for 495º, sketch this angle in standard position.

The reference angle for 420º is
420º − 360º = 60º. Because 420º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60 . See the
Function Values of Special Angles table on page 50.) sin 420

sin 60

cos 420

cos 60

tan 420

tan 60

cot 420

cot 60

sec 420

sec 60

csc 420

csc 60

495º is coterminal with 495º − 360º = 135º.
The reference angle is 180º − 135º = 45º.
Since 495° lies in quadrant II, the cosine, tangent, cotangent, and secant are negative.

sin 495

3
2
1
2
3

sin 45

cos 495

3
3
2

tan 45 cot 45 sec 45 csc 45

2
2

cos 45

tan 495 cot 495 sec 495 csc 495

2
2

1
1

2
2

25. To find the reference angle for 570º sketch this angle in standard position.

2 3
3

23. To find the reference angle for 480º, sketch this angle in standard position.

480º is coterminal with 480º − 360º = 120º.
The reference angle is 180º − 120º = 60º.
Because 480º lies in quadrant II, the cosine, tangent, cotangent, and secant are negative. sin 480

sin 60

3
2
1
2

cos 480

cos 60

tan 480

tan 60

3

cot 480

cot 60

3
3

570° is coterminal with 570º − 360º = 210º.
The reference angle is 210º − 180º = 30º.
Since 570º lies in quadrant III, the sine, cosine, secant, and cosecant are negative.
1
sin 570 sin 30
2
3 cos 570 cos 30
2
3 tan 570 tan 30
3
cot 570 cot 30
3
2 3 sec 570 sec 30
3
csc 570 csc 30
2

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58

Chapter 2 Acute Angles and Right Triangles

26. To find the reference angle for 750°, sketch this angle in standard position.

cos1305

cos 45

tan1305 cot1305 sec1305 csc1305 tan 45 1 cot 45 1 sec 45 csc 45

sec 60

csc 420

2
2
2
2

cot 60

sec 420

sin 45

tan 60

cot 420

sin1305

cos 60

tan 420

27. 1305º is coterminal with
1305 3 360 1305 1080 225 . The reference angle is 225º − 180º = 45º. Since
1305º lies in quadrant III, the sine, cosine, and secant and cosecant are negative.

sin 60

cos 420

750° is coterminal with 30 because
750 2 360 750 720 30 . Since
750 lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 30 .
1
sin 750 sin 30
2
3 cos 750 cos 30
2
3 tan 750 tan 30
3
cot 750 cot 30
3
2 3 sec 750 sec 30
3
csc 750 csc 30 2

sin 420

csc 60

3
3
2
2 3
3

29. To find the reference angle for 300 , sketch this angle in standard position.

The reference angle for 300 is
300 360 60 . Because 300 lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60°. See the
Function Values of Special Angles table on page 50.)
3
sin 300 sin 60
2
1 cos 300 cos 60
2
tan 300 tan 60
3
cot

300

cot 60

sec

300

sec 60

csc

2
2

3
2
1
2
3

300

csc 60

28. 1500º is coterminal with
1500 4 360 1500 1440 60 .
Because 1500º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60°.

Copyright © 2013 Pearson Education, Inc.

3
3
2
2 3
3

Section 2.2 Trigonometric Functions of Non-Acute Angles

30. −390° is coterminal with
390 2 360
390 720 330 . The reference angle is 360 330 30 . Since
−390° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.
1
sin 390 sin 30
2
3 cos 390 cos 30
2
3 tan 390 tan 30
3
cot 390 cot 30
3

sec

390

csc

390

2 3
3
csc 30
2

sec 30

31. –510° is coterminal with
510 2 360
510 720 210 . The reference angle is 210 180 30 . Since
–510° lies in quadrant III, the sine, cosine, and secant and cosecant are negative.
1
sin 510 sin 30
2
3 cos 510 cos 30
2
3 tan 510 tan 30
3
cot 510 cot 30
3

sec

510

csc

510

2 3
3
2

sec 30 csc 30

32. −1020º is coterminal with
1020 3 360
1020 1080 60 .
Because −1020º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60 .

sin 420

sin 60

cos 420

cos 60

tan 420

tan 60

cot 420

cot 60

sec 420

sec 60

csc 420

csc 60

3
2
1
2
3
3
3
2

59

33. −1290° is coterminal with
1290 4 360
1290 1440 150 .
The reference angle is 180 150 30 .
Since −1290° lies in quadrant II, the cosine, tangent, cotangent, and secant are negative.
1
sin 2670 sin 30
2
3 cos 2670 cos 30
2
3 tan 2670 tan 30
3
cot 2670 cot 30
3
2 3 sec 2670 sec 30
3
csc 2670 csc 30 2
34. −855º is coterminal with
855 3 360
855 1080 225 . The reference angle is 225º − 180º = 45º. Since
−855º lies in quadrant III, the sine, cosine, and secant and cosecant are negative.

2
2
2
2

sin

855

sin 45

cos

855

cos 45

tan cot 855
855

sec

855

sec 45

2

csc

855

csc 45

2

tan 45 cot 45

1
1

35. −1860º is coterminal with
1860 6 360
1860 2160 300 .
The reference angle is 360º − 300º = 60º.
Since −1860º lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.

3
2

sin

1860

cos

1860

tan

1860

tan 60

3

cot

1860

cot 60

3
3

sec

1860

csc

1860

2 3
3

Copyright © 2013 Pearson Education, Inc.

sin 60 cos 60

sec 60 csc 60

1
2

2
2 3
3

60

Chapter 2 Acute Angles and Right Triangles

36. −2205º is coterminal with
2205 7 360
2205 2520 315 .
The reference angle is 360º − 315º = 45º.
Since −2205º lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative.

sin

2205

cos

2205

tan cot 2205
2205

sec

2205

csc

2205

2
2

sin 45 cos 45

2
2

tan 45 cot 45 sec 45

1
1
2

csc 45

2

37. Since 1305 is coterminal with an angle of
1305 3 360 1305 1080 225 , it lies in quadrant III. Its reference angle is
225 180 45 . Since the sine is negative in quadrant III, we have

sin1305

2
.
2

sin 45

38. Since 1500 is coterminal with an angle of
1500 4 360 1500 1440 60 , it lies in quadrant I. Because 1500 lies in quadrant
I, the values of all of its trigonometric functions will be positive, so

sin1500

sin 60

3
.
2

39. Since 510 is coterminal with an angle of
510 2 360
510 720 210 , it lies in quadrant III. Its reference angle is
210 180 30 . Since the cosine is negative in quadrant III, we have

cos

510

cos 30

3
.
2

40. Since 1020 is coterminal with an angle of
1020 3 360
1020 1080 60 , it lies in quadrant I. Because 1020 lies in quadrant I, the values of all of its trigonometric functions will be positive, so tan 1020 tan 60
3.
41. Since −855º is coterminal with
855 3 360
855 1080 225 , it lies in quadrant III. Its reference angle is
225º − 180º = 45º. Since the cosecant is negative in quadrant III, we have. csc 855 csc 45
2.

42. Since −495º is coterminal with an angle of
495 2 360
495 720 225 , it lies in quadrant III. Its reference angle is
225 180 45 . Since the secant is negative in quadrant III, we have sec 495 sec 45
2.
43. Since 3015º is coterminal with
3015 8 360 3015 2880 135 , it lies in quadrant II. Its reference angle is
180º − 135º = 45º. Since the tangent is negative in quadrant II, we have tan 3015 tan 45
1.
44. Since 2280º is coterminal with
2280 6 360 2280 2160 120 , it lies in quadrant II. Its reference angle is
180º − 120º = 60º. Since the cotangent is negative in quadrant II, we have

cot 2280

3
.
3

cot 60

45. sin 2 120

2

3
2

cos 2 120

1
2

3
4

1
4

2
4

2

2
2
2
1
4

1
2

2

2

46. sin 225

cos 225

47. 2 tan 2 120

2

3sin 2 150
3

2 3

3
1
4

3

48. cot 2 135

1

2

1
2

49. sin 2 225

1

2
2

02

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2

23
4

1

cos 2 270

50. cot 2 90

cos 2 180

2

sin 30 4 tan 45
1
1
41 1
4
2
2

2

02

1

2

9
2

tan 2 60

2

3

sec2 180

2
2

2

2
4

3

csc2 135
2

2

1 2 1

7
2

2

Section 2.2 Trigonometric Functions of Non-Acute Angles

51. cos 2 60

sec 2 150
2

1
2
1 4
4 3
52. cot 2 135

2 3
3
29
4
12

3

4

57. sin 2 45

2

tan 4 60
2

1

csc 2 210
2

2

2
2

2
2

04

1 9 10

3
2

cos 60

1
2

3 1
2

3 1
, the statement is false.
2

54. sin 30

sin 60 sin 30 60
Evaluate each side to determine whether this statement is true or false.
1
3 1
3
sin 30 sin 60 and 2
2
2 sin 30 60 sin 90 1
Since

2

sec2 60 is true.

22

4 . Since 4 = 4, the statement

3
2

2

3

3, the statement is false.

?

2 cos 45
Evaluate each side to determine whether this statement is true or false. cos 2 45 cos 90 0 and

?

1 2

3
2

3
1
2
Since

1
2

2
2

2 cos 45

2

Since 0

2, the statement is false.

2

?

60. sin 2 30

2 sin 30 cos 30
Evaluate each side to determine whether this statement is true or false.

2

1 2

3
4

Since

3 and 2

sin 60

3
2

2

1
2

3
2

3
2

3
, the statement is true.
2

61. 150º is in quadrant II, so the reference angle is
180º − 150º = 30º.

cos 30

56. cos 60 2 cos 2 30 1
Evaluate each side to determine whether this statement is true or false.
1
cos 60 and 2

2 cos 2 30

1 3 1 4 and

59. cos 2 45

2 sin 30 cos 30

1 and 2cos 30
2
1
2

2

1, the given statement is false.

55. cos 60 2 cos 30
Evaluate each side to determine whether this statement is true or false.

Since

1

3

sin 2 30

?

cos 60

2
4

1

?

3

2
4

tan 2 60

60 cos 30 cos 60
Evaluate each side to determine whether this statement is true or false. cos 30 60 cos 90 0 and

1

1

2

58. tan 2 60 1 sec 2 60
Evaluate each side to determine whether this statement is true or false.

sin 4 180

53. cos 30

Since 0

2

?

Since 1 = 1, the statement is true.

?

cos 30

cos 2 45

61

x r x

r cos 30

6

3
2

3 3

1 y 3 y r sin 30 6
2
r
Since 150º is in quadrant II, the x- coordinate will be negative. The coordinates of P are and sin 30

3 3, 3 .

1

1
2

1
, the statement is true.
2
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Chapter 2 Acute Angles and Right Triangles

62

62. 225º is in quadrant III, so the reference angle is 225º − 180º = 45º.

x r cos 45

x

r cos 45

10

2
2

69. If

90
180
90
180
180
90
Since 180º is coterminal with
−180º + 360º = 180º and −90º is coterminal lies in quadrant with −90º + 360º = 270º, is negative.
III, and sin

5 2

and
2
y
5 2 y r sin 45 10
2
r
Since 225º is in quadrant III, both the x- and ycoordinates will be negative. The coordinates sin 45

of P are

0.8

2

0.6

2

, sin 2

cos 2

64. For every angle
2 2
3

angle
65. If

90

9
16

, sin 2
4
9

cos 2

145
144

90

2
3

180

45

90
Thus
sec
68. If

90
Thus
cot

2

2

is positive.

is in the interval 90 ,180 , then
180

45

and n 360 are coterminal angles, so the sine of each of these will result in the same value. 72.
3
4

and sin

90 . Thus

2

71.

and n 360 are coterminal angles, so the cosine of each of these will result in the same value.

1 . Since

is in the interval 90 ,180 , then

90 . Thus

.

73. If n is even, and n 180 are coterminal angles, so the tangent of each of these will result in the same value. If n is odd, and n 180 have the same reference angle, but are positioned two quadrants apart. The tangent is positive for angles located in quadrants I and III, and negative for angles located in quadrants II and IV, so the tangent of each angle is the same value.

2

74. The reference angle for 115° is
180°−115° = 65°. Since 115° is in quadrant II decreases on the the cosine is negative. Cos interval (90°, 180°) from 0 to –1. Therefore, cos 115° is closest to −0.4.

180
270
180 360 .
180 lies in quadrant IV, and
180 is positive.

75. The reference angle for 115° is
180°−115° = 65°. Since 115° is in quadrant II the cosine is negative. Sin decreases on the interval (90°, 180°) from 1 to 0. Therefore, sin 115° is closest to 0.9.

2

lies in quadrant I, and sin
67. If

0,

1 , there is no

for which cos

lies in quadrant I, and cos
66. If

1 . Since

0.6 and
0 and sin

is in the interval 90 ,180 , then

90
180
90
180
180
90
Since 180º is coterminal with
−180º + 360º = 180º and −90º is coterminal lies in quadrant with −90º + 360º = 270º, is negative.
III, and cos

0.64 0.36 1 , there is an

for which cos angle sin
0.8 . Since cos lies in quadrant IV.
3 2
4

70. If

5 2, 5 2 .

63. For every angle

is in the interval 90 ,180 , then

2

is positive.

is in the interval 90 ,180 , then

is in the interval 90 ,180 , then
180
270
180 360 .
180 lies in quadrant IV, and
180 is negative.

2
. Sine and
2
cosine are opposites in quadrants II and IV.
Thus, 180
180 45 135 in quadrant II, and 360
360 45 315 in quadrant IV.

76. When

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45 , sin

cos

Section 2.2 Trigonometric Functions of Non-Acute Angles

2
. Sine and
2
cosine are both positive in quadrant I and both negative in quadrant III. Since
180 45 180 225 , 45º is the quadrant I angle, and 225º is the quadrant III angle. 77. When

78. L

45 , sin

2

1

cos

S2

200 h S tan

63

82. sec
2
must lie in
Since sec is negative, quadrants II or III. Since the absolute value of sec is 2, the reference angle, must be
45 . The quadrant II angle equals 180
180 45 135 , and the equals quadrant III angle
180
180 45 225 .

2
2
Since cos is positive, must lie in quadrants I or IV. Since one angle, namely
45 , lies in quadrant I, that angle is also the reference angle, . The angle in quadrant IV will be 360
360 45 315 .

83. cos
(a) Substitute h = 1.9 ft,
2

L

0.9 ,

L

3,

4 , and S = 336 ft:
4 ( 3) 3362
200 1.9 336 tan .9

550 ft

(b) Substitute h = 1.9 ft,
2

1

1.5 ,

1

3,

4 , and S = 336 ft:

3
3

84. cot

4 ( 3) 3362
200 1.9 336 tan1.5

369 ft

(c) Answers will vary.

Since cot is negative, must lie in quadrants II or IV. Since the absolute value of
3
, the reference angle, must 3 equals be 60 . The quadrant II angle
180
180 60 120 , and the equals quadrant IV angle
360
360 60 300 . cot 1
2
must lie in
Since sin is positive, quadrants I or II. Since one angle, namely
30 , lies in quadrant I, that angle is also the reference angle, . The angle in quadrant II will be 180
180 30 150 .

79. sin

3
2
Since cos is positive, must lie in quadrants I or IV. Since one angle, namely
30 , lies in quadrant I, that angle is also the reference angle, . The angle in quadrant IV will be 360
360 30 330 .

80. cos

81. tan
3
Since tan is negative, must lie in quadrants II or IV. Since the absolute value of must be tan is 3, the reference angle, equals 60 . The quadrant II angle
180
180 60 120 , and the equals quadrant IV angle
360
360 60 300 .

is

85. csc
2
Since csc is negative, must lie in quadrants III or IV. The absolute value of csc is 2, so the reference angle, , is 30 .
The angle in quadrant III will be
180
180 30 210 , and the quadrant IV angle is
360
360 30 330 .

3
2

86. sin

Since sin is negative, must lie in quadrants III or IV. The absolute value of
3
, so the reference angle, , is
2
60°. The angle in quadrant III will be
180
180 60 240 , and the quadrant IV angle is
360
360 60 300 . sin is

Copyright © 2013 Pearson Education, Inc.

Chapter 2 Acute Angles and Right Triangles

3
3
must lie in
Since tan is positive, quadrants I or III. Since one angle, namely
30 , lies in quadrant I, that angle is also the reference angle, . The angle in quadrant III will be 180
180 30 210 .

87. tan

88. cos

1
2

Since cos is negative, must lie in quadrants II or III. Since the absolute value of
1
cos is , the reference angle, must be
2
60 . The quadrant II angle equals 180
180 60 120 , and the equals quadrant III angle
180
180 60 240 .
89. csc
2
must lie in
Since csc is negative, quadrants III or IV. Since the absolute value must of csc is 2, the reference angle, equals be 45 . The quadrant III angle
180
180 45 225 . and the quadrant IV angle equals 360
360 45 315 .
90. cot
1
must lie in
Since cot is negative, quadrants II or IV. Since the absolute value of cot is 1 the reference angle, must be equals 45 . The quadrant II angle
180
180 45 135 . and the quadrant IV angle equals 360
360 45 315 .

3. To find values of the cotangent, secant, and cosecant functions with a calculator, it is necessary to find the reciprocal of the reciprocal function value.
4. To determine the cosecant of an angle, we find the reciprocal of the cosine of the angle, but to determine the angle with a given cosecant value, we find the sine inverse of the reciprocal of the value.

For Exercises 5−21, be sure your calculator is in degree mode. If your calculator accepts angles in degrees, minutes, and seconds, it is not necessary to change angles to decimal degrees. Keystroke sequences may vary on the type and/or model of calculator being used. Screens shown will be from a
TI-84 Plus calculator. To obtain the degree (°) and ( ) symbols, go to the ANGLE menu (2nd APPS).
′

64

For Exercises 5−15, the calculation for decimal degrees is indicated for calculators that do not accept degree, minutes, and seconds.
5. sin 38 42

0.62524266

6. cos 41 24

0.75011107

7. sec13 15

1.0273488

Section 2.3 Finding Trigonometric
Function Values Using a
Calculator
1. The CAUTION at the beginning of this section suggests verifying that a calculator is in degree mode by finding sin 90°. If the calculator is in degree mode, the display should be 1.
2. When a scientific or graphing calculator is used to find a trigonometric function value, in most cases the result is an approximate value.

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