chapter 1 trigonometric functions - testbanku.eu · section 1.1 angles 5 copyright © 2011 pearson...

Copyright © 2011 Pearson Education Inc. Publishing as Addison-Wesley. 1

Chapter 1 Trigonometric Functions

1.1 Angles

1.1 Practice Exercises

1. complement: 90º − 67° = 23º supplement: 180º − 67° = 113º

2. a.

( )

37 12 37

24 15 45

61 27 82

61 60 2261 22 61 22

2728 28

αβ

α β

′ ′′= + +′ ′′+ = + + +

′ ′′+ = + +

= + + +′′ ′′= + + =

′′′° °′ ′

b. 37 12 37 37 11 1 3737 11 60 37 37 11 9736 1 11 9736 60 11 97 36 71 97

α ′ ′′ ′ ′ ′′= + + = + + +′ ′′ ′′ ′ ′′= + + + = + +

=

°° °° + ° + +′ ′′

= ° + + + = ° + +′ ′ ′′ ′ ′′

( )36 71 9724 15 45

12 56 52 12 56 52

αβ

α β

= + +′ ′′− = − ° + +′ ′′

− = + + =′ ′′ ′ ′′

3. 9 22

13 9 22 13 13.1660 3600

⎛ ⎞° = + + ° ≈ °′ ′′ ⎜ ⎟⎝ ⎠

4. 41.275 41 0.275(60 ) 41 16.541 16 0.5(60 ) 41 16 30

° = ° + = ° +′ ′= ° + + = °′ ′′ ′ ′′

5.

6. a. Since 765° is a positive angle greater than 720° = (360°)(2), subtract 720° to get an angle between 0° and 360°. 765° – 720° = 45°. Therefore, the 765° and 45° angles are coterminal. All angles that are coterminal with 45° are 45 360 ,n° + ⋅ ° where n is an integer.

b. Adding 2(360°) = 720° to –570° gives an angle between 0° and 360°. –570° + 720° = 150°

Therefore, the –570° and 150° angles are coterminal. All angles that are coterminal with 150° are 150 360 ,n° + ⋅ ° where n is an integer.

7.

In the figure, OP is the diagonal of the 3 3× square PQRO. So OP bisects the right angle

QOR. Therefore 1

180 90 225 .2

θ = ° + ⋅ ° = °

Since there are infinitely many angles with OP as the terminal side, and those angles differ by 360°, 225 360 ,nθ = ° + ⋅ ° n any integer.

8.

Angles 2 and 3 are supplements, and angles 3 and 7 are congruent, so angles 2 and 7 are supplements. Therefore,

( ) ( )( )

2 7 1806 3 8 43 18014 40 180 14 140 10

m mx xx x x

∠ + ∠ = °⇒− ° + + ° = °⇒+ ° = °⇒ = °⇒ = °

Thus, ( )2 6 10 3 57m∠ = ⋅ − ° = ° and

( )7 8 10 43 123 .m∠ = ⋅ + ° = °

1.1 A Exercises: Basic Skills and Concepts

1. The degree measure of one complete revolution is 360°.

2. The sum of two complementary angles is 90°.

3. An angle is in standard position if the initial side is the positive x-axis and the vertex is at the origin.

2 Chapter 1 Trigonometric Functions

Copyright © 2011 Pearson Education Inc. Publishing as Addison-Wesley.

4. For any integer n, an angle of 360nθ° + ⋅ ° has the same terminal side as the angle of θ degrees.

5. True

6. False. An angle is standard position is quadrantal if its terminal side lies on a coordinate axis.

7. complement: 90° – 47° = 43° supplement: 180° – 47° = 133°

8. complement: 90° – 75° = 15° supplement: 180° – 75° = 105°

9. complement: none because the measure of the angle is greater than 90° supplement: 180° – 120° = 60°

10. complement: none because the measure of the angle is greater than 90° supplement: 180° – 160° = 20°

11. complement: none because the measure of the angle is greater than 90°; supplement: none because the measure of the angle is greater than 180°

12. complement: none because the measure of the angle is negative; supplement: none because the measure of the angle is negative

13. 90 θ° −

14. An obtuse angle does not have a complement.

15. 180 θ° −

16. 180 θ° −

17. For an angle θ to be its own complement, we must have 90 2 90θ θ θ= ° − ⇒ = °⇒

45 .θ = °

18. For an angle θ to be its own supplement, we must have 180 2 180θ θ θ= ° − ⇒ = °⇒

90 .θ = °

19. a. 34 12

27 5

61 17 61 17

αβ

α β

= + ′+ = + + ′+ = + = °′ ′

b. 34 12

27 5

7 7 7 7

αβ

α β

= + ′− = − + ′− = + = °′ ′

20. a. 64 37

23 12

87 49 87 49

αβ

α β

= + ′+ = + + ′+ = + = °′ ′

b. 64 37

23 12

41 25 41 25

αβ

α β

= + ′− = − + ′− = + =′ ′

21. a. 47 54

12 14

59 68 59 60 8

59 1 8 60 8 60 8

αβ

α β

= + ′+ = + + ′

′+ = + = ° + +′ ′= + + = + =′ ′ ′

b. 47 5412 1435 40 35 40

αβ

α β ° °

= ° + ′− = − ° + ′− = + =′ ′

22. a. 35 43

15 35

50 78 50 60 18

50 1 18 51 18

51 18

αβ

α β

= + ′+ = + + ′

′+ = + = + +′ ′= + + = +′ ′= ′

b. 35 4315 3520 8 20 8

αβ

α β

= ° + ′− = − ° + ′+ = ° + = °′ ′

23. a. 15 38

13 45

28 83 28 60 23

28 1 23 29 23

29 23

αβ

α β

= + ′+ = + + ′

′+ = + = + +′ ′= + + = +′ ′= ′

b. 15 38 14 1 3814 60 38 14 98

α = ° + = ° + ° +′ ′= ° + + = ° +′ ′ ′

14 98

13 45

1 53 1 53

αβ

α β

= + ′− = − + ′− = + =′ ′

24. a. 28 42

16 56

44 98 44 60 38

44 1 38 45 38

45 38

αβ

α β

= + ′+ = + + ′

′+ = + = + +′ ′= + + = +′ ′= ′

Section 1.1 Angles 3


b. 28 42 27 1 4227 60 42 27 102

α = ° + = ° + ° +′ ′= ° + + = ° +′ ′ ′

27 102

16 56

11 46 11 46

αβ

α β

= + ′− = − + ′− = + =′ ′

25. a. 70 12 15

54 18

124 30 15 124 30 15

αβ

α β

= + +′ ′′+ = + + ′+ = + + =′ ′′ ′ ′′

b. 70 12 15 69 1 12 1569 60 12 15 69 72 15

α = + + = ° + ° + +′ ′′ ′ ′′= ° + + + = ° + +′ ′ ′′ ′ ′′

( )69 72 1554 18

15 54 15 15 54 15

αβ

α β

= ° + +′ ′′− = − ° + ′− = ° + + = °′ ′′ ′ ′′

26. a. 16 15 12

12 23

28 38 12 28 38 12

αβ

α β

= + +′ ′′+ = + + ′+ = + + =′ ′′ ′ ′′

b. 16 15 12 15 1 15 1215 60 15 12 15 75 12

α = + + = ° + ° + +′ ′′ ′ ′′= ° + + + = ° + +′ ′ ′′ ′ ′′

( )15 75 1212 23

3 52 12 3 52 12

αβ

α β

= ° + +′ ′′− = − ° + ′− = ° + + = °′ ′′ ′ ′′

27. a. 12 15 22

8 27 36

20 42 58 20 42 58

αβ

α β

= + +′ ′′+ = + + +′ ′′+ = + + =′ ′′ ′ ′′

b. 12 15 22 12 14 1 2212 14 60 22 12 14 8211 1 14 8211 60 14 82 11 74 82

α = + + = ° + + +′ ′′ ′ ′ ′′= ° + + + = ° + +′ ′′ ′′ ′ ′′= ° + ° + +′ ′′= ° + + + = ° + +′ ′ ′′ ′ ′′

( )11 74 82

8 27 36

3 47 46 3 47 46

αβ

α β

= ° + +′ ′′− = − ° + +′ ′′− = ° + + = °′ ′′ ′ ′′

28. a. 89 45 40

56 35 46

145 80 86

145 60 20 60 26

145 1 20 1 26

146 21 26 146 21 26

αβ

α β

= + +′ ′′+ = + + +′ ′′

+ = + +′ ′′= + + + +′ ′ ′′ ′′= + + + +′ ′ ′′= + + =′ ′′ ′ ′′

b. 89 45 40 89 44 1 40

89 44 60 40 89 44 100

α ′ ′′ ′ ′ ′′= + + = + + +′ ′′ ′′ ′ ′′= + + + = + +

( )89 44 100

56 35 46

33 9 54 33 9 54

αβ

α β

= + +′ ′′− = − + +′ ′′

− = ° + + = °′ ′′ ′ ′′

29. a. 187 56 33

220 34 67

407 90 100

407 60 30 60 40

407 1 30 1 40

408 31 40 408 31 40

αβ

α β

= + +′ ′′+ = + + +′ ′′

+ = + +′ ′′= + + + +′ ′ ′′ ′′= + + + +′ ′ ′′= + + =′ ′′ ′ ′′

b. 220 34 67219 1 34 67219 60 34 67219 94 67

β = ° + +′ ′′= ° + ° + +′ ′′= ° + + +′ ′ ′′= ° + +′ ′′

( )( )

187 56 93219 94 67

32 38 34 32 38 34

αβ

α β

= ° + +′ ′′− = − ° + +′ ′′− = − ° + + = − °′ ′′ ′ ′′

30. a. 240 35 48

335 6 54

575 41 102

575 41 60 42

575 41 1 42

575 42 42 575 42 42

αβ

α β

= + +′ ′′+ = + + +′ ′′+ = + +′ ′′

= + + +′ ′′ ′′= + + +′ ′ ′′= + + =′ ′′ ′ ′′

b. 335 6 54334 1 6 54334 60 6 54334 66 54

β = + +′ ′′= ° + ° + +′ ′′= ° + + +′ ′ ′′= ° + +′ ′′

( )( )

240 35 48334 66 54

94 31 6 94 31 6

αβ

α β

= ° + +′ ′′− = − ° + +′ ′′− = − ° + + = − °′ ′′ ′ ′′

31. 45

70 45 70 70.7560

⎛ ⎞° = + ° = °′ ⎜ ⎟⎝ ⎠

32. 38

38 38 38 38.6360

⎛ ⎞° = + ° ≈ °′ ⎜ ⎟⎝ ⎠

33. 42 30

23 42 30 23 23.7160 3600

⎛ ⎞° = + + ° ≈ °′ ′′ ⎜ ⎟⎝ ⎠

34. 50 50

45 50 50 45 45.8560 3600

⎛ ⎞° = + + ° ≈ °′ ′′ ⎜ ⎟⎝ ⎠



35. 42 57

15 42 57 15 15.7260 3600

⎛ ⎞− ° = − + + ° ≈ − °′ ′′ ⎜ ⎟⎝ ⎠

36. 18 13

70 18 13 70 70.3060 3600

⎛ ⎞− ° = − + + ° ≈ − °′ ′′ ⎜ ⎟⎝ ⎠

37. 27.32 27 0.32(60 ) 27 19.227 19 0.2(60 ) 27 19 12

° = ° + = ° +′ ′= ° + + = °′ ′′ ′ ′′

38. 120.64 120 0.64(60 ) 120 38.4120 38 0.4(60 ) 120 38 24

° = ° + = ° +′ ′= ° + + = °′ ′′ ′ ′′

39. 13.347 13 0.347(60 ) 13 20.8213 20 0.82(60 ) 13 20 49

° = ° + = ° +′ ′= ° + + = °′ ′′ ′ ′′

40. 110.433 110 0.433(60 ) 110 25.98110 25 0.98(60 ) 110 25 59

° = ° + = ° +′ ′= ° + + = °′ ′′ ′ ′′

41. 19.0511 19 0.0511(60 ) 19 3.06619 3 0.066(60 ) 19 3 4

° = ° + = ° +′ ′= ° + + = °′ ′′ ′ ′′

42. 82.7272 82 0.7272(60 ) 82 43.63282 43 0.632(60 ) 82 43 38

° = ° + = ° +′ ′= ° + + = °′ ′′ ′ ′′

43.

Quadrant II

44.

Quadrant IV

45.

Quadrant I

46.

Quadrant II

47.

Quadrant IV

48.

Quadrant III

49.

Quadrant II

50.

Quadrant I

51. Since 400° is a positive angle greater than 360°, subtract 360° to get an angle between 0° and 360°: 400° – 360° = 40°. Therefore, the 400° and 40° angles are coterminal. All angles that are coterminal with 40° are 40 360 ,n° + ⋅ ° where n is an integer.



52. Since 700° is a positive angle greater than han 360°, subtract 360° to get an angle between 0° and 360°: 700° – 360° = 340°. Therefore, the 700° and 340° angles are coterminal. All angles that are coterminal with 340° are 340 360 ,n° + ⋅ ° where n is an integer.

53. Since 1785° is a positive angle greater than 1440° = (360°)(4), subtract 1440° to get an angle between 0° and 360°. 1785° – 1440° = 345°. Therefore, the 1785° and 345° angles are coterminal. All angles that are coterminal with 345° are 345 360 ,n° + ⋅ ° where n is an integer.

54. Since 2064° is a positive angle greater than 1800° = (360°)(5), subtract 1800° to get an angle between 0° and 360°. 2064° – 1800° = 264°. Therefore, the 2064° and 264° angles are coterminal. All angles that are coterminal with 264° are 264 360 ,n° + ⋅ ° where n is an integer.

55. Adding 360° to –50° gives an angle between 0° and 360°: –50° + 360° = 310° Therefore, the –50° and 310° angles are coterminal. All angles that are coterminal with 310° are 310 360 ,n° + ⋅ ° where n is an integer.

56. Adding 360° to –225° gives an angle between 0° and 360°: –225° + 360° = 135° Therefore, the –225° and 135° angles are coterminal. All angles that are coterminal with 135° are 135 360 ,n° + ⋅ ° where n is an integer.

57. Adding 2(360°) = 720° to –400° gives an angle between 0° and 360°: –400° + 720° = 320° Therefore, the –400° and 320° angles are coterminal. All angles that are coterminal with 320° are 320 360 ,n° + ⋅ ° where n is an integer.

58. Adding 2(360°) = 720° to –700° gives an angle between 0° and 360°: –700° + 720° = 20° Therefore, the –700° and 20° angles are coterminal. All angles that are coterminal with 20° are 20 360 ,n° + ⋅ ° where n is an integer.

For exercises 59–66, review the explanations in Example 7 and Practice Problem 7.

59. (3, 3) lies in Quadrant I, so a line through that point forms a 45° angle with the positive x-axis. Thus, 45 360 ,nθ = ° + ⋅ ° n any integer.

60. (4, –4) lies in Quadrant IV, so a line through that point forms a 270° + 45° = 315° angle with the positive x-axis. Thus, 315 360 ,nθ = ° + ⋅ ° n any integer.

61. (–5, 5) lies in Quadrant II, so a line through that point forms a 90° + 45° = 135° angle with the positive x-axis. Thus, 135 360 ,nθ = ° + ⋅ ° n any integer.

62. (–2, –2) lies in Quadrant III, so a line through that point forms a 180° + 45° = 225° angle with the positive x-axis. Thus, 225 360 ,nθ = ° + ⋅ ° n any integer.

63. (1, 0) lies on the positive x-axis, so a line through that point forms a 0° angle with the positive x-axis. Thus, 0 360 ,nθ = ° + ⋅ ° n any integer.

64. (0, 2) lies on the positive y-axis, so a line through that point forms a 90° angle with the positive x-axis. Thus, 90 360 ,nθ = ° + ⋅ ° n any integer.

65. (–3, 0) lies on the negative x-axis, so a line through that point forms a 180° angle with the positive x-axis. Thus, 180 360 ,nθ = ° + ⋅ ° n any integer.

66. (0, –4) lies on the negative y-axis, so a line through that point forms a 270° angle with the positive x-axis. Thus, 270 360 ,nθ = ° + ⋅ ° n any integer.

1.1 B Exercises: Applying the Concepts

67. 60 sec

20 min 20 min 1200 secmin

3606 rev 6 rev 2160

rev

= ⋅ =

°= ⋅ = °

Now use a proportion. 10 sec 10 2160

181200 sec 2160 1200

xx

° ⋅= ⇒ = = °°

68. 360

500 rev 500 rev 180,000rev

°= ⋅ = °

Now use a proportion. 1 sec 180,000

300060 sec 180,000 60

xx

°= ⇒ = = °°

69. There are 50 minutes or 56

of an hour from

4:20 pm to 5:10 pm. The minute hand travels

360° in one hour, so it travels 56

360 300⋅ ° = °

in the given time period.



70. There are 4 hours 30 minutes from 3:15 pm to 7:45 pm. The hour hand travels 360° in 12

hours, so it travels 4.5

360 13512

⋅ ° = ° in the

given time period.

71.

The sum of the angles is 180°, so we have ( ) ( )30 75 15 1802 120 180 2 60 30x xx x x+ + + + = ⇒+ = ⇒ = ⇒ =

72.

The sum of the angles is 180°, so we have

( ) ( )3 15 40 2 20 1805 75 180 5 105 21

x xx x x+ + + + = ⇒+ = ⇒ = ⇒ =

The marked angles are (3x + 15) = 3(21) + 15 = 78°, and (2x + 20) = 2(21) + 20 = 62°.

Use the figure below for exercises 73 and 74.

73. 3 ; 180 3 180

4 180 45

45 180 135

α β α β β ββ β

α α

= + = °⇒ + = °⇒= °⇒ = °

+ ° = °⇒ = °

Since α and γ are vertical angles, they are

equal and 135 .γ = ° Since β and θ are

vertical angles, they are equal and 45 .θ = °

74. ( ) ( )( ) ( )

( )( )

4 2 ; 3 3

180 4 2 3 3 180

7 5 180 7 175 25

4 25 2 102

3 25 3 78

x x

x x

x x x

α βα β

αβ

= + ° = + °+ = °⇒ + + + = °⇒+ = °⇒ = °⇒ = °= ⋅ + ° = °= ⋅ + ° = °

Since α and γ are vertical angles, they are

equal and 102 .γ = ° Since β and θ are

vertical angles, they are equal and 78 .θ = °


75. The angle vertical with α is the interior angle

on the same side of the transversal as ,β so the

angles are supplementary.

( ) ( )

( ) ( )

( )( )

5 70 , 4 29180

5 70 4 29 1809 99 180 9 81 9

5 9 70 1154 9 29 65

x x

x xx x x

α βα β

αβ

= + ° = + °+ = °⇒+ + + = °⇒+ = °⇒ = °⇒ = °= ⋅ + ° = °= ⋅ + ° = °

76. The angle vertical with α is the interior angle on the same side of the transversal as ,β so the

angles are supplementary. ( ) ( )

( ) ( )

( )( )

7 94 , 5 10180

7 94 5 10 18012 84 180 12 96 8

7 8 94 1505 8 10 30

x x

x xx x x

α βα β

αβ

= + ° = − °+ = °⇒+ + − = °⇒+ = °⇒ = °⇒ = °

= ⋅ + ° = °= ⋅ − ° = °

77.

m BOF α∠ = and ,m COF β∠ = so

40 30 70m BOC α β∠ = + = ° + ° = °

F



78.

6030

m BCD m BCDm DCEα = ∠ ⇒ ∠ = °⇒∠ = °

Angles CEF and DCE are interior angles on the same side of the transversal, so they are supplementary.

30 180 150m CEF m CEF∠ + ° = °⇒ ∠ = °

1.1 C Exercises: Beyond the Basics

79.

The sum of the measures of angles AOC, COE, and BOE is 180°, so we have 2 7

90 180 3 90 30.3 3

x x x x+ + = ⇒ = ⇒ =

Thus, 2

30 203

m AOC∠ = ⋅ ° = ° and

730 70 .

3m BOE∠ = ⋅ ° = ° Angles AOC and

DOB are vertical angles, so their measures are equal. 20 .m DOB y∠ = = ° Angles DOB and

DOA are supplements, so 180

20 180 160 .m DOA m DOBz z∠ + ∠ = °⇒+ ° = °⇒ = °

80.

a. From exercise 77, we know that

45 35 80m BPD m ABP m CDP∠ = ∠ + ∠

= ° + ° = °

b. ABQ ABP PBQCDQ CDP PDQ

∠ = ∠ +∠∠ = ∠ + ∠

Adding the two equations and BQD∠ , we

have

( )

( )( )

( )

ABQ BQD CDQABP PBQ BQD

CDP PDQPBQ BQD PDQ

ABP CDP

∠ + ∠ + ∠= ∠ + ∠ + ∠

+ ∠ +∠= ∠ + ∠ + ∠

+ ∠ + ∠

Recall that the sum of the interior angles of a quadrilateral is 360°, so

360

80 360280

PBQ BQD PDQ BPDPBQ BQD PDQPBQ BQD PDQ

∠ + ∠ + ∠ + ∠ = °⇒∠ + ∠ + ∠ + ° = °⇒∠ + ∠ + ∠ = °

From part (a), we know that 80 ,ABP CDP∠ +∠ = ° so

( )( ) 280 80

360

PBQ BQD PDQABP CDP

∠ +∠ + ∠+ ∠ + ∠ = ° + °

= °

81.

MP bisects angle AMN, MQ bisects angle BMN, NQ bisects angle DNM, and NP bisects angle CNM. Therefore, ,AMP NMP∠ = ∠

, ,BMQ NMQ MNQ DNQ∠ = ∠ ∠ = ∠ and

.MNP CNP∠ = ∠ Since the sum of the interior angles on the same side of a transversal is 180°, we have

( )( )

( )( )

( )

180

180

1802 180

90 (1)

BMN DNMBMQ NMQ

DNQ MNQNMQ NMQ

MNQ MNQNMQ MNQ

NMQ MNQ

∠ + ∠ = °⇒∠ + ∠

+ ∠ + ∠ = °⇒∠ + ∠

+ ∠ + ∠ = °⇒∠ + ∠ = °⇒

∠ + ∠ = °

Similarly, we can show that 90 .MNP PNM∠ + ∠ = ° (2)

In triangle MQN, we have 180

90 180 90NMQ MNQ Q

Q Q

∠ + ∠ + ∠ = °⇒° + ∠ = °⇒ ∠ = °

In triangle MPN, we have 180

90 180 90MNP PNM P

P P∠ + ∠ + ∠ = °⇒° + ∠ = °⇒ ∠ = °

(continued on next page)



(continued from page 7)

Now we must show that angles PMQ and PNQ are also right angles. We know that angles AMN and DNM are equal since they are alternate interior angles. Each angle is bisected, so we can conclude that angle PMN = angle MNQ. Similarly, we can deduce that angle QMN = angle MNP. Adding equations (1) and (2) then substituting, we have

( ) ( )180

1802 2 180

9090

NMQ MNQ MNPPNM

NMQ PNM MNQ MNP

NMQ MNPNMQ MNP NMQ NMPPMQ

∠ + ∠ + ∠+∠ = °⇒

∠ + ∠ + ∠ + ∠= °⇒

∠ + ∠ = °⇒∠ + ∠ = ∠ + ∠ = °⇒∠ = °

Similarly, we can show that angle PNQ = 90°. Thus, MPNQ is a rectangle.

82.

From exercise 81, we know that MPNQ is a rectangle. Since ,MN AB⊥ angles AMN and

BMN each measure 90°. MP bisects angle AMN, so angle PMN = 45°. MQ bisects angle BMN, so angle QMN = 45°. Thus

QMN PMN≅△ △ by AAS, and therefore,

MP = MQ. From geometry, we know that a rectangle with two consecutive equal sides is a square, so MPNQ is a square.

83.

Since ,AB CD ABC BCD∠ = ∠ ⇒

70 .α β+ = ° Since ,CD EF

180 160 18020 .

CEF ECD ββ∠ + ∠ = °⇒ ° + = °⇒= °

20 70 50 .α α+ = °⇒ = ° Therefore, 50 20 30 .α β− = ° − ° = °

1.1 Critical Thinking

84. The maximum number of points of intersection is 6.

You may want to investigate the formula for

determining the maximum number of points of

intersection of n distinct lines, ( )1

.2

n n −

1.2 Triangles

1.2 Practice Problems

1.

( ) ( )

( )

33 11 3 110 18044 2 70 2 26 1333 13 20; 11 3 13 50

x xx x x

− + + + = ⇒+ = ⇒ = ⇒ =− = + =

The three angles are 20°, 50°, and 110°.

2.

( ) ( )2 22

2 2

3 10 5

6 9 100 10 256 109 10 25

84 4 21

x x

x x x xx x

x x

+ + = ++ + + = + +

+ = += ⇒ =

The sides of the triangle are 10, 3 + 21 = 24, and 5 + 21 = 26.

Section 1.2 Triangles 9


3.

3420 3420 2

2 342022

1710 2 2418 ft

x x

x

= ⇒ = = ⇒

= ≈

4.

AC is the leg opposite the 60° angle, so

6 6 36 3 2 3 3.5 ft

332 4 3 6.9 ft

a a

c a

= ⇒ = = = ≈

= = ≈

5.

2 36 122 3

2 18 962

24 6 46 12

y

x

x xxy

y y

= =

= ⇒ = ⇒ =

= ⇒ = ⇒ =

6.

a. Because BC is parallel to DE, AB and AC are transversals. Thus, ADE ABC∠ = ∠ and

.AED ACB∠ = ∠

Both triangles contain A, so triangles ADE and ABC are similar because they have equal corresponding angles.

b. To find x, we use the proportion

( )

9 89 12

108 8 9 108 72 8

36 936 8

8 2

AE AD

AC AB xx x

x x

= ⇒ = ⇒+

= + ⇒ = + ⇒

= ⇒ = =

To find y, we use the proportion 8

128 1212 16

210

3

AD DE yy

AB BC

y

= ⇒ = ⇒ = ⇒

=


1. The sum of the measures of the three angle of a triangle is 180°.

2. In an isosceles triangle, the two angles opposite equal sides are equal or congruent.

3. In a 30°-60°-90° triangle, the hypotenuse is two time the length of the shortest side and the side

opposite the 60° angle is 3 time the length of

the shortest side.

4. In similar triangles, the lengths of the corresponding sides are proportional.

5. True

6. False. The corresponding sides of two similar triangles are proportional while the corresponding sides of two congruent triangles are equal. Note that congruent triangles are also similar.

7. 180 50 72 180122 180 58A B C C

C C+ + = °⇒ ° + ° + = °⇒° + = °⇒ = °

8. 180 64 48 180112 180 68

A B C AA A+ + = °⇒ + ° + ° = °⇒+ ° = °⇒ = °

9. 180 48 15 98 180146 15 180 33 45A B C B

B B+ + = °⇒ ° + + ° = °⇒′° + = °⇒ = °′ ′

10. 18034 67 45 180101 45 180 78 15

A B CC

C C

+ + = °⇒° + ° + = °⇒′° + = °⇒ = °′ ′

11. 18046.72 65 180111.72 180 68.28

A B CBB B

+ + = °⇒° + + ° = °⇒° + = °⇒ = °



12. 18069 54.67 180123.67 180 56.33

A B CC

C C

+ + = °⇒° + ° + = °⇒

° + = °⇒ = °

13. ( ) ( )

18060 57 25 3 180142 2 180 2 38 19

57 19 3825 3 19 82

A B Cx x

x x xBC

+ + = ° ⇒° + − ° + + ° = °⇒° + = °⇒ = °⇒ = °= − = °= + ⋅ = °

14. ( ) ( )

18050 4 50 80 2 180

180 2 180 2 0 050 ; 80

A B Cx xx x x

A C

+ + = ° ⇒+ ° + ° + − ° = °⇒° + = °⇒ = °⇒ = °= ° = °

15. 1802 3 180 6 180 3030 ; 2 30 60 ; 3 30 90

A B Cx x x x xA B C

+ + = °⇒+ + = °⇒ = °⇒ = °= ° = ⋅ ° = ° = ⋅ ° = °

16. 1802 3 4 180 9 180 20

2 20 40 ; 3 20 604 20 800

A B Cx x x x x

A BC

+ + = °⇒+ + = °⇒ = °⇒ = °= ⋅ ° = ° = ⋅ ° = °= ⋅ ° = °

17. 2 2 2 2 2 2

2 25 12

25 144 169 13

a b c c

c c c

+ = ⇒ + = ⇒+ = ⇒ = ⇒ =

18. 2 2 2 2 2 2

2 27 24

49 576 625 25

a b c c

c c c

+ = ⇒ + = ⇒+ = ⇒ = ⇒ =

19. 2 2 2 2 2 2

2 25 13

25 169 144 12

a b c b

b b b

+ = ⇒ + = ⇒+ = ⇒ = ⇒ =

20. 2 2 2 2 2 2

2 220 29

400 841 441 21

a b c b

b b b

+ = ⇒ + = ⇒+ = ⇒ = ⇒ =

21. ( ) ( )

( )( )

2 22 2 2 2

2 2

2 2

2

8 9 13

64 16 81 18 169

2 2 145 169 2 2 24 0

12 0 4 3 04, 3

a b c x x

x x x x

x x x x

x x x xx

+ = ⇒ − + + = ⇒− + + + + = ⇒+ + = ⇒ + − = ⇒+ − = ⇒ + − = ⇒= −

If x = –4, then a = 8 – (–4) = 12 and b = 9 + (–4) = 5. If x = 3, then a = 8 – 3 = 5 and b = 9 + 3 =12.

22.

( ) ( )2 2 2

2 22

2 2

2 2

17 2 24 15 2

289 68 4 576 225 60 4

4 68 865 4 60 225640 128 5

a b c

x x

x x x x

x x x xx x

+ = ⇒− + = + ⇒− + + = + + ⇒− + = + + ⇒= ⇒ =

17 2 5 7; 15 2 5 25a c= − ⋅ = = + ⋅ =

23.

( ) ( ) ( )

( )( )

2 2 2

2 2 2

2 2

2

2 2

2 2

5 6 3 5 4

25 10 36 36 9

25 40 16

10 46 61 16 40 25

6 6 36 0 6 03 2 0 3, 2

a b c

x x x

x x x x

x x

x x x x

x x x xx x x

+ = ⇒+ + + = + ⇒+ + + + +

= + + ⇒+ + = + + ⇒

− + + = ⇒ − − = ⇒− + = ⇒ = −

If x = –2, then b = 6 + 3(–2) = 0, which is not possible, so reject x = –2. If x = 3, then a = 5 + 3 = 8, b = 6 + 3(3) = 15, and c = 5 + 4(3) = 17.

24.

( ) ( ) ( )

( )( )

2 2 2

2 2 2

2 2

2

2 2

2 2

2 1 10 6 6 7

4 4 1 100 120 36

36 84 49

40 116 101 49 84 36

9 32 65 0 9 32 65 013

5 9 13 0 5,9

a b c

x x x

x x x x

x x

x x x x

x x x x

x x x

+ = ⇒− + + = + ⇒− + + + +

= + + ⇒+ + = + + ⇒

− + + = ⇒ − − = ⇒

− + = ⇒ = −

If 13

,9

x = − then a and c are negative, so reject

this solution. If x = 5, then a = 2(5) – 1 = 9, b = 10 + 6(5) = 40, and c = 6 + 7(5) = 41.

In exercises 25–32, recall that in a 45°-45°-90° triangle, both legs are equal and the length of the

hypotenuse is 2 times the length of a leg.

25. Since each leg has length 4, the hypotenuse has

length 4 2.

26. Since each leg has length 5, the hypotenuse has

length 5 2.

27. Since each leg has length 12

, the hypotenuse

has length 22

.

28. Since each leg has length 35

, the hypotenuse

has length 3 25

.

29. Since the hypotenuse has length 3 2, each leg

has length 3 22

3.=

30. Since the hypotenuse has length 6 2, each leg

has length 6 22

6.=



31. Since the hypotenuse has length 4, each leg has

length 4 2422

2 2.= =

32. Since the hypotenuse has length 6, each leg has

length 6 2622

3 2.= =

In exercises 33–382, recall that in a 30°-60°-90° triangle, if the length of the shorter leg (the leg opposite the 30° angle) is x, then the length of the

longer leg (the angle opposite the 60° angle) is 3,x and the length of the hypotenuse is 2x.

33. The length of the shortest side is 4, so the

length of the longer leg is 4 3, and the length of the hypotenuse is 8.

34. The length of the shortest side is 6, so the

length of the longer leg is 6 3, and the length of the hypotenuse is 12.

35. The length of the side opposite the 60° angle (the longer leg) is 4, so the length of the shorter

leg is 4 3433

,= and the length of the

hypotenuse is 8 33

.

36. The length of the side opposite the 60° angle (the longer leg) is 6, so the length of the shorter

leg is 6 3633

2 3,= = and the length of the

hypotenuse is 4 3.

37. The length of the hypotenuse is 4, so the length of the shorter leg (the leg opposite the 30° angle) is 2, and the length of the longer leg (the

leg opposite the 60° angle) is 2 3.

38. The length of the hypotenuse is 6, so the length of the shorter leg (the leg opposite the 30° angle) is 3, and the length of the longer leg (the

leg opposite the 60° angle) is 3 3.

39.

, ,A Q B P C R

AB AC BC

QP QR PR

∠ = ∠ ∠ = ∠ ∠ = ∠

= =

40.

, ,P N Q L R M

PQ PR QR

NL NM LM

∠ = ∠ ∠ = ∠ ∠ = ∠

= =

41.

, ,R M P S Q T

RP RQ PQ

MS MT ST

∠ = ∠ ∠ = ∠ ∠ = ∠

= =

42.

, ,A D B C AOB DOC

AB AO BO

DC DO CO

∠ = ∠ ∠ = ∠ ∠ = ∠

= =

43.

, ,A A B D C E

AB AC BC

AD AE DE

∠ = ∠ ∠ = ∠ ∠ = ∠

= =



44.

, ,A A B D C E

AB AC BC

AD AE DE

∠ = ∠ ∠ = ∠ ∠ = ∠

= =

45.

18064 43 180 73

A B CA A

∠ + ∠ + ∠ = °⇒∠ + ° + ° = °⇒ ∠ = °

Similarly, 73 .D∠ = ° Thus, the triangles are similar since the corresponding angles are equal.

, ,A D B E C FAB AC BC

DE DF EF

∠ = ∠ ∠ = ∠ ∠ = ∠

= =

46.

180

35 110 180 35P Q R

Q Q∠ + ∠ + ∠ = °⇒° + ∠ + ° = °⇒ ∠ = °

Similarly, 35 .M∠ = ° Thus, the triangles are similar since the corresponding angles are equal.

, ,P L Q M R NPQ PR QR

LM LN MN

∠ = ∠ ∠ = ∠ ∠ = ∠

= =

47.

Since B B∠ = ∠ ′ and ,C C∠ = ∠ ′ it follows

that 180A B CA B C

A B C A B CA B C A B CA A

∠ + ∠ + ∠ = °= ∠ + ∠ +∠ ⇒′ ′ ′

∠ + ∠ + ∠ = ∠ + ∠ + ∠ ⇒′ ′ ′∠ + ∠ + ∠ = ∠ + ∠ + ∠ ⇒′∠ = ∠ ′

Thus, the triangles are similar since the corresponding angles are equal.

, ,A A B B C CAB AC BC

A B A C B C

∠ = ∠ ∠ = ∠ ∠ = ∠′ ′ ′

= =′ ′ ′ ′ ′ ′

48. If two angles of one triangle are equal to two angles of another triangle, then we can show that the third angles of the triangles are equal. (See exercise 47.) Thus, the triangles are similar.

49.

,AB DC A D B C⇒∠ = ∠ ∠ = ∠

AOB DOC∠ = ∠ since they are vertical angles. Thus, the triangles are similar, and

.AO BO AB

DO CO DC= =

45

15 123.7 4

11.112

xx

yy

= ⇒ =

= ⇒ =



50.

,AB DC A OCD B ODC⇒∠ = ∠ ∠ = ∠

.O O∠ = ∠ Thus, the triangles are similar, and

.AO BO AB

CO DO CD= =

26 2 10 4 10 2.5

5 63 2

2 18 96

xx x x x

x

y yy

= ⇒ = + ⇒ = ⇒ =+

= ⇒ = ⇒ =

51.

,BA DC A C OBA ODC⇒ ∠ = ∠ ∠ = ∠

.O O∠ = ∠ Thus, the triangles are similar, and

.AO BO AB

CO DO CD= =

10 1545

30x

x= ⇒ =

We must use the Pythagorean theorem to find BO.

2 2 2 2

210 15 100 225

325 5 13

5 13 15

455 13

75 13 15 225 13 15 150 13

10 13

BO BO

BO BO

BO AB

DO CD y

y y

y

+ = ⇒ + = ⇒= ⇒ =

= ⇒ = ⇒+

+ = ⇒ = ⇒=

52.

,AB DC A D B C⇒∠ = ∠ ∠ = ∠

COD BOA∠ = ∠ because they are vertical angles. Thus, the triangles are similar and

.AO AB BO

DO DC CO= =

We can find CO = z using the Pythagorean

theorem. 2 2 2 23 4 25 5 .z z z+ = ⇒ = ⇒ = 16

123 4

1620

5 4

AO AB xx

DO DCBO AB y

yCO DC

= ⇒ = ⇒ =

= ⇒ = ⇒ =


53.

Let BA and CD represent the two trees.

AC = 17, AE = 47 – 39 = 8 ft 2 2 2 2 2 2

217 8

289 64 225 15

AC AE CE CE

CE CE

= + ⇒ = += − = ⇒ =

So, the distance between the trees is 15 ft.

54.

CB = Distance the first car traveled after 2

hours = 30 × 2 = 60 miles





AC = Distance the second car traveled after 2 hours = 40 × 2 = 80 miles

2 2 2 2 2 2

260 80

3600 6400 10,000 100

AB AC BC AB

BC AB

= + ⇒ = + ⇒= + = ⇒ =

The distance between two cars is 100 miles.

55.

Denote the flagpole by BC and its shadow by

AC. 3,BC AC= 20 3BC = ≈ 34.6 ft

56.

Let AB represent the ladder leaning against the

wall BC. By the properties of 45-45-90 triangle, AC = BC = 9 ft.

2 2 2 2 29 9 182

9 2 12.7

AB AC BC

AB

= + = + = ⇒= ≈

So, the ladder is approximately 12.7 ft long.

57.

Let B denote the location of the balloon and let

AB represent the cable.

3 200 3100 3 173.2 m

2 2

ABAC = = = ≈

So, the balloon is 173.2 meters above the ground.

58.

Let BC represent the building and let AC

represent the shadow of the building. Then,

BC = 150 ft and 3AC AB= ⇒

150 3 259.8AC = ≈ The shadow is approximately 259.8 ft long.

59.

In triangle ABC, 1

32

BC AC= = and

3 3 3.AB BC= = Thus, the area of the

rectangle is 3 3 3 9 3 sq cm.⋅ =

60.

In triangle ABC, 2 6 2AC BC BC= ⇒ = ⇒

6 6 23 2.

22BC = = = In a 45°-45°-90°

triangle, the legs are equal, so

3 2.AB BC= = Thus, the area of the

rectangle is 3 2 3 2 9 2 18 sq cm.⋅ = ⋅ =

61.

Since the angle of elevation of the sun is the

same in each triangle, the triangles are similar. Let AC be the height of the building.


Not drawn to scale




Then we have 32

48.6 4

AC BC ACAC

DF EF= ⇒ = ⇒ =

The building is 48 feet tall.

62.

Not drawn to scale

Since the angle of elevation of the sun is the same in each triangle, the triangles are similar. Let AC be the height of the tower. Then we

have 60

30.5.7 11.4

AC BC ACAC

DF EF= ⇒ = ⇒ =

The tower is 30 feet tall.

63.

39

93 13

hh= ⇒ =

The ball must be hit from a height of 9 feet.

64.

First, we must find the distance the player is

standing from the net. 7.5 15

112.5 45 3 67.5 33 15

22.5

xx x

x

+= ⇒ = + ⇒ = ⇒

=The length of the base of the larger triangle is 15 + 22.5 = 37.5 ft. Now use the Pythagorean theorem to find d.

2 2 2 237.5 7.5 1462.5 38.2d d d+ = ⇒ = ⇒ ≈ The ball traveled approximately 38.2 feet.


65.

Apply the Pythagorean theorem to triangles

ABD and ACD.

2 2 2

2 2 2(1)

(2)

AB AD BD

AC AD CD

= += +

Now subtract equation (2) from equation (1). 2 2 2 2

2 2 2 2AB AC BD CD

AB AC BD CD

− = − ⇒= + −

Substitute BC – CD for BD, then expand.

( )22 2 2

2 2 2 2

2 2 22

2

AB AC BC CD CD

AC BC BC CD CD CD

AB AC BC BC CD

= + − −= + − ⋅ + −= + − ⋅

66. Let BC = a, AC = b, and AB = c. If C = 60°, then 30CAD∠ = ° and 2AC b CD= = ⇒

.2

bCD = From exercise 65, we have

2 2 2 2 2 22 .2

bc b a a c b a ab= + − ⋅ ⇒ = + −

67.

Apply the Pythagorean theorem to triangles

ABD and ACD. 2 2 2

2 2 2(1)

(2)

AB AD BD

AC AD CD

= += +

Now subtract equation (2) from equation (1). 2 2 2 2

2 2 2 2AB AC BD CD

AB AC BD CD

− = − ⇒= + −

Substitute BC + CD for BD, then expand.

( )22 2 2

2 2 2 2

2 2 22

2

AB AC BC CD CD

AC BC BC CD CD CD

AB AC BC BC CD

= + + −= + + ⋅ + −= + + ⋅

68. Let BC = a, AC = b, and AB = c. If 120 ,ACB∠ = ° then 60ACD∠ = ° and

30 .CAD∠ = ° So, 2AC b CD= = ⇒ .2

bCD =

From exercise 67, we have

2 2 2 2 2 22 .2

bc b a a c b a ab= + + ⋅ ⇒ = + +



69.

Given right triangle PMN, with legs a and b

and hypotenuse x. Thus, 2 2 2. (1)a b x+ =

Right triangle: If 2 2 2 ,c a b= + then 2 2 .c x c x= ⇒ = Therefore, PMN ABC≅△ △

by SSS. C N∠ ≅ ∠ because corresponding parts of congruent triangles are congruent, so angle C is a right angle.

Obtuse triangle: If 2 2 2 ,c a b> + then using

equation (1), 2 2 .c x c x> ⇒ > From geometry, we know that in two triangles if two sides from one triangle are equal to two corresponding sides from another triangle, then the included angle opposite the longer side is greater than the included angle opposite the shorter side. That is, .c x C N> ⇒ ∠ > ∠ Therefore, angle C is obtuse and triangle ABC is an obtuse triangle.

Acute triangle: If 2 2 2 ,c a b< + then using

equation (1), 2 2 .c x c x< ⇒ < From geometry, we know that in two triangles if two sides from one triangle are equal to two corresponding sides from another triangle, then the included angle opposite the longer side is greater than the included angle opposite the shorter side. That is, .x c N C> ⇒ ∠ > ∠ Therefore, angle C is actue and triangle ABC is an acute triangle.

70. We will use the results from exercise 69. The longest side of the triangle is 16 cm, so

c = 16, a = 13, and b = 10. 2 256,c = 2 169,a = and 2 100.b =

2 2 2256 169 100 ,c a b< + ⇒ < + so the triangle is acute.

71.

Because triangle ABC is equilateral, we know

that 60B C∠ = ∠ = ° and 30 .BAD CAD∠ = ∠ = ° In right triangle ABD,

we have

( ) ( )22 2 2

3, 22

32 3

2

2 3 4 3

ABAD BD AB BD BD

ABAD AD AB

AD AB AD AB

= = ⇒ = ⇒

= ⇒ = ⇒

= ⇒ =

72.

ABC DAB∼△ △ by AA. Therefore,

2

2 2 2

1 1.

a c a a

b p bc p b c p= ⇒ = ⇒ = Since

2 2 2 ,a b c= + this becomes 2 2

2 2 2

1b c

b c p

+ = ⇒

2 2

2 2 2 2 2 2 2 2

1 1 1 1.

b c

b c b c p c b p+ = ⇒ + =

73.

D is the midpoint of AB and E is the midpoint

of AC. Then, 12

AD AB= and 12

.AE AC=

Since ,A A∠ = ∠ ABC ADE∼△ △ by SAS. Because corresponding sides of similar

triangles are proportional, 12

.DE BC=

Because corresponding angles of similar triangles are equal, .D B∠ = ∠ These are corresponding angles formed by a transversal (AB) cutting BC and DE, so .BC DE



74.

D is the midpoint of AB, E is the midpoint of

AC, and F is the midpoint of BC. Then, 12

,AD BD AB= = 12

,AE EC AC= = and

12

.BF FC BC= = Since ,A A∠ = ∠

ABC ADE∼△ △ by SAS. Since ,B B∠ = ∠ ABC DBF∼△ △ by SAS.

Since ,C C∠ = ∠ ABC EFC∼△ △ by SAS.

Since each side in triangle DEF is 12

the

corresponding side in triangle ABC, ABC FDE∼△ △ by SSS.


75. We are given that .ABC XYZ∼△ △ Therefore,

.BC AC AB

YZ XZ XY= = and .C Z∠ = ∠

ADC XWZ∼△ △ by AA, so

.AD AC AD BC AC AB

XW XZ XW YZ XZ XY= ⇒ = = =

76. The area of triangle ABC is ( )( )12

,BC AD and

the area of triangle XYZ is ( )( )12

.YZ XW Using

the results from exercise 74, we have

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )

1212

2

2

BC AD BC AD AD AD

YZ XW XW XWYZ XW

AD

XW

= =

=

77. Using the result of exercises 75 and 76, we have

2 22

2

49 35 35 812025 45.

81 49x x

x

⋅= ⇒ = = ⇒ =

The length of the side is 45 cm.

78. Using the result of exercise 75, we have 2 2

22

12 3.2 3.2 4840.96 6.4.

48 12x x

x

⋅= ⇒ = = ⇒ =

The height of the larger triangle is 6.4 ft.

79.

By observation, we can conclude that the

number of triangles formed by the diagonals from one vertex of an n-gon is n – 2. Since the sum of the angles of a triangle is 180°, the sum of the interior angles of an n-gon is 180(n – 2). In a regular polygon, the interior angles are equal, so the measure of each interior angle is

( )1802n

n− degrees.

80.

From exercise 79, we know that the angles 1

and 2 measure ( )1805 2 108

5− = ° each. Then,

angles 3 and 4 measure 180 108 72° − ° = ° each. The sum of the measures of the angles in a triangle is 180°, so 72 72 180V° + ° + ∠ = °⇒

36 .V∠ = °

81. From exercise 79, we know that the measure of an interior angle of a regular n-gon is

( )1802 .n

n− ° Then the measure of each

exterior angle is ( )180180 2 .n

n° − − ° Following

the reasoning in exercise 80, the measure of the angle at each point of the star is

( )180 720180 2 180 2 180 .n

n n

°⎛ ⎞° − ° − − ° = ° −⎜ ⎟⎝ ⎠



82. From exercise 81, we know that the measure of

each angle of the star is 720

180 ,n

°° − so the

sum of the angles is 720

180 180 720.n nn

°⎛ ⎞° − = −⎜ ⎟⎝ ⎠ Now solve

180 720 540 180 1260 7.n n n− = ⇒ = ⇒ = The polygon has 7 sides.


83. Statement (iv) is always true. This is a restatement of the Triangle Inequality, which states that the sum of the lengths of two sides of a triangle is greater than the length of the third side.

1.3 Trigonometric Functions


1. x = –5, y = –12

( )22( 5) 12 13r = − + − =

12 13sin csc

13 125 13

cos sec13 512 12 5 5

tan cot5 5 12 12

y r

r yx r

r xy x

x y

θ θ

θ θ

θ θ

= = − = = −

= = − = = −

− −= = = = = =− −

2. 2 22, 5 2 ( 5) 29x y r= = − ⇒ = + − =

5 5 29 29sin csc

29 5292 2 29 29

cos sec29 229

5 5 2tan cot

2 2 5

y r

r y

x r

r xy x

x y

θ θ

θ θ

θ θ

−= = = − = = −

= = = = =

−= = = − = = −

3. a.

( )2 2

180 1, 0,

1 0 1

x y

r

θ = °⇒ = − =

= − + =

0sin 0

11

csc , undefined0

1cos 1

11

sec 11

y

rr

yx

rr

x

θ

θ

θ

θ

= = =

= =

−= = = −

= = = −−

0

tan 011

cot , undefined0

y

xx

y

θ

θ

= = =−−= =

b.

( )22

270 0, 1,

0 1 1

x y

r

θ = °⇒ = = −

= + − =

1sin 1

11

csc 11

0cos 0

11

sec , undefined0

1tan , undefined

00

cot 01

y

rr

yx

rr

xy

xx

y

θ

θ

θ

θ

θ

θ

−= = = −

= = = −−

= = =

= =

−= =

= = =−

4. 1170 90 3 360 ,° = ° + ⋅ ° so 1170º is coterminal with 90º. Thus, sin1170 sin 90 1csc1170 csc90 1cos1170 cos 90 0sec1170 sec90 , undefinedtan1170 tan 90 , undefinedcot1170 cot 90 0

° = ° =° = ° =° = ° =° = °° = °° = ° =

5. 630 630 2 360 90 ,− ° = − ° + ⋅ ° = ° so –630º is coterminal with 90º. Thus,

( )( )( )( )( )( )

sin 630 sin 90 1csc 630 csc90 1cos 630 cos 90 0sec 630 sec90 , undefinedtan 630 tan 90 , undefinedcot 630 cot 90 0

− ° = ° =− ° = ° =− ° = ° =− ° = °− ° = °− ° = ° =

6.

3 2 3sin 60

1 21 2 2 3

csc 6033 2 3

y

r

r

y

° = = =

° = = = =


Section 1.3 Trigonometric Functions 19



1 2 1

cos 604 1 2

1sec 60 2

1 2

3 2tan 60 3

1 2

1 2 1 3cot 60

33 2 3

x

r

x

y

x

x

y

° = = =

° = = =

° = = =

° = = = =

7. 405 45 360 ,° = ° + ° so 405º is coterminal with 45º. Thus,

2sin 405 sin 45

2csc 405 csc 45 2

2cos 405 cos 45

2sec 405 sec 45 2tan 405 tan 45 1cot 405 cot 45 1

° = ° =

° = ° =

° = ° =

° = ° =° = ° =° = ° =

8. Since sin 0θ > and cos 0θ < , then θ lies in quadrant II.

9. Because tan 0θ < and cos 0θ > , θ lies in

quadrant IV. 4

tan5

θ = − , so let x = 5 and

y = −4. 2 25 ( 4) 41= + − =r

4 4 41sin ;

4141θ = = − = −y

r

41sec

5θ = =r

x

10. 20 0

2 2sin 16 , 140 ft/sec, 45

140 sin 45 16 70 2 16

θ θ= − = = °= ° − = −

h v t t v

h t t t t

The graph of h is a parabola with the x-coordinate of the vertex

70 23.09

2 2( 16)= − = − ≈

−b

a.

2(3.09) 70 2(3.09) 16(3.09) 153.12 ft= − ≈h

( )2( ) 70 2 16 0

70 2 16 0 0 or 6.19

= − = ⇒− = ⇒ = ≈

h t t t

t t t t

The ball remains in flight for about 6.19 seconds, and 6.19(140) cos 45 613 ft= ° ≈d .

Thus, the ball reaches a maximum height of about 153 ft and has a range of about 613 ft.


1. For a point P(x, y) on the terminal side of an angle θ in standard position, we let r =

2 2 .x y+

2. sin , cos ,y x

r rθ θ= = tan .

y

xθ =

3. csc , sec ,r r

y xθ θ= = cot .

x

yθ =

4. If 1θ and 2θ are coterminal angles, then

1 2sin equals sin .θ θ

5. False. The value of a trigonometric function for any angle is the same for any point on the terminal side of .θ

6. False. In each quadrant, cosine and secant are either both positive or both negative, and sine and cosecant are either both positive or both negative.

For exercises 7–22, recall that

sin cos tan

csc sec cot

y x y

r r xr r x

y x y

θ θ θ

θ θ θ

= = =

= = =

Also, 2 2 .r x y= +

7. ( )2 23, 4 3 4 5x y r= − = ⇒ = − + =

4 3 4sin , cos , tan

5 5 35 5 3

csc , sec , cot4 3 4

θ θ θ

θ θ θ

= = − = −

= = − = −

8. ( )224, 3 4 3 5x y r= = − ⇒ = + − =

3 4 3sin , cos , tan

5 5 45 5 4


θ θ θ

θ θ θ

= − = = −

= − = = −

9. 2 25, 12 5 12 13x y r= = ⇒ = + =

12 5 12sin , cos , tan ,

13 13 513 13 5


θ θ θ

θ θ θ

= = =

= = =



10. 2 212, 5 ( 12) 5 13x y r= − = ⇒ = − + =

5 12 5sin , cos , tan ,

13 13 1213 13 12


θ θ θ

θ θ θ

= = − = −

= = − = −

11. 2 27, 24 7 24 25x y r= = ⇒ = + =

24 7 24sin , cos , tan ,

25 25 725 25 7


θ θ θ

θ θ θ

= = =

= = =

12. 2 224, 7 ( 24) 7 25x y r= − = ⇒ = − + =

7 24 7sin , cos , tan ,

25 25 2425 25 24


θ θ θ

θ θ θ

= = − = −

= = − = −

13. 2 224, 7 ( 24) ( 7) 25x y r= − = − ⇒ = − + − =

7 24 7sin , cos , tan ,

25 25 2425 25 24


θ θ θ

θ θ θ

= − = − =

= − = − =

14. 2 27, 24 ( 7) 24 25x y r= − = ⇒ = − + =

24 7 24sin , cos , tan ,

25 25 725 25 7


θ θ θ

θ θ θ

= = − = −

= = − = −

15. 2 21, 1 1 1 2x y r= = ⇒ = + =

1 2 1 2sin , cos , tan 1

2 22 2csc 2, sec 2, cot 1

θ θ θ

θ θ θ

= = = = =

= = =

16. 2 23, 3 ( 3) ( 3) 3 2x y r= − = − ⇒ = − + − =

3 2 3 2sin , cos ,

2 23 2 3 23 2 3 2

csc 2, sec 2,3 3

3 3tan 1, cot 1

3 3

θ θ

θ θ

θ θ

= − = − = − = −

= − = − = − = −

− −= = = =− −

17. ( ) ( )2 22, 2 2 2 2x y r= = ⇒ = + =

2 2sin , csc 2,

2 22 2

cos , sec 2,2 2

2 2tan 1, cot 1

2 2

θ θ

θ θ

θ θ

= = =

= − = =

= = = =

18. ( )223, 3 3 3 2 3x y r= − = ⇒ = + =

3 1sin , csc 2

22 33 3 2 3

cos , sec ,2 32 3

3 3tan , cot 3

3 3

θ θ

θ θ

θ θ

= = =

= − = = −

= − = − = −

19. ( )2 23, 1 3 ( 1) 2x y r= = − ⇒ = + − =

1sin , csc 2 ,

23 2 2 3

cos , sec ,2 33

1 3tan

333

cot 33

θ θ

θ θ

θ

θ

= − = −

= = =

= − = −

= − = −

20. ( ) ( )2 213, 3 13 3 4x y r= = ⇒ = + =

3 4 4 3sin , csc ,

4 3313 4 4 13

cos , sec ,4 13133 39 13 39

tan , cot13 313 3

θ θ

θ θ

θ θ

= = =

= = =

= = = =

21. 2 25, 2 5 ( 2) 29x y r= = − ⇒ = + − =

2 2 29 29sin , csc ,

29 2295 5 29 29

cos , sec ,29 529

2 5tan , cot

5 2

θ θ

θ θ

θ θ

= − = − = −

= = =

= − = −



22. 2 23, 5 ( 3) 5 34= − = ⇒ = − + =x y r

5 34 3 34 5sin , cos , tan ,

34 34 334 34 3


θ θ θ

θ θ θ

= = − = −

= = − = −

23. 450° is coterminal with 450° − 360° = 90°. Thus, sin 450° = sin 90° = 1.

24. 450° is coterminal with 450° − 360° = 90°. Thus, cos 450° = cos 90° = 0.

25. −90° is coterminal with 360° − 90° = 270°. Thus, cos (−90°) = cos 270° = 0.

26. −90° is coterminal with 360° − 90° = 270°. Thus sin (−90°) = sin 270° = –1.

27. 450° is coterminal with 450° − 360° = 90°. Thus, tan 450° = tan 90°, which is undefined.

28. 540° is coterminal with 540° − 360° = 180°. Thus, cot 540° = cot 180°, which is undefined.

29. –540° is coterminal with –540° + 2(360°) = 180°. Thus, tan (–540°) = tan 180° = 0.

30. 1080° is coterminal with 1080° − 3(360°) = 0°. Thus, sec 1080° = sec 0° = 1.

31. 900° is coterminal with 900° − 2(360°) = 180°. Thus, csc 900° = csc 180°, which is undefined.

32. 1080° is coterminal with 1080° − 3(360°) = 0°. Thus, csc 1080° = csc 0°, which is undefined.

33. −1530° is coterminal with 5(360°) − 1530° = 270°. Thus, sin (−1530°) = sin 270° = −1.

34. −2610° is coterminal with 8(360°) − 2610° = 270°. Thus, cos (−2610°) = cos 270° = 0.

35. sin 60 sin 30° + ° = 3 1 3 1

2 2 2

++ =

36. 1 3 3 1

cos 60 cos 302 2 2

+° + ° = + =

37. sin 60 cos 60° − ° = 3 1 3 1

2 2 2

−− =

38. sin 30 cos 30° − ° = 1 3 1 3

2 2 2

−− =

39. 2 2

sin 45 cos 452 2

1

2° ° = =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

40. 1 1

sin 30 cos 602 2

1

4° ° = =⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

41. 3 3 1

cos 30 tan 302 3 2

° ° = =⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

42. 3 3 1

sin 60 cot 602 3 2

⎛ ⎞ ⎛ ⎞° ° = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

43. ( )2

2 1 3sin 30 cos 30

2 2

1 2 3 34 4 4

31

2

⎛ ⎞° + ° = +⎜ ⎟⎝ ⎠

= + +

= +

44. ( )2

2 1 3sin 30 cos 30

2 2

1 2 3 34 4 4

31

2

⎛ ⎞° − ° = −⎜ ⎟⎝ ⎠

= − +

= −

45. Quadrant III 46. Quadrant III

47. Quadrant II 48. Quadrant III

49. Quadrant IV 50. Quadrant III

51. Quadrant II 52. Quadrant III

53.

( )2 2

2 2 2 2

2

5sin ;

13

13 5 13 25

144 12

yr x y

r

x x

x x

θ = = − = +

= + − ⇒ = + ⇒

= ⇒ = ±

Since θ lies in quadrant III, x = –12.

54. If θ lies in quadrant IV, x = 12.

55. 2 2

2 2 2 2

2

7cos ;

25

25 7 25 49

576 24

xr x y

r

y y

y y

θ = = = +

= + ⇒ = + ⇒= ⇒ = ±

Since θ lies in quadrant I, y = 24.

56. Since θ lies in quadrant IV, y = –24.



57.

2 2 2

5cos , in Quadrant III 0, 0

135, 13 13 ( 5) 12

θ θ= − ⇒

= − = ⇒ = − + ⇒ = −

x y

x r y y

< <

12 5 12sin , cos , tan ,

13 13 55 13 13

cot , sec , csc12 5 12

θ θ θ

θ θ θ

= − = − =

= = − = −

58.

2 2

3tan , in Quadrant IV 0, 0

4

3, 4 4 ( 3) 5

θ θ= − ⇒

= − = ⇒ = + − =

x y

y x r

> <

3 4 3sin , cos , tan ,

5 5 44 5 5


θ θ θ

θ θ θ

= − = = −

= − = = −

59.

2 2

3cot , in Quadrant II 0, 0

4

3, 4 ( 3) 4 5

θ θ= − ⇒

= − = ⇒ = − + =

x y

x y r

< >

4 3 4sin , cos , tan ,

5 5 33 5 5


θ θ θ

θ θ θ

= = − = −

= − = − =

60.

( )22 2

4sec , in Quadrant IV 0, 0

7

7, 4 4 7 3

θ θ= ⇒

= = ⇒ = + ⇒ = −

x y

x r y y

> <

3 7 3 7sin , cos , tan ,

4 4 77 4 7 4


θ θ θ

θ θ θ

= − = = −

= − = = −

61. 3

sin , tan 05

θ θ θ= ⇒< is in Quadrant II and

x < 0. 2 2 23, 5 5 3 4= = ⇒ = + ⇒ = −y r x x

3 4 3

sin , cos , tan ,5 5 4

4 5 5cot , sec , csc

3 4 3

θ θ θ

θ θ θ

= = − = −

= − = − =

62. 3

cot , sec 02

θ θ θ= ⇒> is in Quadrant I and

x > 0, y > 0. 2 23, 2 3 2 13= = ⇒ = + =x y r

2 13 3 13 2

sin , cos , tan ,13 13 3

3 13 13cot , sec , csc

2 3 2

θ θ θ

θ θ θ

= = =

= = =

63. sec 3, sin 0θ θ θ= ⇒< is in Quadrant IV and

y < 0. 2 2 23, 1 3 1 2 2= = ⇒ = + ⇒ = −r x y y

2 2 1sin , cos , tan 2 2,

3 32 3 2

cot , sec 3, csc4 4

θ θ θ

θ θ θ

= − = = −

= − = = −

64. tan 2, sin 0θ θ θ= − ⇒> is in Quadrant II and x < 0, y > 0.

( )2 22, 1 1 2 5= = − ⇒ = − + =y x r

2 5 5sin , cos , tan 2,

5 51 5

cot , sec 5, csc2 2

θ θ θ

θ θ θ

= = − = −

= − = − =


Use the figure below for exercises 65−72.

65. ( )2

2

144sin 30 7.5625 ft

6444sin 30

1.375 sec16

44 sin 30 cos 3052.39 ft

16

H

t

R

= ° =

°= =

° °= ≈

66. ( )2

2

144sin 45 15.125 ft

6444sin 45

1.94 sec16

44 sin 45 cos 4560.5 ft

16

H

t

R

= ° =

°= ≈

° °= =



67. ( )2

2

144sin 60 22.6875 ft

6444sin 60

2.38 sec16

44 sin 60 cos 6052.39 ft

16

H

t

R

= ° =

°= ≈

° °= ≈

68. ( )2

2

144sin 90 30.25 ft

6444sin 90

2.75 sec16

44 sin 90 cos 900 ft

16

H

t

R

= ° =

°= =

° °= =

69. a.

( ) ( )

22

2

2

2 22

16sec 45tan 45

80

16 2 11

20080

y x x

x x x x

°= ° −

= − = −

b. ( )21100 100 50 ft

200y = − =

70. ( )2180sin 45 50 ft

64H = ° =

71. 80sin 45

3.54 sec16

t°= ≈

72. 280 sin 45 cos 45

200 ft16

R° °= =

73. ( )

2

3 2 cos 600.99 ft

7 9cos 60A

°= ≈

+ °

74. a. 0θ = ° for a point due east of the sound source. 25 15cos 0 40 dBD = + ° =

b. 180θ = ° for a point due west of the sound source. 25 15cos180 10 dBD = + ° =

c. 270θ = ° for a point due south of the sound source. 25 15cos 270 25 dBD = + ° =

75. a. 100sin 30 50 fth = ° =

b. 100sin 60 86.6 fth = ° ≈


76. 0 90θ θ° < < °⇒ is in quadrant I. Therefore, x and y are positive.

2 2 2

2

5, 3 5 34

16 4 sin5

r x y

y y θ

= = ⇒ = + ⇒

= ⇒ = ⇒ =

77. 90 180θ θ° < < ° ⇒ is in quadrant II. Therefore, x is negative and y is positive.

( )22 2

2

5, 3 5 34

16 4 sin5

r x y

y y θ

= = − ⇒ = − + ⇒

= ⇒ = ⇒ =

78. 90 180θ θ° < < ° ⇒ is in quadrant II. Therefore, x is negative and y is positive

2 2 2

2

13, 5 13 512

144 12 cos13

r y x

x x θ

= = ⇒ = + ⇒

= ⇒ = − ⇒ = −

79. 270 360θ θ° < < °⇒ is in quadrant IV. Therefore, x is positive and y is negative.

22 2

2

1 11, 1

2 23 34 2

3 2 3sin

1 2

r x y

y y

θ

⎛ ⎞= = ⇒ = + ⇒⎜ ⎟⎝ ⎠

= ⇒ = − ⇒

−= = −

For exercises 80−83, refer to the following figure.

80. Since ( )sin sin ,θ θ− = − the y-value of the

point on the unit circle on the terminal side of θ− equals the opposite of y-value of the point

on the unit circle on the terminal side of .θ

Since ( )cos cos ,θ θ− = the x-value of the point

on the unit circle on the terminal side of θ− equals the x-value of the point on the unit circle on the terminal side of .θ Thus, the coordinates of the point on the unit circle are (x, −y), and the point on the unit circle is S. The triangle we are seeking is .SOM



81. Since sin sin(180 )θ θ= ° − , the y-value of the

point on the unit circle on the terminal side of θ equals the y-value of the point on the unit circle on the terminal side of 180 θ° − . Since cos cos(180 )θ θ= − ° − , the x-value of the

point on the unit circle on the terminal side of θ is the opposite of the x-value of the point on the unit circle on the terminal side of 180 θ° − . Thus, the coordinates of the point on the unit circle are (−x, y), and the point on the unit circle is Q. The triangle we are seeking is

.QON

82. Since sin sin(180 )θ θ= − ° + , the y-value of the

point on the unit circle on the terminal side of θ is the opposite the y-value of the point on the unit circle on the terminal side of 180 θ° + . Since cos cos(180 )θ θ= − ° + , the x-value of the

point on the unit circle on the terminal side of θ is the opposite of the x-value of the point on the unit circle on the terminal side of 180 θ° + . Thus, the coordinates of the point on the unit circle are (−x, −y), and the point on the unit circle is R. The triangle we are seeking is

.RON

83. Since sin sin(360 )θ θ= − ° − , the y-value of the

point on the unit circle on the terminal side of θ is the opposite of the y-value of the point on the unit circle on the terminal side of 360 θ° − . Since cos cos(360 )θ θ= ° − , the x-value of the

point on the unit circle on the terminal side of θ is equals the x-value of the point on the unit circle on the terminal side of 360 θ° − . Thus, the coordinates of the point on the unit circle are (x, −y), and the point on the unit circle is S. The triangle we are seeking is .SOM

84. ( )( )

2sin 45 sin 45

2tan 60 tan 60 3

− ° = − ° = −

− ° = − ° = −

85. 2

sin135 sin(180 45 ) sin 452

2cos135 cos(180 45 ) cos 45

2tan120 tan(180 60 ) tan 60 3

° = ° − ° = ° =

° = ° − ° = − ° = −

° = ° − ° = − ° = −

86. 2

sin 225 sin(180 45 ) sin 4521

cos 240 cos(180 60 ) cos 602

3tan 210 tan(180 30 ) tan 30

3

° = ° + ° = − ° = −

° = ° + ° = − ° = −

° = ° + ° = ° =

87.

( )

2sin 315 sin(360 45 ) sin 45

21

cos 300 cos(360 60 ) cos 602

3tan 330 tan 360 30 tan 30

3

° = ° − ° = − ° = −

° = ° − ° = ° =

° = ° − ° = − ° = −

88.

For acute angle θ ,

22

2 22 2sin sinθ θ= = ⇒ =

++

y y y

r x yx y.

For any real numbers x and y, 2 2 2x y y+ ≥ , so 2

22 2

1 sin 1y

x yθ≤ ⇒ ≤ ⇒

+

sin 1 1 sin 1.θ θ≤ ⇒ − ≤ ≤

Similarly, we can show that

2 2cosθ = = ⇒

+

x x

r x y

22

2 2cos θ = ⇒

+x

x y

2cos 1 cos 1 1 cos 1.θ θ θ≤ ⇒ ≤ ⇒ − ≤ ≤

89.

( )180 90 90m QON θ θ∠ = ° − ° + = ° −

Thus, .m Q θ∠ = N M∠ ≅ ∠ and

,OQ OP r= = so POM NOQ≅△ △ by AAS.


Section 1.4 Reference Angles 25



a. Q corresponds to O. The coordinates of M are (x, 0), so the y-coordinate of Q is x. Since the coordinates of Q must satisfy the

equation 2 2 2,x y r+ = the x-coordinate of

Q must by −y. Thus, the coordinates of Q are (−y, x).

b. sin( 90 ) cos

cos( 90 ) sin

tan( 90 ) cot

x

ry y

r rx x

y y

θ θ

θ θ

θ θ

+ ° = =

−+ ° = = − = −

+ ° = = − = −−

90. a. 2 2

2 2 2 2 2 2

tan cot (1)y x

x yx y

r x y x r y

θ θ= ⇒ = ⇒ =

= + ⇒ = −

Substituting from (1), we have 2

2 2 2 2 22

12

21 2 2

cos 452 2 2

xx r x x r

rx

rθ θ

= − ⇒ = ⇒ = ⇒

= = ⇒ = ⇒ = °

b. For 180 270 , tan cotθ θ θ° < < ° = ⇒

225 .θ = °

91. ( )

( )

1sin sin 30

21

cos cos 602

A B

A B

− = = °

+ = = °

From these two equations, we have the system

{ 302 90 45 , 15

60A B

A A BA B− = ° ⇒ = °⇒ = ° = °+ = °

92. In triangle ABC, 180A B C+ + = ° ⇒ 180 .A B C+ = ° − Substituting we have

( ) ( )tan tan tan 180 tantan tan 0

A B C C CC C

+ + = ° − += − + =

93. Since the largest value for cos 1,θ = the smallest value for sec 1.θ =

94. Since the smallest negative value for sin 1,θ = − the largest negative value for csc 1.θ = −

95. cos 105° = cos 60° + cos 45° is false because cos 105° < 0, while cos 60° > 0 and cos 45° > 0. The sum of two positive numbers is positive.

96. sin 260° = 2 sin 130° because sin 260° < 0 while sin 130° > 0.

97. tan 123° = tan 61° + tan 62° is false because tan 123° < 0, while tan 61° > 0 and tan 62° > 0. The sum of two positive numbers is positive.

98. sec 380° = sec 185° + sec 195° is false because sec 380° > 0 while sec 185° < 0 and sec 195° < 0. The sum of two negative numbers is negative.


99. False. For example, if 90 ,θ = ° then we have

( )cos sin 90 cos1 0.9998° ° = ° ≈ while

( )sin cos 90 sin 0 0° ° = ° =

100. False. secθ and tanθ are both undefined for 90 .θ = °

1.4 Reference Angles


1. a. 175 180 175 5θ θ= °⇒ = ° − ° = °′

b. 210 30 210 30 180 30 30θ θ= ° ⇒ = ° − ° = °′ ′ ′ ′

2. 2025 225 5 360 ,° = ° + ⋅ ° so the 2025° angle is coterminal with 225°. Since 225° lies in quadrant III, the reference angle is determined by 225° – 180° = 45°.

3. –70° + 360° = 290° Since 290° lies in quadrant IV, the reference angle is determined by 360° – 290° = 70°.

4. 1025 305 2 360 ,° = ° + ⋅ ° so 1025º is coterminal with 305º. Because 305º lies in quadrant IV,

360 305 55θ = ° − ° = °′ . In quadrant IV, cosθ is positive, so cos1025 cos 55 0.5736.° = ° ≈

5. Because 120° lies in quadrant II, its reference angle is 180° – 120° = 60°. In quadrant II, cotθ is negative, so

3cot120 cot 60 .

3° = − ° = −

6. Since (–510°) = 2(360°) – 510° = 210°, the angles –510° and 210° are coterminal. Because 210° lies in quadrant III, the reference angle is 210° – 180° = 30°. In quadrant III, sinθ is

negative, so ( ) 1sin 510 sin 30 .

2− ° = − ° = −



7. 570 360 210 ,° = ° + ° so 570° is coterminal with 210°. Because 210° lies in quadrant III, the reference angle is 210° – 180° = 30°. In quadrant III, sin , cos , sec ,θ θ θ and cscθ are

negative, and tanθ and cotθ are positive. 1

sin 570 sin 210 sin 302

3cos 570 cos 210 cos 30

23

tan 570 tan 210 tan 303

csc570 csc 210 csc30 2

2 3sec570 sec 210 sec30

3cot 570 cot 210 cot 30 3

° = ° = − ° = −

° = ° = − ° = −

° = ° = ° =

° = ° = − ° = −

° = ° = − ° = −

° = ° = ° =

8. 600 240 360 .θ = ° = ° + ° 240° lies in quadrant III, thus, 60 ,θ = °′ and cos 0.θ <

1 1cos 60 cos 600 cos 240 .

2 2° = ⇒ ° = ° = −

167.8 67.2 101.4 m

2h

⎛ ⎞= − − =⎜ ⎟⎝ ⎠

9. tan 3θ θ= ⇒ lies in quadrant I or in

quadrant III. In quadrant I, tan 3θ = ⇒ 60 .θ = ° In quadrant III,

2 180 60 240 .θ = ° + ° = ° Thus, all values of θ

are given by 60 360n° + ⋅ ° or 240 360 .n° + ⋅ °


1. The reference angle θ ′ for a nonquadrantal angle θ in standard position is the acute angle formed by the terminal side of θ and the x-axis.

2. If θ is in quadrant II, then 180 .θ θ= ° −′

3. If θ is in quadrant III, then 180 .θ θ= − °′

4. If θ is in quadrant IV, then 360 .θ θ= ° −′

5. False. Depending on which quadrant θ lies in, some of the trigonometric function values of the angle will be the same as those of its reference angle.

6. True

7. Because 46° lies in quadrant I, the reference angle is 46°.

8. Because 87° lies in quadrant I, the reference angle is 87°.

9. Because 96° lies in quadrant II, the reference angle is 180° – 96° = 84°.

10. Because 126° lies in quadrant II, the reference angle is 180° – 126° = 54°.

11. Because 192° lies in quadrant III, the reference angle is 192° – 180° = 12°.

12. Because 220° lies in quadrant III, the reference angle is 220° – 180° = 40°.

13. Because 290° lies in quadrant IV, the reference angle is 360° – 290° = 70°.

14. Because 305° lies in quadrant IV, the reference angle is 360° – 305° = 55°.

15. –145° is coterminal with 360° – 145° = 215°. Because 215° lies in quadrant III, the reference angle is 215° – 180° = 35°.

16. –190° is coterminal with 360° – 190° = 170°. Because 170° lies in quadrant II, the reference angle is 180° – 170° = 10°.

17. –260° is coterminal with 360° – 260° = 100°. Because 100° lies in quadrant II, the reference angle is 180° – 100° = 80°.

18. –320° is coterminal with 360° – 320° = 40°. Because 40° lies in quadrant I, the reference angle is 40°.

19. 1570° is coterminal with 1570° – 4(360°) = 130°. Because 130° lies in quadrant II, the reference angle is 180° – 130° = 50°.

20. 1900° is coterminal with 1900° – 5(360°) = 100°. Because 100° lies in quadrant II, the reference angle is 180° – 100° = 80°.

21. –1360° is coterminal with –1360° + 4(360°) = 80°. Because 80° lies in quadrant I, the reference angle is 80°.

22. –2040° is coterminal with –2040° + 6(360°) = 120°. Because 120° lies in quadrant II, the reference angle is 180° – 120° = 60°.

23. reference angle = 60°

3 2 3sin120 , tan120 3, csc

2 3θ° = ° = − =


2cos135 , cot135 1, sec135 2

2° = − ° = − ° = −




1 3sin150 , cos150 , cot150 3

2 2° = ° = − ° = −

26. tan180 0, cot180° = ° is undefined,

sec 180° = –1


3 3cos 210 , tan 210 ,

2 32 3

sec 2103

° = − ° =

° = −


2sin 225 , tan 225 1, sec 225 2

2° = − ° = ° = −


3 3sin 240 , cot 240 ,

2 32 3

csc 2403

° = − ° =

° = −

30. tan 270° is undefined, sec 270° is undefined, csc270° = –1


3tan 300 3, cot 300 ,

32 3

csc3003

° = − ° = −

° = −


2sin 315 , tan 315 1, csc315 2

2° = − ° = − ° = −


3 3cos 330 , tan 330 ,

2 32 3

sec3303

° = ° = −

° =

34. cot 360° is undefined, sec 360° = 1, csc 360° is undefined

35. –300° is coterminal with –300° + 360° = 60°, which is the reference angle.

( ) 3sin 300 sin 60

2− ° = ° =


( ) 3cos 330 cos30

2− ° = ° =


( )tan 315 tan 45 1− ° = ° =


( )csc 330 csc30 2− ° = ° =

39. –240° is coterminal with –240° + 360° = 120°, which is in quadrant II. The reference angle is 180° – 120° = 60°.

( )sec 240 sec120 sec 60 2− ° = ° = − ° = −


( ) 3cot 240 cot120 cot 60

3− ° = ° = − ° = −


( ) 2sin 225 sin135 sin 45

2− ° = ° = ° =


( ) 3cos 210 cos150 cos 30

2− ° = ° = − ° = −

43. –150° is coterminal with –150° + 360° = 210°, which is in quadrant III. The reference angle is 210° – 180° = 30°.

( ) 3tan 150 tan 210 tan 30

3− ° = ° = ° =

44. –135° is coterminal with –135° + 360° = 225°, which is in quadrant III. The reference angle is 225° – 180° = 45°.

( )sec 135 sec 225 sec 45 2− ° = ° = − ° = −

45. –60° is coterminal with –60° + 360° = 300°, which is in quadrant IV. The reference angle is 360° – 300° = 60°.

( ) 3sin 60 sin 300 sin 60

2− ° = ° = − ° = −

46. –45° is coterminal with –45° + 360° = 315°, which is in quadrant IV. The reference angle is 360° – 315° = 45°.

( ) 2cos 45 cos 315 cos 45

2− ° = ° = ° =



47. 1470° = 30° + 4(360°), so 1470° is coterminal with 30°, which is the reference angle.

1sin1470 sin 30

2° = ° =

48. 1860° = 60° + 5(360°), so 1860° is coterminal with 60°, which is the reference angle.

1cos1860 cos 60

2° = ° =

49. 1125° = 45° + 3(360°), so 1125° is coterminal with 45°, which is the reference angle. tan1125 tan 45 1° = ° =

50. 2190° = 30° + 6(360°), so 2190° is costerminal with 30°, which is the reference angle.

2 3sec 2190 sec30

3° = ° =

51. –2130° is coterminal with –2130° + 6(360°) = 30°, which is the reference angle.

( )csc 2130 csc30 2− ° = ° =


( ) 2 3sec 1410 sec30

3− ° = ° =


( ) 3tan 690 tan 30

3− ° = ° =


( ) 3cot 2100 cot 60

3− ° = ° =

55. Because cosθ is positive, θ lies in quadrant I or quadrant IV. In quadrant I, 1 60 .θ = °

In quadrant IV, 2 360 60 300 .θ = ° − ° = °

56. Because sinθ is negative, θ lies in quadrant

III or quadrant IV. 3

sin 602

θ θ= ⇒ = °′ ′

In quadrant III, 1 180 60 240 .θ = ° + ° = °

In quadrant IV, 2 360 60 300 .θ = ° − ° = °

57. Because tanθ is negative, θ lies in quadrant

II or quadrant IV. tan 3 60θ θ= ⇒ = °′ ′

In quadrant II, 1 180 60 120 .θ = ° − ° = °

In quadrant IV, 2 360 60 300 .θ = ° − ° = °

58. Because secθ is negative, θ lies in quadrant

II or quadrant III. sec 2 60θ θ= ⇒ = °′ ′

In quadrant II, 1 180 60 120 .θ = ° − ° = °

In quadrant III, 2 180 60 240 .θ = ° + ° = °

59. 3 2 3

csc sin 1,2 3

θ θ= ⇒ = > which is not

possible.

60. 3 2 3

sec cos 1,2 3

θ θ= − ⇒ = − < − which is

not possible.

61. Because cotθ is positive, θ lies in quadrant I or quadrant III. In quadrant I, 1 30 .θ = °

In quadrant III, 2 180 30 210 .θ = ° + ° = °

62. Because cosθ is negative, θ lies in quadrant

II or quadrant III. 1

cos 602

θ θ= ⇒ = °′ ′

In quadrant II, 1 180 60 120 .θ = ° − ° = °

In quadrant III, 2 180 60 240 .θ = ° + ° = °

63. sin 1 90θ θ= ⇒ = °

64. cos 1 0θ θ= ⇒ = °

65. cos 1 180θ θ= − ⇒ = °

66. sin 1 270θ θ= − ⇒ = °

67. cos 0 90 , 270θ θ= ⇒ = ° °

68. sin 0 0 , 180θ θ= ⇒ = ° °

69. There is no value of θ for which sin 2.θ =

70. There is no value of θ for which sin 2.θ = −

71. Because sinθ is negative, θ lies in quadrant

III or quadrant IV. 2

sin 452

θ θ= ⇒ = °′ ′

In quadrant III, 1 180 45 225 .θ = ° + ° = °

In quadrant IV, 360 45 315 .θ = ° − ° = ° All values of θ are given by 225 360nθ = ° + ⋅ ° or

315 360 ,nθ = ° + ⋅ ° n any integer.

72. Because cosθ is negative, θ lies in quadrant

II or quadrant III. 3

cos 302

θ θ= ⇒ = °′ ′

In quadrant II, 1 180 30 150 .θ = ° − ° = °

In quadrant III, 2 180 30 210 .θ = ° + ° = ° All

values of θ are given by 150 360nθ = ° + ⋅ ° or 210 360 ,nθ = ° + ⋅ ° n any integer.



73. Because tanθ is positive, θ lies in quadrant I

or quadrant III. 3

tan 303

θ θ= ⇒ = °′ ′

In quadrant I, 1 30 .θ = °

In quadrant III, 2 180 30 210 .θ = ° + ° = ° All


74. Because secθ is positive, θ lies in quadrant I

or quadrant IV. sec 2 60θ θ= ⇒ = °′ ′

In quadrant I, 1 60 .θ = °

In quadrant IV, 2 360 60 300 .θ = ° − ° = ° All


75. 2

sin 452

θ θ= ⇒ = °′ ′

Since θ lies in quadrant III, 180 45 225 .θ = ° + ° = °

76. 2

sin 452

θ θ= ⇒ = °′ ′

Since θ lies in quadrant IV, 360 45 315 .θ = ° − ° = °

77. 1

cos 602

θ θ= ⇒ = °′ ′

Since θ lies in quadrant II, 180 60 120 .θ = ° − ° = °

78. 1

cos 602

θ θ= ⇒ = °′ ′


79. tan 3 60θ θ= ⇒ = °′ ′


80. tan 3 60θ θ= ⇒ = °′ ′


81. 2

sin 452

θ θ= ⇒ = °′ ′


82. 3

cos 302

θ θ= ⇒ = °′ ′


83. tan 3 60θ θ= ⇒ = °′ ′


84. sec 2 60θ θ= ⇒ = °′ ′


1.4B Exercises: Applying the Concepts

85.

From geometry, we know that an altitude

drawn from the vertex of an isosceles triangle bisects the vertex angle and also bisects the base of the triangle. Therefore,

3sin 60 8sin 60 8 4 3.

8 2

dd

⎛ ⎞° = ⇒ = ° = =⎜ ⎟⎝ ⎠

The length of the chain is

2 4 3 8 3 13.86 ft.⋅ = ≈

86.

cos 60 12cos 60 612

dd° = ⇒ = ° =

The buoy is 6 feet from the cliff.

1.4C Exercises: Beyond the Basics

87. tan 1 45 ; tanθ θ θ= ⇒ = °′ ′ positive θ⇒ lies

in quadrant I or in quadrant III. Thus, 1 45θ = °

or 2 225 .θ = ° To find the negative angles,

subtract 360°. 1 45 360 315 ,θ = ° − ° = − °

2 225 360 135 .θ = ° − ° = − °



88. For 180 90 , 180 .θ θ θ− ° < < − ° = ° +′

89. For ( )270 180 , 180 .θ θ θ− ° < < − ° = − ° +′

90. For 360 270 , 360 .θ θ θ− ° < < − ° = ° +′

Use the figure below for exercises 91–93.

91. a. ( )sin sinb

rθ θ− = − = −

b. ( )cos cosa

rθ θ− = =

c. ( )tan tanb

aθ θ− = − = −

92. a. ( )sin 180 sinb

rθ θ° − = =

b. ( )cos 180 cosa

rθ θ° − = − = −

c. ( )tan 180 tanb

aθ θ° − = − = −

93. a. ( )sin 180 sinb

rθ θ° + = − = −

b. ( )cos 180 cosa

rθ θ° + = − = −

c. ( )tan 180 tanb b

a aθ θ−° + = = =

−

94. sin 300 cos150 cos 300 sin150

3 3 1 1 3 11

2 2 2 2 4 4

° ° + ° °⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − + = + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

95.

( ) ( )

cos 270 cos120 sin 270 sin120

1 3 30 1

2 2 2

° ° + ° °⎛ ⎞⎛ ⎞= − + − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

96. cos 690 cos 780 sin 690 sin 780cos 330 cos 60 sin 330 sin 60

3 1 1 3 3 3 32 2 2 2 4 4 2

° ° − ° °= ° ° − ° °⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − − = + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

97.

( )( )( )( )( )

33

33

tan 780 tan 330 tan 60 tan 330

1 tan 780 tan 330 1 tan 60 tan 330

3

1 3

3 3 3

3 3 3

2 3 3

6 3

° + ° ° + °=− ° ° − ° °

+ −=

− −

−=− −

= =

98.

( )( )( )( )( )

3 33 3

3 33 3

tan1470 tan1230 tan 30 tan150

1 tan1470 tan1230 1 tan 30 tan150

1

3 3 3 3

9 3 3

6 33

6

° − ° ° − °=+ ° ° + ° °

− −=

+ −

+=+ −

= =

Section 1.5 Fundamental Trigonometric Identities 31


99. ( )( )

tan 1 45 or 2251

sin 30 or 1502

α β α β α β

α β α β α β

+ = ⇒ + = ° + = °

− = ⇒ − = ° − = °

Now solve the systems.

{{{{

452 75

3037.5 , 7.5

452 195

15097.5 , 52.5 307.5225

2 25530

127.5 , 97.5225

2 375150

187.5 , 37.5

α β αα βα β

α β αα βα β

α β αα βα β

α β αα βα β

+ = °⇒ = °⇒− = °= ° = °

+ = ° ⇒ = °⇒− = °= ° = − ° = °

+ = °⇒ = °⇒− = °= ° = °

+ = °⇒ = °⇒− = °= ° = °

100. ( )( )

sec 2 60 or 300csc 2 30 or 150

α β α β α βα β α β α β+ = ⇒ + = ° + = °− = ⇒ + = ° + = °

Now solve the systems.

{{{{

602 90

3045 , 15300

2 33030

165 , 13560

2 210150

105 , 45 315300

2 450150

225 , 75

α β αα βα β

α β αα βα β

α β αα βα β

α β αα βα β

+ = °⇒ = °⇒− = °= ° = °

+ = °⇒ = °⇒− = °= ° = °

+ = ° ⇒ = °⇒− = °= ° = − ° = °

+ = ° ⇒ = °⇒− = °= ° = °


101. csc sec 1

tan 1 135 , 315

r r y

y x xθ θ

θ θ

= − ⇒ = − ⇒ = − ⇒

= − ⇒ = ° °

102. sec tanr y

r yx x

θ θ= ⇒ = ⇒ = ⇒ there is no

solution.

1.5 Fundamental Trigonometric Identities


1. a. 1 1

sincsc 5

θθ

= =

b. 1 1

sec 2cos 1 2

θθ

= = = −−

c. 1 1 11

cottan 5 11 5

θθ

= = =

2. cos 12 17 12

cotsin 5 17 5

θθθ

= = = −−

3. 1 1 1 1 5

tan ; sincot 2 csc 55

sin 1 5 5 2 5tan cos

cos 2 cos 5

1 5 5sec

cos 22 5

θ θθ θθθ θθ θ

θθ

= = = = = −−

−= ⇒ = ⇒ = −

= = − = −

4. 2

cos and sin 0 is in quadrant II.3

θ θ θ= − > ⇒

In quadrant II, sinθ is positive, while tanθ is negative.

22 2 2

2

2sin cos 1 sin 1

35 5

sin sin sin9 3sin 5 3 5

tancos 2 3 2

θ θ θ

θ θ θ

θθθ

⎛ ⎞+ = ⇒ + − = ⇒⎜ ⎟⎝ ⎠

= ⇒ = =

= = = −−

5.

2 2

2

tan 1 and 90 1801 1

cot 1tan 1

1 tan sec

1 1 2 sec sec 2

θ θ

θθ

θ θθ θ

= − ° < < °

= = = −−

+ = ⇒+ = = ⇒ = ±

Since θ is in quadrant II, sec 2.θ = −

1 1 2cos

sec 22θ

θ= = − = −

sin sin 2tan 1 sin

cos 22 21 1 2

csc 2sin 2 2 2

θ θθ θθ

θθ

= ⇒ − = ⇒ =−

= = = =



6. ( ) ( )

( ) ( )

2

2 2 2

sin sinsin 1 cos sin 1 costan tan sin sincos cos

1 1sec 1 sec 1 1 cos 1 cos 1 cos1 1cos cossin 1 cos 1 cos 2sin 2sin 2

sin1 cos 1 cos sin

θ θθ θ θ θθ θ θ θθ θ

θ θ θ θ θθ θθ θ θ θ θ

θθ θ θ

− + ++ = + = + =

+ − + − −+ −

⎡ − + + ⎤⎣ ⎦= = = =− −

7. If cos 1 sinθ θ= − is an identity, then it must be true for all values of .θ When 30 ,θ = ° we

have 3

cos 30 ,2

° = while

1 11 sin 30 1 .

2 2− ° = − = Therefore, the equation

is not an identity.


1. Trigonometric functions sec , csc ,θ θ and

cotθ are, respectively, the reciprocals of cos , sin ,θ θ and tan .θ

2. The quotient identities are sin

tancos

θ θθ= and

coscot .

sin

θ θθ=

3. The identity 2 2sin cos 1θ θ+ = can also be

written in equivalent forms 2 21 cos sin ,θ θ− =

2 21 sin cos ,θ θ− = 2sin 1 cos ,θ θ= ± − and

2cos 1 sin .θ θ= ± −

4. Equivalent forms of the identity 2 21 tan secθ θ+ = are 2 2sec tan 1 ,θ θ− =

2 2 2sec 1 tan , sec 1 tan ,θ θ θ θ− = = ± + and

2tan sec 1.θ θ= ± −

5. Equivalent forms of the identity 2 21 cot cscθ θ+ = are 2 2csc cot 1 ,θ θ− =

2 2 2csc 1 cot , csc 1 cot ,θ θ θ θ− = = ± + and

2cot csc 1.θ θ= ± −

6. True

7. 1 1 3

cscsin 2 3 2

θθ

= = =

8. 1 1 4

seccos 3 4 3

αα

= = = −−

9. 1 1

cossec 5

ββ

= =

10. 1 1

sincsc 5

ββ

= =

11. 1 1 7

cottan 2 7 2

θθ

= = = −−

12. 1 1 5

tancot 3 5 3

θθ

= = =

13. sin 5 13 5

tancos 12 13 12

θθθ

= = =

14. cos 6 61 6

cotsin 55 61

θθθ

−= = = −

15. cos 3 cos

cotsin 2 2 13

3 3 13cos

1313

θ θθθ

θ

= ⇒ = ⇒

= =

16. sin sin

tan 3cos 1 10

3 3 10sin

1010

α ααα

α

= ⇒ − = ⇒

= − = −

17. sin 2 2 13

tancos 3 cos

3 3 13cos

1313

θθθ θ

θ

−= ⇒ = ⇒

= − = −

18. cos 5 5 29

cotsin 2 sin

2 2 29sin

2929

θθθ θ

θ

= ⇒ − = ⇒

= − = −



19. 1 1 4

cossec 17 4 17

sin 1 sintan

cos 4 4 17

4 17 1 17sin

4 1717

ααα ααα

α

= = =

= ⇒ = ⇒

= = =

20. 1 1 3

sincsc 34 3 34

cos 5 coscot

sin 3 3 34

5 5 34cos

3434

ααα ααα

α

= = =

= ⇒ = ⇒

= =

21. Since θ is in quadrant III, cosθ is negative. 2

2 2 2

2 2

12sin cos 1 cos 1

13144 25

cos 1 cos169 169

25 5cos

169 13

θ θ θ

θ θ

θ

⎛ ⎞+ = ⇒ − + = ⇒⎜ ⎟⎝ ⎠

+ = ⇒ = ⇒

= − = −

22. Since θ is in quadrant II, sinθ is positive. 2

2 2 2

2 2

2sin cos 1 sin 1

134 9

sin 1 sin13 13

9 3 3 13sin

13 1313

θ θ θ

θ θ

θ

⎛ ⎞+ = ⇒ + − = ⇒⎜ ⎟⎝ ⎠

+ = ⇒ = ⇒

= = =

23. Since θ is in quadrant III, secθ is negative. 2 2 2 21 tan sec 1 3 10 sec

sec 10

θ θ θθ

+ = ⇒ + = = ⇒= −

24. Since θ is in quadrant IV, tanθ is negative. 2

2 2 25 9sec 1 tan 1 tan

4 169 3

tan16 4

θ θ θ

θ

⎛ ⎞− = ⇒ − = = ⇒⎜ ⎟⎝ ⎠

= − = −

25. Since θ is in quadrant III, cscθ is negative. 2

2 2 21 51 cot csc 1 csc

2 4

5 5csc

4 2

θ θ θ

θ

⎛ ⎞+ = ⇒ + = = ⇒⎜ ⎟⎝ ⎠

= − = −

26. Since θ is in quadrant II, cotθ is negative. 2 2 2 2csc 1 cot 3 1 8 cot

cot 8 2 2

θ θ θθ

− = ⇒ − = = ⇒= − = −

27. Since θ is in quadrant III, sin , cos , sec ,θ θ θ

and cscθ are negative, and tanθ and cotθ are positive.

22 2 2

2 2

1 5csc

sin 3

3sin cos 1 cos 1

59 16

cos 1 cos25 25

16 4 1 5cos ; sec

25 5 cos 4sin 3 5 3 1 4

tan ; cotcos 4 5 4 tan 3

θθ

θ θ θ

θ θ

θ θθ

θθ θθ θ

= = −

⎛ ⎞+ = ⇒ − + = ⇒⎜ ⎟⎝ ⎠

+ = ⇒ = ⇒

= − = − = = −

−= = = = =−

28. Since θ is in quadrant II, cos , tan , cot ,θ θ θ

and secθ are negative, and sinθ and cscθ are positive.

22 2 2

2 2

1 13sec

cos 12

12sin cos 1 sin 1

13144 25

sin 1 sin169 169

25 5 1 13sin ; csc

169 13 sin 5sin 5 13 5

tancos 12 13 12

1 12cot

tan 5

θθ

θ θ θ

θ θ

θ θθ

θθθ

θθ

= = −

⎛ ⎞+ = ⇒ + − = ⇒⎜ ⎟⎝ ⎠

+ = ⇒ = ⇒

= = = =

= = = −−

= = −

29. Since θ is in quadrant I, all the trigonometric functions of θ are positive.

2 2 2 2

22 2 2

1 1cot

tan 2

1 tan sec sec 1 2 5

1 1 5sec 5; cos

sec 55

1 51 cot csc 1 csc

2 4

5 5 1 2 2 5csc ; sin

4 2 csc 55

θθ

θ θ θ

θ θθ

θ θ θ

θ θθ

= =

+ = ⇒ = + = ⇒

= = = =

⎛ ⎞+ = ⇒ + = = ⇒⎜ ⎟⎝ ⎠

= = = = =





22 2 2

2 2

1 1cos

sec 3

1sin cos 1 sin 1

31 8

sin 1 sin9 9

8 2 2sin

9 3

1 3 3 2csc

sin 42 2

sin 2 2 3tan 2 2

cos 1 3

1 1 2cot

tan 42 2

θθ

θ θ θ

θ θ

θ

θθθθθ

θθ

= = −

⎛ ⎞+ = ⇒ + − = ⇒⎜ ⎟⎝ ⎠

+ = ⇒ = ⇒

= =

= = =

= = = −−

= = − = −

31. Since θ is in quadrant III, sin , cos , sec ,θ θ θ

and cscθ are negative, and tanθ and cotθ are positive.

22 2 2

2 2

1 2sin

csc 3

2sin cos 1 cos 1

34 5

cos 1 cos9 9

5 5cos

9 3

1 3 3 5sec

cos 55

sin 2 3 2 2 5tan

cos 55 3 5

1 5cot

tan 2

θθ

θ θ θ

θ θ

θ

θθθθθ

θθ

= = −

⎛ ⎞+ = ⇒ − + = ⇒⎜ ⎟⎝ ⎠

+ = ⇒ = ⇒

= − = −

= = − = −

−= = = =−

= =



2 2

22

1 1tan ; 1 tan sec

cot 2

1 5 5sec 1 sec

2 4 2

θ θ θθ

θ θ

= = − + = ⇒

⎛ ⎞= + − = ⇒ = −⎜ ⎟⎝ ⎠

1 2 2 5

cossec 55

θθ

= = − = −

( )22 2 21 cot csc 1 2 5 csc

1 1 5csc 5; sin

csc 55

θ θ θ

θ θθ

+ = ⇒ + − = = ⇒

= = = =

33. cos 0θ < and sin 0θ θ> ⇒ is in quadrant II. Thus, cos , tan , cot ,θ θ θ and secθ are

negative, and sinθ and cscθ are positive. 1 5 1 5

sec ; csccos 3 sin 4sin 4 5 4

tancos 3 5 3

1 3cot

tan 4

θ θθ θθθθ

θθ

= = − = =

= = = −−

= = −

34. cos 0θ > and sin 0θ θ< ⇒ is in quadrant IV. Thus, sin , tan , cot ,θ θ θ and cscθ are

negative, and cosθ and secθ are positive. 1 5 1 5

sec ; csccos 3 sin 4sin 4 5 4

tancos 3 5 3

1 3cot

tan 4

θ θθ θθθθ

θθ

= = = = −

−= = = −

= = −

35. sin 0 and cos 0θ θ θ> > ⇒ is in quadrant I. Thus, all trigonometric functions of θ are positive.

1 3csc 3

sin 3

1 3 6sec

cos 26

sin 3 3 1 2tan

cos 26 3 21 2

cot 2tan 2

θθ

θθθθθ

θθ

= = =

= = =

= = = =

= = =


negative, and sinθ and cscθ are positive. 1 3

csc 3sin 3

1 3 6sec

cos 26

sin 3 3 1 2tan

cos 26 3 21 2

cot 2tan 2

θθ

θθθθθ

θθ

= = =

= = − = −

= = = − = −−

= = − = −



37. tan 0 and sec 0α θ θ> < ⇒ is in quadrant III. Thus, sin , cos , sec ,α α α and cscθ are

negative, and tanα and cotα are positive.

1 2 2 5cos

sec 55sin 1 sin 5

tan sincos 2 52 5 5

1 5 1csc 5; cot 2

sin tan5

ααα αα αα

α αα α

= = − = −

= ⇒ = ⇒ = −−

= = − = − = =

38. tan 0 and sec 0α α θ> > ⇒ is in quadrant I. Thus, all trigonometric functions of α are positive.

1 2 2 5cos

sec 55sin 1 sin 5

tan sincos 2 52 5 5

1 5 1csc 5; cot 2

sin tan5

ααα αα αα

α αα α

= = =

= ⇒ = ⇒ =

= = = = =

39. sec 0 and cot 0β β θ> > ⇒ is in quadrant I.

Thus, all trigonometric functions of β are

positive.

22 2 2

2

1 1 1 4cos ; tan 2 2

sec 3 cot 2

1sin cos 1 sin 1

38 8 2 2

sin sin9 9 31 3 3 2

cscsin 42 2

β βα β

β β β

β β

αα

= = = = =

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = =

= = =

40. sec 0 and cot 0β β θ> < ⇒ is in quadrant IV.

Thus, sin , tan , cot ,β β β and csc β are

negative, and cos β and sec β are positive.

22 2 2

2

1 1cos

sec 31 4

tan 2 2cot 2

1sin cos 1 sin 1

38 8 2 2

sin sin9 9 31 3 3 2

cscsin 42 2

βα

ββ

β β β

β β

αα

= =

= = − = −

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = − = −

41. cot 0 and sin 0θ θ θ> < ⇒ is in quadrant III. Thus, sin , cos , sec ,θ θ θ and cscθ are

negative, and tanθ and cotθ are positive.

22 2 2

2

1 5 1 13tan ; csc

cot 12 sin 5

5sin cos 1 cos 1

13

144 144 12cos cos

169 169 131 13

seccos 12

θ θθ θ

θ θ θ

θ θ

θθ

= = = = −

⎛ ⎞+ = ⇒ − + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = −

42. cot 0 and sin 0θ θ θ< < ⇒ is in quadrant IV. Thus, sin , tan , cot ,θ θ θ and cscθ are

negative, and cosθ and secθ are positive.

22 2 2

2

1 5 1 13tan ; csc

cot 12 sin 5

5sin cos 1 cos 1

13

144 144 12cos cos

169 169 131 13

seccos 12

θ θθ θ

θ θ θ

θ θ

θθ

= = − = = −

⎛ ⎞+ = ⇒ − + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = =

= =


negative, and sinθ and cscθ are positive.

22 2 2

2

1 4csc

sin 3

3sin cos 1 cos 1

4

7 7 7cos cos

16 16 4

4 4 7sec

77

sin 3 4 3 3 7tan

cos 77 4 7

cos 7 4 7cot

sin 3 4 3

θθ

θ θ θ

θ θ

θ

θθθθθθ

= =

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= − = −

= = = − = −−−= = = −



44. cos 0 and tan 0θ θ θ> < ⇒ is in quadrant IV. Thus, sin , tan , cot ,θ θ θ and cscθ are


22 2 2

2

1 5sec

cos 4

4sin cos 1 sin 1

5

9 9 3sin sin

25 25 51 5

cscsin 3sin 3 5 3

tancos 4 5 4

cos 4 5 4cot

sin 3 5 3

θθ

θ θ θ

θ θ

θθθθθθθθ

= =

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = −

−= = = −

= = = −−

45. tan 0 and sin 0θ θ θ> < ⇒ is in quadrant III. Thus, sin , cos , sec ,θ θ θ and cscθ are

negative, and tanθ and cotθ are positive.

2 2 2 2

22 2 2

1 1cot

tan 2

tan 1 sec 2 1 5 sec

1 1 5sec 5; cos

sec 55

1 5cot 1 csc 1 csc

2 4

5 5csc

4 2

1 2 2 5sin

csc 55

θθ

θ θ θ

θ θθ

θ θ θ

θ

θθ

= =

+ = ⇒ + = = ⇒

= − = = − = −

⎛ ⎞+ = ⇒ + = = ⇒⎜ ⎟⎝ ⎠

= − = −

= = − = −

46. cot 0 and sec 0θ θ θ< > ⇒ is in quadrant IV. Thus, sin , tan , cot ,θ θ θ and cscθ are negative,

and cosθ and secθ are positive.

( )

22 2 2

22 2 2

1 1tan

cot 2

1 5tan 1 sec 1 sec

2 4

5 5 1 2 2 5sec ; cos

4 2 sec 55

cot 1 csc 2 1 5 csc

1 1 5csc 5; sin

csc 55

θθ

θ θ θ

θ θθ

θ θ θ

θ θθ

= = −

⎛ ⎞+ = ⇒ − + = = ⇒⎜ ⎟⎝ ⎠

= = = = =

+ = ⇒ − + = = ⇒

= − = = − = −

47. sec 0θ > and sin 0θ θ< ⇒ is in quadrant IV. Thus, sin , tan , cot ,θ θ θ and cscθ are negative,

and cosθ and secθ are positive.

22 2 2

2

1 2cos

sec 5

2sin cos 1 sin 1

5

21 21 21sin sin

25 25 5

1 5 5 21csc

sin 2121

sin 21 5 21tan

cos 2 5 2

cos 2 5 2 2 21cot

sin 2121 5 21

θθ

θ θ θ

θ θ

θθθθθθθθ

= =

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = − = −

−= = = −

= = = − = −−

48. csc 0 and tan 0θ θ θ> < ⇒ is in quadrant II. Thus, cos , tan , cot ,θ θ θ and secθ are

negative, and sinθ and cscθ are positive.

22 2 2

2

1 1sin

csc 2

1sin cos 1 cos 1

2

3 3 3cos cos

4 4 2

1 2 2 3sec

cos 33

sin 1 2 1 3tan

cos 33 2 3

cos 3 2cot 3

sin 1 2

θθ

θ θ θ

θ θ

θθθθθθθθ

= =

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = − = −

= = = − = −−−= = = −

49. ( )( ) 2

2 2

1 sin 1 sin sin

1 sin sin 1

θ θ θ

θ θ

− + +

= − + =

50. ( )( )( )2

2 2 2 2

cos 1 cos 1 sin

cos 1 sin sin cos 1

1 1 0

θ θ θ

θ θ θ θ

+ − +

= − + = + −

= − =

51. ( )( )( )2

2 2 2 2

2 2

1 tan 1 tan sec

1 tan sec 1 sec tan

tan tan 0

θ θ θ

θ θ θ θ

θ θ

+ − +

= − + = + −

= − =



52. ( )( )( )2

2 2 2 2

sec 1 sec 1 tan

sec 1 tan sec tan 1

1 1 0

θ θ θ

θ θ θ θ

− + −

= − − = − −

= − =

53. ( )( )2 2

sec tan sec tan

sec tan 1

θ θ θ θ

θ θ

+ −

= − =

54. ( )( )2 2

csc cot csc cot

csc cot 1

θ θ θ θ

θ θ

+ −

= − =

55.

( )( )

2sec 4sec

sec 2sec 2 sec 2

secsec 2

sec 2 sec 2

θ θθ

θ θθ

θθ θ

− −−

− += −

−= + − =

56.

( )( )

29 csccsc

3 csc3 csc 3 csc

csc3 csc

3 csc csc 3

θ θθ

θ θθ

θθ θ

− ++

+ −= +

+= − + =

57. ( )

2 2

sin cos tan cot

sin cossin cos

cos sin

sin cossin cos sin cos

cos sin

sin cos 1

θ θ θ θθ θθ θθ θθ θθ θ θ θθ θ

θ θ

+

⎛ ⎞= +⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + =

58. ( )sec csc sin cos

sec csc1 1

sec csccsc sec

sec csc1 1

sec csc sec csccsc sec

sec cscsec csc

1sec csc

θ θ θ θθ θ

θ θθ θ

θ θ

θ θ θ θθ θθ θ

θ θθ θ

++

⎛ ⎞+⎜ ⎟⎝ ⎠=

+⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=+

+= =+

59.

( ) ( )( )( )

2 2

1 12sec

sec tan sec tansec tan sec tan

2secsec tan sec tan

2sec2sec

sec tan2sec

2sec 01

θθ θ θ θ

θ θ θ θθ

θ θ θ θθ θ

θ θθ θ

+ −− +

+ + −= −

+ −

= −−

= − =

60.

( ) ( )( )( )

2 2

1 12cot

csc cot csc cotcsc cot csc cot

2cotcsc cot csc cot

2cot2cot

csc cot2cot

2cot 01

θθ θ θ θ

θ θ θ θθ

θ θ θ θθ θ

θ θθ θ

− −− +

+ − −= −

+ −

= −−

= − =

61. ( )( )2 tan 3 tan 1tan 2 tan 3

tan 1 tan 1tan 3

θ θθ θθ θ

θ

− +− − =+ +

= −

62.

( )( )

2 2

2

tan sec 1 sec 1 sec 1

sec 1 sec 1

sec sec 2

sec 1sec 2 sec 1

sec 1sec 2

θ θ θ θθ θ

θ θθ

θ θθ

θ

+ − − + −=− −

+ −=−

+ −=

−= +

63. If ( )2 2 2sin cos sin cosθ θ θ θ+ = + is an

identity, then it must be true for all values of .θ When 30 ,θ = ° we have

( )2

2 1 3sin 30 cos 30

2 2

1 3 3 3 2 32 1

4 4 4 2 2

⎛ ⎞° + ° = +⎜ ⎟⎝ ⎠⎛ ⎞ += + + = + =⎜ ⎟⎝ ⎠

while

2 2sin cos 1.θ θ+ = Therefore, the equation is not an identity.

64. If ( )2 21 cos 1 cosθ θ− = − is an identity, then it

must be true for all values of .θ When 30 ,θ = ° we have

( )2

2 31 cos30 1 0.0179

2

⎛ ⎞− ° = − ≈⎜ ⎟⎝ ⎠

while

22 3 1

1 cos 1 .2 4

θ⎛ ⎞

− = − =⎜ ⎟⎝ ⎠ Therefore, the

equation is not an identity.

65. If ( )sin 45 sin sin 45θ θ+ ° = + ° is an identity,

then it must be true for all values of .θ When

45 ,θ = ° we have ( )sin 45 45 sin 90 1° + ° = ° =

while sin 45 sin 45 2.° + ° = Therefore, the equation is not an identity.



66. If ( )cos 45 cos cos 45θ θ+ ° = + ° is an identity,

then it must be true for all values of .θ When

45 ,θ = ° we have ( )cos 45 45 cos 90 0° + ° = ° =

while cos 45 cos 45 2.° + ° = Therefore, the equation is not an identity.

67. If sin 2 2sinθ θ= is an identity, then it must be true for all values of .θ When 45 ,θ = ° we

have ( )sin 2 45 sin 90 1⋅ ° = ° = while

2sin 45 2.° = Therefore, the equation is not an identity.

68. If cos 2 2cosθ θ= is an identity, then it must be true for all values of .θ When 45 ,θ = ° we

have ( )cos 2 45 cos90 0⋅ ° = ° = while

2cos 45 2.° = Therefore, the equation is not an identity.

69. If 2 2tan 1 secθ θ− = is an identity, then it must be true for all values of .θ When

45 ,θ = ° we have ( )2tan 45 1 0° − = while

2sec 45 2.° = Therefore, the equation is not an identity.

70. If 2 2cot 1 secθ θ+ = is an identity, then it must be true for all values of .θ When

90 ,θ = ° we have ( )2cot 90 1 1° + = while

2sec 90° is undefined. Therefore, the equation is not an identity.

71. If 2sec 1 sec 1θ θ− = − is an identity, then it must be true for all values of .θ When

45 ,θ = ° we have 2sec 45 1 1° − = while

sec 45 1 2 1.° − = − Therefore, the equation is not an identity.

72. If 2csc 1 csc 1θ θ+ = + is an identity, then it must be true for all values of .θ When

45 ,θ = ° we have 2csc 45 1 3° + = while

csc 45 1 2 1.° + = + Therefore, the equation is not an identity.


73. 20

20cscsin

x θθ

= =

74. 60

60cottan

x θθ

= =

75. ( )( )180

2 2180

sin 180tan

cosn

n

A nr nrn

°

°°⎛ ⎞= = ⎜ ⎟⎝ ⎠

76. ( )2 1804 3 tan 36 tan 45 36 sq ft

4A

°⎛ ⎞= = ° =⎜ ⎟⎝ ⎠

77. ( )( ) ( )( ) ( )

( )

1 2 1 2

1 2 1 2

1 2 1 2

1 2 1 2

1 2

1 2

1 sin cos cos

1 sin cos

1 sin cos

cos cos1 tan

tan1

m m m m

m m m m

m m m m

m m m m

m m

m m

θ θ θθ θθ θ

θ θθ

θ

+ = −+ = −+ −

=

+ = −−=

+

78.

( )

1 2

351 23

1 2 5

34,

54 17 5

tan 11 17 51 4

tan 1 45 or 135

m m

m m

m mθ

θ θ θ

= =

−−= = = =+ +

= ⇒ = ° = °



1.5 C Beyond the Basics

79. ( ) ( )

( )

2 2

2 2 2 2

2 2 2 2

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2

sin cos cos sin cos cos sin sin

sin cos 2sin cos cos sin cos sin

cos cos 2cos cos sin sin sin sin


sin cos sin sin cos sin cos

α β α β α β α βα β α β α β α β

α β α β α β α βα β α β α β α βα β α β α β α

+ + −= + +

+ − += + + += + + +( )

( ) ( )2

2 2 2 2 2 2 2 2

cos

sin cos sin cos sin cos sin cos 1

β

α β β α β β α α= + + + = + =

80. ( ) ( )

( )

2 2

2 2 2 2

2 2 2 2

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2


sin cos 2sin cos cos sin cos sin

cos cos 2cos cos sin sin sin sin


sin cos sin sin cos sin cos

α β α β α β α βα β α β α β α β

α β α β α β α βα β α β α β α βα β α β α β α

− + += − +

+ + += + + += + + +( )

( ) ( )2

2 2 2 2 2 2 2 2

cos

sin cos sin cos sin cos sin cos 1

β

α β β α β β α α= + + + = + =

81. ( ) ( )( ) ( )( )( ) ( )

4 4 6 6

22 2 2 2 2 2 4 2 2 4

2 2 4 2 2 4

2 2 4 2 2 4

4 2 2

3 sin cos 2 sin cos

3 sin cos 2sin cos 2 sin cos sin sin cos cos

3 1 2sin cos 2 sin sin cos cos

3 6sin cos 2sin 2sin cos 2 cos

3 2sin 4sin cos 2 co

α α α α

α α α α α α α α α α

α α α α α α

α α α α α αα α α

+ − +

⎡ ⎤ ⎡ ⎤= + − − + − +⎢ ⎥ ⎣ ⎦⎣ ⎦= − − − +

= − − + −= − − −

( )( )

4

4 2 2 4

22 2

s

3 2 sin 2sin cos cos

3 2 sin cos 3 2 1

αα α α α

α α

= − + +

= − + = − =

82.

( )( )

( )( )

6 6 2 4

2 2 4 2 2 4 2 4

4 2 2 4 2 4

4 2 2 4 2

2 2 2 2 2

2 2 2 2

sin cos 3cos 3cos

sin cos sin sin cos cos 3cos 3cos

sin sin cos cos 3cos 3cos

sin sin cos 2cos 3cos

sin 2cos sin cos 3cos

sin 2cos 3cos sin c

α α α α

α α α α α α α α

α α α α α αα α α α αα α α α α

α α α α

+ + −

= + − + + −

= − + + −

= − − +

= − + +

= − + = + 2os 1α =

83. ( ) ( )

( )

2 2 22

2 2

2 2 2

2 2

2 2 2 2

2 2 2

22 2

2 2 2

1 cos 1 cos1 cos 1 cos 4cos4cot

1 cos 1 cos 1 cos sin

1 2cos cos 1 2cos cos 4cos

1 cos sin

2 2cos 4cos 2 2cos 4cos

sin sin sin

2 1 cos2 2cos 2sin

sin sin sin

θ θθ θ θθθ θ θ θ

θ θ θ θ θθ θ

θ θ θ θθ θ θ

θθ θθ θ

− + +− ++ − = −+ − −

− + + + += −−

+ + −= − =

−−= = = 2θ

=



84.

( )( ) ( )( )

( ) ( )( )

3 3 3 3

2 2 2 2

2 2 2 2

2 2 2 2

cos sin cos sin

cos sin cos sin

cos sin cos cos sin sin cos sin cos cos sin sin

cos sin cos sin

cos cos sin sin cos cos sin sin

2cos 2sin 2 cos sin 2

θ θ θ θθ θ θ θ

θ θ θ θ θ θ θ θ θ θ θ θθ θ θ θ

θ θ θ θ θ θ θ θ

θ θ θ θ

+ −++ −

+ − + − + += +

+ −

= − + + + +

= + = + =

85.

2

2 2 2 2 22 2

2 2 2 2 2 2 2

2

1 2sinsec 2 tan 1 2sin 1 2sin 1 2sincos cos 1

1 3 tan 3sin cos 3sin 1 sin 3sin 1 2sin1

cos

θθ θ θ θ θθ θ

θ θ θ θ θ θ θθ

++ + + += = = = =+ + − + ++

86.

2

2 2 2 2 2 2 22 2 2

2 2 2 2 2 2 2 2

2 2 2

1 1 sin1

csc sec 1 tan cos sin cos sinsin cos cos 01 1csc sec 1 tan sin cos sin cos sin

1sin cos cos

αα α α α α α αα α αα α α α α α α α

α α α

+ ++ + + +− = − = − =− − − −− −

87. ( ) ( )( ) ( )

( )( ) ( )

2 22 2 2 2

2 2 4 2 2 4

2 4 6 2 4 6

2 2 4 4 6 6

2 2 4 4

cos 3 4cos sin 3 4sin

cos 9 24cos 16cos sin 9 24sin 16sin

9cos 24cos 16cos 9sin 24sin 16sin

9cos 9sin 24cos 24sin 16cos 16sin

9 cos sin 24 cos sin 16 cos

α α α α

α α α α α α

α α α α α α

α α α α α α

α α α α

− + −

= − + + − +

= − + + − +

= + − − + +

= + − + + ( )( )( )

( )( )

6 6

4 4 2 2 4 2 2 4

4 4 4 2 2 4

4 2 2 4 4 2 2 4

22 2

sin

9 24cos 24sin 16 cos sin cos cos sin sin

9 24cos 24sin 16cos 16cos sin 16sin

9 8cos 16cos sin 8sin 9 8 cos 2cos sin sin

9 8 cos sin 9 8 1

α α

α α α α α α α α

α α α α α αα α α α α α α α

α α

+

= − − + + − +

= − − + − +

= − − − = − + +

= − + = − =

88. ( ) ( )

( ) ( ) ( )

2 2 2 2

2 2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2 2

sin csc cos sec tan cot

sin 2sin csc csc cos 2cos sec sec tan cot

sin 2 csc cos 2 sec tan cot

sin cos sec tan csc cot 4 1 1 1 4 7

θ θ θ θ θ θ

θ θ θ θ θ θ θ θ θ θθ θ θ θ θ θ

θ θ θ θ θ θ

+ + + − −

= + + + + + − −

= + + + + + − −

= + + − + − + = + + + =

89. Start by squaring both sides of tan cot 2.θ θ+ =

( )2 2

2 2

2 2 2 2

tan cot 2

tan 2 tan cot cot 4

tan 2 cot 4 tan cot 2

θ θθ θ θ θθ θ θ θ

+ = ⇒+ + = ⇒+ + = ⇒ + =

90. Start by squaring both sides of sec cos 2.θ θ+ =

( )2 2

2 2

2 2 2 2

sec cos 2

sec 2sec cos cos 4

sec 2 cos 4 sec cos 2

θ θθ θ θ θθ θ θ θ

+ = ⇒+ + = ⇒+ + = ⇒ + =

Chapter 1 Review Exercises 41


91. Start by dividing the numerator and denominator by cos .θ

5sin 3cos5sin 3cos cos cos

5sin 2cos5sin 2coscos cos

5 tan 3 4 3 1

5 tan 2 4 2 6

θ θθ θ θ θ

θ θθ θθ θθθ

−− =+ +

− −= = =+ +

92. Start by dividing the numerator and denominator by sin .θ

3434

sin cossin cos 1 cotsin sin

sin cossin cos 1 cotsin sin

34cot 3 cot

411 cot 1 4 1

1 cot 7 4 71

θ θθ θ θθ θ

θ θθ θ θθ θ

θ θ

θθ

−− −= =+ ++

= ⇒ =

−− = = =+ +

93.

( )

2 2 2 2

2 2

2 2

2 2

2 2

2 2

2 2

2

2 2

17

1 1csc sec sin cos

1 1csc secsin cos

cos sin

cos sin

cos sin

cos 1 cos

2cos 12 2

1 1sec 1 tan

2 2 7 31 1 1

8 7 4 41

θ θ θ θθ θ

θ θθ θθ θθ θ

θ θ

θ

θ θ

−− =+ +

−=+

= −

= − −

= −

= − = −+

= − = − = − =+

94. 2 2

2 2

1 cos sin

2 sin 2 sin

θ θθ θ

− =+ +

Now divide the numerator and denominator by sin .θ

2

2

2 2

22 2

sin1 1sin

22 sin 2csc 11sinsin sin

θθ

θ θθθ θ

= =+++

( ) ( ) 112 133

1 1 1 3

112 1 12 1 cot 1θ= = = =

+ ++ +


95. a. The equation is an identity for 0 90θ≤ ≤ ° because cos 0θ ≥ in the given interval.

b. The equation is not an identity for 0 180θ≤ ≤ ° because the left side is negative for 90 180θ° < ≤ ° while the right side is positive.

96. False. We can show that

( )cos cos cosα β α β− = − is not an identity

by letting 60α = ° and 30 .β = ° Then

( ) ( )cos cos 60 30α β− = ° − ° 3

cos 30 ,2

= ° =

while cos cosα β− =

1 3cos 60 cos30 .

2 2° − ° = −

97. No, sin , cos , and tanθ θ θ cannot all be

negative for an angle .θ Note that if two of the

functions in sin

tancos

θθθ

= are negative, then

the third has to be positive.

Chapter 1 Review Exercises

1. 90° – 67.8° = 22.2°

2. 180° – 123.4° = 56.6°

3. 15 30

64 15 30 64 64.2660 3600

⎛ ⎞° = + + ° ≈ °′ ′′ ⎜ ⎟⎝ ⎠

4. 34.742 34 0.742(60 ) 34 44.5234 44 0.52(60 ) 34 44 31

° = ° + = ° +′ ′= ° + + ≈ °′ ′′ ′ ′′

5. a. Quadrant III



b. Quadrant III

c. Quadrant III

d. Quadrant IV

6. a. 580 220 360 ,° = ° + ° so 550° is coterminal with 220°.

b. 1460 20 4 360 ,° = ° + ⋅ ° so 1460° is coterminal with 20°.

c. 675 45 2 360 ,− ° = ° − ⋅ ° so –675° is coterminal with 45°.

d. 1345 95 4 360 ,− ° = ° − ⋅ ° so –1345° is coterminal with 95°.

7. a. (–2, 0) lies on the negative x-axis, so a line through that point forms a 180° angle with the positive x-axis. Thus, 180 360 ,nθ = ° + ⋅ ° n any integer.

b. (3, –3) lies in Quadrant IV, so a line through that point forms a 270° + 45° = 315° angle with the positive x-axis. Thus,

315 360 ,nθ = ° + ⋅ ° n any integer.

8. In a 45°-45°-90° triangle, both legs are equal

and the length of the hypotenuse is 2 times

the length of a leg. Therefore, the length of the other leg is 6 cm, and the length of the

hypotenuse is 6 2 cm.

9. In a 30°-60°-90° triangle, if the length of the shorter leg (the leg opposite the 30° angle) is x, then the length of the longer leg (the angle

opposite the 60° angle) is 3,x and the length of the hypotenuse is 2x. The shortest side is

6 cm, so the other leg has length 6 3 cm, and the hypotenuse has length 12 cm.

10. ( ) ( ) ( )

1802 10 3 20 3 30 180

8 20 180 8 160 20

A B Cx x x

x x x

+ + = °⇒+ ° + − ° + + ° = °⇒+ ° = °⇒ = °⇒ = °

The measures of the angles are 2 20 10 50 , 3 20 20 40 ,A B= ⋅ + = ° = ⋅ − = ° and

3 20 30 90 .C = ⋅ + = °

11. True

12. False. A scalene triangle is a triangle whose sides have different lengths. A scalene triangle may be acute, right, or obtuse.

13. ( )22 23, 4 3 4 5

4 3 4sin cos tan

5 5 35 5 3

csc sec cot4 3 4

x y r r

θ θ θ

θ θ θ

= − = ⇒ = − + ⇒ =

= = − = −

= = − = −

14. ( ) ( )2 225, 12 5 12

1312 5 12

sin cos tan13 13 513 13 5

csc sec cot12 5 12

x y r

r

θ θ θ

θ θ θ

= − = − ⇒ = − + − ⇒=

= − = − =

= − = − =

15. ( ) ( )2 222, 3 2 3 13

3 3 13 13sin csc

13 313

2 2 13 13cos sec

13 2133 2

tan cot2 3

x y r r

θ θ

θ θ

θ θ

= = − ⇒ = + − ⇒ =

= − = − = −

= = =

= − = −



16. ( ) ( )2 223, 5 3 5 34

5 5 34 34sin csc

34 534

3 3 34 34cos sec

34 3345 3

tan cot3 5

x y r r

θ θ

θ θ

θ θ

= = ⇒ = + ⇒ =

= = − =

= = =

= =

17. ( ) ( )2 222, 0 2 0 2

0 2sin 0 csc , undefined

2 02 2

cos 1 sec 12 20 2

tan 0 cot , undefined2 0

x y r r

θ θ

θ θ

θ θ

= = ⇒ = + ⇒ =

= = =

= = = =

= = =

18. ( ) ( )2 220, 3 0 3 3

3 3sin 1 csc 1

3 30 3

cos 0 sec , undefined3 03 0

tan , undefined cot 00 3

x y r r

θ θ

θ θ

θ θ

= = ⇒ = + ⇒ =

= = = =

= = =

= = =

19. ( ) ( )2 224, 0 4 0 4

0 4sin 0 csc , undefined

4 04 4

cos 1 sec 14 4

0 4tan 0 cot , undefined

4 0

x y r r

θ θ

θ θ

θ θ

= − = ⇒ = − + ⇒ =

= − = = −

= − = − = − = −

−= = =−

20. ( ) ( )2 220, 5 0 5 5

5 5sin 1 csc 1

5 50 5

cos 0 sec , undefined5 0

5 0tan , undefined cot 0

0 5

x y r r

θ θ

θ θ

θ θ

= = − ⇒ = + − ⇒ =

= − = − = − = −

= = =

−= = =−

21. tan 0 and sin 0θ θ θ< > ⇒ is in quadrant II.

22. cot 0 and csc 0θ θ θ> < ⇒ is in quadrant III.

23. cot 0 and sec 0θ θ θ> < ⇒ is in quadrant III.

24. sec 0 and csc 0θ θ θ< > ⇒ is in quadrant II.

25. sec 0 and tan 0θ θ θ> < ⇒ is in quadrant IV.

26. sin 0 and cot 0θ θ θ< < ⇒ is in quadrant IV.

27.

2 2 2

5cot , in Quadrant II 0, 0

125, 12 ( 5) 12 13

x y

x y r r

θ θ= − ⇒

= − = ⇒ = − + ⇒ =

< >

12 5 12sin cos tan

13 13 513 13 5

csc sec cot12 5 12

θ θ θ

θ θ θ

= = − = −

= = − = −

28.

( )22 2

5cot , in Quadrant IV 0, 0

125, 12 5 12 13

x y

x y r r

θ θ= − ⇒

= = − ⇒ = + − ⇒ =

> <

12 5 12sin cos tan

13 13 513 13 5

csc sec cot12 5 12

θ θ θ

θ θ θ

= − = = −

= − = = −

29. 3

sin and cos 05

θ θ θ= > ⇒ is in quadrant I.

Thus, x > 0 and y > 0. 2 2 23, 5 5 3 4y r x x= = ⇒ = + ⇒ =

3 4 3sin cos tan

5 5 45 5 4

csc sec cot3 4 3

θ θ θ

θ θ θ

= = =

= = =

30. 13

sec and sin 012

θ θ θ= < ⇒ is in

quadrant IV. Thus, x > 0 and y < 0. 2 2 212, 13 13 12 5x r y y= = ⇒ = + ⇒ = −

5 12 5sin cos tan

13 13 1213 13 12

csc sec cot5 12 5

θ θ θ

θ θ θ

= − = = −

= − = = −

31. 260 260 180 80θ θ= °⇒ = ° − ° = °′

32. 530 170 360 ,° = ° + ° so the 530° angle is coterminal with 170°. Since 170° lies in quadrant II, the reference angle is determined by 180° – 170° = 10°.

33. –275° + 360° = 85° Since 85° lies in quadrant I, the reference angle is 85°.

34. –1315° + 4(360°) = 125° Since 125° lies in quadrant II, the reference angle is determined by 180° – 125° = 55°.



35. 390 360 30 ,° = ° + ° so 390° is coterminal with 30°, which is the reference angle. In quadrant I, all trigonometric functions of θ are positive.

1sin 390 sin 30

23

cos 390 cos3023

tan 390 tan 303

csc390 csc30 2

2 3sec390 sec30

3cot 390 cot 30 3

° = ° =

° = ° =

° = ° =

° = ° =

° = ° =

° = ° =

36. –390° is coterminal with –390° + 2(360°) = 330°. 330° lies in quadrant IV, so the reference angle is 360° – 330° = 30°. In quadrant IV, sin , tan , cot ,θ θ θ and cscθ are negative, and

cosθ and secθ are positive.

( )

( )

( )( )

( )

( )

1sin 390 sin 30

23

cos 390 cos 302

3tan 390 tan 30

3csc 390 csc30 2

2 3sec 390 sec30

3

cot 390 cot 30 3

− ° = − ° = −

− ° = ° =

− ° = − ° = −

− ° = − ° = −

− ° = ° =

− ° = − ° = −

37. –495° is coterminal with –495° + 2(360°) = 225°. 225° lies in quadrant III, so the reference angle is 225° – 180° = 45°. In quadrant III, sin , cos , sec ,θ θ θ and cscθ are negative, and

tanθ and cotθ are positive.

( ) 2sin 495 sin 45

2− ° = − ° = −

( )( )( )( )( )

2cos 495 cos 45

2tan 495 tan 45 1

csc 495 csc 45 2

sec 495 sec 45 2

cot 495 cot 45 1

− ° = − ° = −

− ° = ° =

− ° = − ° = −

− ° = − ° = −− ° = ° =

38. 1020 300 2 360 ,° = ° + ⋅ ° so 1020° is coterminal with 300°. 300° lies in quadrant IV, so the reference angle is 360° – 300° = 60°. In quadrant IV, sin , tan , cot ,θ θ θ and cscθ are


3sin1020 sin 60

21

cos1020 cos 602

tan1020 tan 60 3

2 3csc1020 csc60

3sec1020 sec 60 2

3cot1020 cot 60

3

° = − ° = −

° = ° =

° = − ° = −

° = − ° = −

° = ° =

° = − ° = −

39. 5 12

sin 0 and cos 013 13

θ θ θ= > = > ⇒ is in

quadrant I. Thus, all trigonometric functions of θ are positive.

1 13csc

sin 51 13

seccos 12sin 5 13 5

tancos 12 13 12

1 12cot

tan 5

θθ

θθθθθ

θθ

= =

= =

= = =

= =

40. 5 5

csc 0 and sec 03 4

θ θ θ= > = > ⇒ is in

quadrant I. Thus, all trigonometric functions of θ are positive.

1 3sin

csc 51 4

cossec 5sin 3 5 3

tancos 4 5 4

1 4cot

tan 3

θθ

θθθθθ

θθ

= =

= =

= = =

= =



41. 1

sin and 2

θ θ= in quadrant II. Thus,

cos , tan , cot ,θ θ θ and secθ are negative,

and sinθ and cscθ are positive.

22 2 2

2

1csc 2

sin

1sin cos 1 cos 1

2

3 3 3cos cos

4 4 2

1 2 2 3sec

cos 33

sin 1 2 1 3tan

cos 33 2 3

cos 3 2cot 3

sin 1 2

θθ

θ θ θ

θ θ

θθθθθθθθ

= =

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = − = −

= = = − = −−−= = = −

42. 1

cos and 2

θ θ= in quadrant IV. Thus,

sin , tan , cot ,θ θ θ and cscθ are negative, and


22 2 2

2

1sec 2

cos

1sin cos 1 sin 1

2

3 3 3sin sin

4 4 2

1 2 2 3csc

sin 33

sin 3 2tan 3

cos 1 2

cos 1 2 1 3cot

sin 33 2 3

θθ

θ θ θ

θ θ

θθθθθθθθ

= =

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = − = −

−= = = −

= = = − = −−

43. tan 4 and θ θ= in quadrant III. Thus, sin , cos , sec ,θ θ θ and cscθ are negative, and

tanθ and cotθ are positive.

2 2 2 2

1 1cot

tan 4

tan 1 sec 4 1 17 sec

1 1 17sec 17; cos

sec 1717

θθ

θ θ θ

θ θθ

= =

+ = ⇒ + = = ⇒

= − = = − = −

2

2 2 21 17cot 1 csc 1 csc

4 16

17 17csc

16 4

1 4 4 17sin

csc 1717

θ θ θ

θ

θθ

⎛ ⎞+ = ⇒ + = = ⇒⎜ ⎟⎝ ⎠

= − = −

= = − = −

44. csc 3 and θ θ= in quadrant II. Thus, cos , tan , cot ,θ θ θ and secθ are negative,


22 2 2

2

1 1sin

csc 3

1sin cos 1 cos 1

3

8 8 2 2cos cos

9 9 3

1 3 3 2sec

cos 42 2

sin 1 3 1 2tan

cos 42 2 3 2 2

cos 2 2 3cot 2 2

sin 1 3

θθ

θ θ θ

θ θ

θθθθθθθθ

= =

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = − = −

= = = − = −−−= = = −

45. cot 2 and θ θ= − in quadrant IV. Thus, sin , tan , cot ,θ θ θ and cscθ are negative, and


( )

22 2 2

2

22 2 2

2

1 1tan

cot 2

1sec 1 tan sec 1

2

5 5 5sec sec

4 4 2

1 2 2 5cos

sec 55

csc 1 cot csc 1 2

csc 5 csc 5

1 1 5sin

csc 55

θθ

θ θ θ

θ θ

θθ

θ θ θ

θ θ

θθ

= = −

⎛ ⎞= + ⇒ = + − ⇒⎜ ⎟⎝ ⎠

= ⇒ = =

= = =

= + ⇒ = + − ⇒

= ⇒ = −

= = − = −



46. 7

tan and 4

θ θ= − in quadrant II. Thus,

cos , tan , cot ,θ θ θ and secθ are negative,


22 2 2

2

22 2 2

2

1 4cot

tan 7

7sec 1 tan sec 1

4

65 65 65sec sec

16 16 4

1 4 4 65cos

sec 6565

4csc 1 cot csc 1

7

65 65csc csc

49 7

1 7 7 65sin

csc 6565

θθ

θ θ θ

θ θ

θθ

θ θ θ

θ θ

θθ

= = −

⎛ ⎞= + ⇒ = + − ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

= = − = −

⎛ ⎞= + ⇒ = + − ⇒⎜ ⎟⎝ ⎠

= ⇒ =

= = =

47. ( )( )

( )

2

2 2

2 2

1 cos 1 cos sin

1 cos sin

1 cos sin 1 1 0

θ θ θθ θθ θ

− + −= − −= − + = − =

48. ( )( )

( )

2

2 2

2 2 2 2

csc 1 csc 1 cot

csc 1 cot

csc 1 cot csc csc 0

θ θ θθ θθ θ θ θ

− + −= − −= − + = − =

49.

( )( )

2cot cot 2cot

cot 1cot 2 cot 1

cotcot 1

cot 2 cot 2

θ θ θθθ θ

θθ

θ θ

+ − −−+ −

= −−

= + − =

50.

( ) ( )

2

22

2 22

2

22 2 2

2

sin sin2 tan

1 sin 1 sinsin 1 sin sin 1 sin

2 tan1 sin

sin sin sin sin2 tan

1 sin

2sin2 tan 2 tan 2 tan 0

cos

θ θ θθ θθ θ θ θ

θθ

θ θ θ θ θθ

θ θ θ θθ

− −− +

+ − −= −

−+ − += −

−

= − = − =

Chapter 1 Test

1. 180 61 31 179 60 61 31 118 29° − ° = ° − ° = °′ ′ ′ ′

2. 0 360 , 90 360180 360 , 270 360

n nn n

° + ⋅ ° ° + ⋅ °° + ⋅ ° ° + ⋅ °

3.

1 2 3 260∠ +∠ + ∠ = ° Angles 1 and 3 are vertical angles, so they are equal. Angles 1 and 2 are supplements, so

2 180 1.∠ = ° − ∠ Now substitute 1 180 1 1 2601 180 260 1 802 100 , 3 80 , 4 100

∠ + ° − ∠ + ∠ = °⇒∠ + ° = °⇒ ∠ = °∠ = ° ∠ = ° ∠ = °

4. ( ) ( ) ( )15 30 45 1803 90 180 3 90 30x x xx x x+ ° + + ° + + ° = °⇒+ ° = °⇒ = °⇒ = °

The angle measures are 30° + 15° = 45°, 30° + 30° = 60°, and 30° + 45° = 75°.

5. 2 3 5 180 10 180 18x x x x x+ + = °⇒ = °⇒ = ° The measures of the angles are 2(18°) = 36°, 3(18°) = 54°, and 5(18°) = 90°.

6. In an isosceles right triangle, both legs are

equal and the length of the hypotenuse is 2 times the length of a leg. Therefore, the length

of legs = 20

10 2 14.1 cm.2= ≈

7. True. Two congruent triangles are similar because the corresponding angles are equal and the lengths of the corresponding sides are proportional.

8.

AD represents the length of the shadow.

312 3 13.5

4.5 129 13.5 1.5

ADAD AD

ADAD AD

= ⇒ = + ⇒+= ⇒ =

The shadow is 1.5 feet long.

Chapter 1 Test 47


9. ( ) ( )2 222, 1 2 1 5

1 5sin

55

x y r r

y

rθ

= = − ⇒ = + − ⇒ =

= = − = −

10. ( ) ( )2 222, 3 2 3

13

13csc

3

x y r

r

r

yθ

= − = − ⇒ = − + − ⇒

=

= = −

11. sin 0 and sec 0θ θ θ> < ⇒ lies in quadrant II.

12. cot 0 and csc 0θ θ< < ⇒ lies in quadrant IV.

13. ( )845 2 360 125° = ° + °

125° lies in quadrant II, so the reference angle is 180° – 125° = 55°.

14. ( )640 2 360 80− ° + ° = °

Since 80° lies in quadrant I, it is the reference angle.

15. θ is in quadrant II, so cosθ is negative. 2

2 2 2

2

4sin cos 1 cos 1

733 33 33

cos cos49 49 7

θ θ θ

θ θ

⎛ ⎞+ = ⇒ + = ⇒⎜ ⎟⎝ ⎠

= ⇒ = − = −

16. θ is in quadrant IV, so secθ is positive. 2

2 2 2

2

5sec 1 tan sec 1

12169 169 13

sec sec144 144 12

θ θ θ

θ θ

⎛ ⎞= + ⇒ = + − ⇒⎜ ⎟⎝ ⎠

= ⇒ = =

17. If 1 tan secθ θ+ = is an identity, then it must be true for all values of .θ When 45 ,θ = ° we

have ( )tan 45 1 2° + = while sec 45 2.° =

Therefore, the equation is not an identity.

18. If 2csc 1 cotθ θ= + is an identity, then it must be true for all values of .θ When

225 ,θ = ° we have csc 225 2° = − while

( )21 cot 225 2.+ ° = Therefore, the equation

is not an identity.

19. 2

2

sin cot sec coscos 1

sin cossin cos

cos cos 0

θ θ θ θθθ θθ θ

θ θ

−⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − =

20.

( )( )

22sin sin 1sin

2sin 12sin 1 sin 1

sin2sin 1

sin 1 sin 1

θ θ θθθ θ

θθ

θ θ

+ − −−− +

= −−

= + − =

chapter 1 trigonometric functions - testbanku.eu · section 1.1 angles 5 copyright © 2011 pearson...

Documents