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### Transcript of Chapter 2 Trigonometric Functions40p6zu91z1c3x7lz71846qd1- ... Chapter 2 Trigonometric Functions...

• Exercise 2.1 Degrees and Radians

18

Chapter 2 Trigonometric Functions

4. Since 45° corresponds to a radian measure of π/4 rad, we have:

90° = 2 · 45° corresponds to 2 · π/4 or π/2 rad. 315° = 7 · 45° corresponds to 7 · π/4 or 7π/4 rad. 135° = 3 · 45° corresponds to 3 · π/4 or 3π/4 rad. 360° = 8 · 45° corresponds to 8 · π/4 or 2π rad. 180° = 4 · 45° corresponds to 4 · π/4 or π rad. 225° = 5 · 45° corresponds to 5 · π/4 or 5π/4 rad. 270° = 6 · 45° corresponds to 6 · π/4 or 3π/2 rad.

6. The angle of degree measure 20 is smaller, since

!

1

2 radian corresponds to a degree measure of

!

180

2"

which is approximately 28.6°.

8.

Coterminal angles have measures 135° + 360° = 495° and 135° – 360° = –225°

10.

Coterminal angles have measures 60° + 360° = 420° and 60° – 360° = –300°

12.

Coterminal angles have measures –140° + 360° = 220° and –140° – 360° = –500°

14.

Coterminal angles have measures

!

"

3 + 2π =

!

7"

!

"

3 – 2π = –

!

5"

• Chapter 2 Trigonometric Functions

19

16.

Coterminal angles have measures

!

7"

4 + 2π =

!

"

!

7"

4 – 2π = –

!

15"

18.

Coterminal angles have measures

!

6"

5 + 2π =

!

16"

!

6"

5 – 2π = –

!

4"

22. 1° corresponds to

!

"

180 rad; 5° corresponds to 5 ·

!

"

180 =

!

"

24. 190° corresponds to 190 ·

!

"

!

19"

26. 108° corresponds to 108 ·

!

"

!

3"

!

180

"

#

\$ %

&

' ( °; 3 rad corresponds to

!

3 " 180

#

\$

% &

'

( ) ° =

!

540

"

#

\$ %

&

' ( ° ≈ 171.9°

!

(1.4)(180)

"

#

\$ %

&

' ( ° =

!

252

"

#

\$ %

&

' ( ° ≈ 80.21°

32.

!

2"

!

2"

9 # 180

"

\$

% &

'

( ) ° = 40° (Exact)

34. Set calculator in degree mode. (A) (0.750) rad is 42.972° (B) (1.5) rad is 85.944° (C) (3.80) rad is 217.724° (D) (–7.21) rad is –413.103°

38. Since θ = s

r , we have

(A) θ =

!

15cm

40 cm = 0.375 rad; 21.5° (B) θ =

!

65cm

40 cm = 1.625 rad; 93.1°

(C) θ =

!

123.5cm

40 cm = 3.0875 rad; 176.9° (D) θ =

!

210 cm

40 cm = 5.25 rad; 300.8°

• Exercise 2.1 Degrees and Radians

20

40. (A) s = rθ (B) s = rθ

= (5)(0.228) = 1.14 in = (5)

!

37° " #

180°

\$

% &

'

( ) = 3.23 in

(C) s = rθ (D) s = rθ

= (5)(1.537) = 7.685 in = (5)

!

175° " #

180°

\$

% &

'

( ) = 15.27 in

42. Since θr =

!

"

180 θd , if both sides of this equation are cut in half, then

!

1

2 θr =

!

"

180 ·

!

1

2 θd

Thus if θd is cut in half, θr is also cut in half.

44. Since θr =

!

s

r , and if s (the numerator) is doubled while r is held constant, then θr will be doubled.

46. (A) A =

!

1

2 r2θ (B) A =

!

1

2 r2θ

=

!

1

2 (10.5)2 (0.150) ≈ 8.27 ft2 =

!

1

2 (10.5)2

!

15.0° " #

180°

\$

% &

'

( ) ≈ 14.43 ft2

(C) A =

!

1

2 r2θ (D) A =

!

1

2 r2θ

=

!

1

2 (10.5)2 (1.74) ≈ 95.92 ft2 =

!

1

2 (10.5)2

!

105° " #

180°

\$

% &

'

( ) ≈ 101.02 ft2

48. Since θ =

!

s

r , θ in radian measure, θ =

!

s

1 = s.

50.

!

9"

4 = 2π +

!

"

4 . So

!

9"

4 is coterminal with 2π +

!

"

4 – 2π =

!

"

4 . Since

!

"

4 is between 0 and

!

"

2 , its

terminal side lies in quadrant I.

52. 696° is coterminal with (696 – 360)° = 336°. Since 336° is between 270° and 360°, its terminal side lies in quadrant IV.

54. –

!

11"

3 is coterminal with –

!

11"

3 + 2π = –

!

5"

3 . Since –

!

5"

3 is between –

!

3"

2 and –2π, its terminal side

56. –672° is coterminal with (–672 + 360)° = –312°. Since –312° is between –270° and –360°, its terminal side lies in quadrant I.

58. –20 rad is approximately –1146° which is coterminal with –1146° + 360° = –786° and with –786° + 360° = –426° and with –426° + 360° = –66°. Since –66° is between 0° and –90°, its terminal side lies in quadrant IV.

60. 116.9853° = (116.9853) ·

!

"

62. 2.562 rad = (2.562) ·

!

180

" = 146.7918°

• Chapter 2 Trigonometric Functions

21

64. 261°15'45" =

!

261+ 15

60 +

45

3,600

"

# \$

%

& ' ° = 261.2625° = (261.2625) ·

!

"

66.

!

43"

!

43"

11 ·

!

180

" = 703.6364°

68. Since θ =

!

s

r , we have

!

"

2 =

!

s

2r =

!

5

70. Since θ =

!

180

" # s

r

\$

% &

'

( ) °, we have

!

"

2 =

!

180

2" # s

r

\$

% &

'

( ) ° =

!

180

2" # 34

15

\$

% &

'

( ) ° = 65°

72. s = rθ, where r = 52 km and θ = 2(27°) = 54°, s = (52)

!

(54°) " #

180°

\$

% &

'

( ) ≈ 49 km

74. θ = 2(0.25) = 0.5

r =

!

s

" =

!

26

0.5 = 52 cm

76. At 4:30, the minute hand has moved

!

1

2 of a circumference from its position at the top of the clock.

The hour hand has moved 4

!

1

2 twelfths of a circumference from the same position. Therefore, they

form an angle of

!

1

2 C –

!

4 1

2

where C = a total circumference or 2π radians. Thus, the desired angle is

!

1

2 (2π) –

!

4 1

2

12 (2π) = π –

!

9"

12 =

!

3"

12 =

!

"

78. We are to find a central angle θ, in a circle of radius 22 cm, when the arc length is 9.5 cm.

s =

!

"

180 rθ ; θ =

!

180s

"r =

!

180(9.5)

"(22) ≈ 25°

80. Since s = rθ we have

r =

!

s

" =

!

18

21° " # 180°

≈ 49 in

82. Since s = rθ , we have

s = (102 cm)

!

2"

3 ≈ 214 cm

84. Since s = rθ, we have

diameter = s = (5.000 km)

!

2.44° " #

180°

\$

% &

'

( ) ≈ 213 m

86. Since s =

!

"

180 rθ, we have

width of field ≈ s =

!

"

180 (865 ft)(2.5) ≈ 38 ft

• Exercise 2.1 Degrees and Radians

22

88. ! = 5