Chapter 2 Trigonometric Functions40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com/...Chapter 2...

Exercise 2.1 Degrees and Radians

18

Chapter 2 Trigonometric Functions

EXERCISE 2.1 Degrees and Radians

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 rad and

!

"

3 – 2π = –

!

5"

3 rad


19

16.


–

!

7"

4 + 2π =

!

"

4 rad and

–

!

7"

4 – 2π = –

!

15"

4 rad

18.


!

6"

5 + 2π =

!

16"

5 rad and

!

6"

5 – 2π = –

!

4"

5 rad

22. 1° corresponds to

!

"

180 rad; 5° corresponds to 5 ·

!

"

180 =

!

"

36 rad ≈ 0.08727 rad

24. 190° corresponds to 190 ·

!

"

180 rad =

!

19"

18 rad ≈ 3.316 rad

26. 108° corresponds to 108 ·

!

"

180 rad =

!

3"

5 rad ≈ 1.885 rad

28. 1 rad corresponds to

!

180

"

#

$ %

&

' ( °; 3 rad corresponds to

!

3 "180

#

$

% &

'

( ) ° =

!

540

"

#

$ %

&

' ( ° ≈ 171.9°

30. 1.4 rad corresponds to

!

(1.4)(180)

"

#

$ %

&

' ( ° =

!

252

"

#

$ %

&

' ( ° ≈ 80.21°

32.

!

2"

9 rad corresponds to

!

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°

36. Set calculator in radian mode. (A) 35° is 0.611 rad (B) 187° is 3.264 rad (C) 437° is 7.627 rad (D) –175.3° is –3.060 rad

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°


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 =

!

"

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

!

1

2θr =

!

"

180 ·

!

1

2θd

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

2r2θ (B) A =

!

1

2r2θ

=

!

1

2 (10.5)2 (0.150) ≈ 8.27 ft2 =

!

1

2 (10.5)2

!

15.0° "#

180°

$

% &

'

( ) ≈ 14.43 ft2

(C) A =

!

1

2r2θ (D) A =

!

1

2r2θ

=

!

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

lies in quadrant I.

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) ·

!

"

180 rad = 2.0418 rad

62. 2.562 rad = (2.562) ·

!

180

" = 146.7918°


21

64. 261°15'45" =

!

261+15

60+

45

3,600

"

# $

%

& ' ° = 261.2625° = (261.2625) ·

!

"

180 = 4.560 rad

66.

!

43"

11 rad =

!

43"

11 ·

!

180

" = 703.6364°

68. Since θ =

!

s

r, we have

!

"

2 =

!

s

2r =

!

5

2(4) = 0.625 rad

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

2C –

!

41

2

12C radians,

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

!

1

2 (2π) –

!

41

2

12 (2π) = π –

!

9"

12 =

!

3"

12 =

!

"

4 rad ≈ 0.79 rad

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 =

!

"

180rθ ; 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 =

!

"

180rθ, we have

width of field ≈ s =

!

"

180 (865 ft)(2.5) ≈ 38 ft


22

88. ! = 5 " 10 rad

– 7

r = 155 mis

Since s = rθ, we have width of object ≈ s = (155 mi)(5 × 10–7 rad) = 7.75 × 10–5 mi 7.75 × 10–5 mi = (7.75 × 10–5 mi)(1,609 m/mi) ≈ 0.125 m 7.75 × 10–5 mi = (0.125 m)(39.37 in) ≈ 4.921 in

90. Assuming that an angle corresponding to an entire circumference is swept out in 1 year (52 weeks), and that the amount swept out in 13 weeks is proportional to the time, we can write

!

angle

time =

!

angle

time

!

"

13 =

!

2"

52

θ =

!

13

52(2π) =

!

"

2rad ≈ 1.57 rad

92. We use the proportion error in distance

error in time = actual distance

actual time . Let x = error in distance. The actual time,

1 year = (365 days) 24hours

day

!

" #

$

% & 3,600

seconds

hour

! " # $

% & .

Thus,

!

x

365 seconds =

!

2"r

365 #24 #3,600 seconds

x = 365 ·

!

2"(6.7#107miles)

365 $24 $3,600

=

!

2"(6.7#107miles)

24 $3,600

≈ 4,900 miles

94. Since A = 1

2r2θ and P = s + 2r = rθ + 2r, we can eliminate θ between the two equations and write

2A = r2θ, θ =

!

2A

r2

.

P = r

!

2A

r2

"

# $

%

& ' + 2r =

!

2A

r + 2r .

Thus, P =

!

2(145.7)

8.4 + 2(8.4) ≈ 51 cm

96. (A) The radius of the pulley is 5 cm and u meters of rope is the same as an arc length of

s = u meters. Use θ =

!

s

r, where s and r are in the same unit.

(B) Yes. In the formula θ =

!

s

r, r changes when the diameter changes.

(C) Since s = 575 cm and r = 5 cm, then θ =

!

575

5 = 115 rad.


23

98. Since the two wheels are coupled together, the distance (arc length) that the drive wheel turns is equal to the distance that the shaft turns. Thus, s = r1θ1 s = r2θ2 r1θ1 = r2θ2

r2 =

!

"1

"2r1 =

!

7

3(12) = 28 mm

100. In view of the explanation in Problem 98 above,

θ1 =

!

r2

r1

θ2 =

!

30

12(12) = 30 rad.

102. The minimum distance is achieved when the smallest front gear and the largest rear gear are used.

For one complete revolution at the pedals, the large rear gear will have

!

20

32 revolution. Thus:

the minimum distance = (28π)

!

20

32

"

# $

%

& ' ≈ 55 in.

EXERCISE 2.2 Linear and Angular Velocity

6. V = rω = 125(0.07) = 8.75 mm/sec 8. V = rω = 0.567(145) = 82.215 km/hr

10. ω = Vr =

!

90

8 = 11.25 rad/sec 12. ω =

Vr =

!

355

125 = 2.85 rad/sec

14. ω =

!

"

t =

!

5"

22.0 = 0.714 rad/sec 16. ω =

!

"

t =

!

4.75

12.9 = 0.368 rad/sec

20. 500 revolutions per second = 500 · 2π rad/sec = 1,000π rad/sec

V = rω =

!

1

2"6

#

$ %

&

' ( (1,000π)cm/sec = 3,000π cm/sec

= 3,000π cm

sec·1

100

m

cm = 94 m/sec

22. ω = Vr =

700 cm/sec

32 cm ≈ 22 rad/sec

24. ω = Vr =

8.11! 106

5.00 ! 10–9 ≈ 1.62 × 1015 rad/sec

26. The earth rotates once every 23.93 hours or 2π radian in 23.93 hours. Thus,

ω = θt =

2!

23.93

rad

hr

V = rω = (3,964 mi)2!

23.93

rad

hr

" # $ %

& ' ≈ 1,041 mi/hr

Exercise 2.2 Linear and Angular Velocity

24

28. (A) The sun makes a full rotation, or 2π radian in 27 days, or 24.27 hours. Thus,

ω = θt =

2!

24.27 ≈ 0.0097 rad/sec

(B) Note that r = 12 × diameter. Thus,

V = rω =

!

1

2" 865,400miles

#

$ %

&

' ( (0.0097 rad/sec) ≈ 4,200 mph

30. The Neptune travels 1 revolution, or 2π radian, in 164 years, or 164 · 365 · 24 hours. Thus,

ω = θt =

!

2"

164 #365 # 24

!

rad

hr

V = rω = (2.795 × 109 mi)

!

2"

164 #365 #24

rad

hr

$

% &

'

( ) ≈ 12,200 mph

32. Using subscript M to denote quantities associated with the moon of Jupiter, and subscript J to denote quantities associated with Jupiter, we can write:

ωM =

!

"M

t ωJ =

!

"J

t θM = ωMt θJ = ωJt

Using the hint (given in Problem 31), 2π = θJ – θM , hence, 2π = ωJt – ωMt = (ωJ – ωM)t

t =

!

2"

#J –#M

Thus, t =

!

2"2"9.9–2"42.5

≈ 12.9 hr

(9 hr 55 min ≈ 9.9 hr; 42 hr 30 min = 42.5 hr)

34. (A) Using right-triangle trigonometry, c = 15 sec θ. Since 1 rps corresponds to an angular velocity of 2π rad/sec, at the end of t sec, θ = 2πt. Substituting the latter into the equation for c, we obtain c = 15 sec 2πt.

(B) The rate of change of the length of the light beam seems to be increasing as time increases from 0.00 to 0.24. When t = 0.25, c = 15 sec(π/2), which is not defined. The light has made one quarter turn and the spot is no longer on the wall.

TABLE 2 t sec 0.00 0.04 0.08 0.12 0.16 0.20 0.24 c ft 15.00 15.49 17.12 20.58 27.99 48.54 238.89

36. In view of the above formula (Problem 35), we have

!

(28")53

11

#

$ %

&

' ( (r)(60)

63,360 = 26.5,

where r is the number of revolutions per minute. Solving for r we obtain

r =

!

(26.5)(63,360)

(28")53

11

#

$ %

&

' ( (60)

≈ 66 rpm.


25

EXERCISE 2.3 Trigonometric Functions

2. Q(a, b) =

!

5

13,12

13

"

# $

%

& ' , r =

!

a2

+ b2 =

!

25

169+144

169 = 1, so

Q

!

5

13,12

13

"

# $

%

& ' is on the unit circle.

sin x =

!

12

13, cos x =

!

5

13, tan x =

!

12

5, cot x =

!

5

12, sec x =

!

13

5, csc x =

!

13

12.

4. Q

!

3

2,1

2

"

# $ $

%

& ' ' =

!

3

2,1

2

"

# $ $

%

& ' ' , r =

!

a2

+ b2 =

!

3

4+1

4 = 1, so Q

!

3

2,1

2

"

# $ $

%

& ' ' is on the unit circle.

sin x =

!

1

2, cos x =

!

3

2, tan x =

!

1

3

, cot x =

!

3 , sec x =

!

2

3

, csc x = 2.

6. Q

!

"1

2,"

1

2

#

$ %

&

' ( =

!

"1

2,"

1

2

#

$ %

&

' ( , r =

!

a2

+ b2 =

!

1

2+1

2 = 1, so Q

!

"1

2,"

1

2

#

$ %

&

' ( is on the unit

circle.

sin x = –

!

1

2

, cos x = –

!

1

2

, tan x = 1, cot x = 1, sec x = –

!

2 , csc x = –

!

2 .

8. P(–12, 9), r =

!

144 + 81 = 15 and Q

!

–4

5,3

5

"

# $

%

& ' . Thus

sin x =

!

3

5, cos x = –

!

4

5, tan x = –

!

3

4, cot x = –

!

4

3, sec x = –

!

5

4, csc x =

!

5

3.

10. P(–5, –12), r =

!

25+144 = 13; Q

!

–5

13,–12

13

"

# $

%

& ' . Thus,

sin x = –

!

12

13, cos x = –

!

5

13, tan x =

!

12

5, cot x =

!

5

12, sec x = –

!

13

5, csc x = –

!

13

12.

12. P(24, 7), r =

!

576+ 49 = 25; Q

!

24

25,7

25

"

# $

%

& ' . Thus,

sin x =

!

7

25, cos x =

!

24

25, tan x =

!

7

24, cot x =

!

24

7, sec x =

!

25

24, csc x =

!

25

7.

14. We sketch a reference triangle and label what we know.

Since cos θ =

!

a

r =

!

12

13, we know that a = 12 and r = 13.

Use the Pythagorean theorem to find b: 122 + b2 = 132 b2 = 169 – 144 = 25 b = 5 b is positive since P(a, b) is in quadrant I. We can now find the other five functions using Definition 1:

sin θ =

!

5

13, tan θ =

!

5

12, cot θ =

!

12

5, sec θ =

!

13

12, csc θ =

!

13

5.

a

b

5 10

5

10

P(a, b)terminal sideof !

12

Exercise 2.3 Trigonometric Functions

26

16. We sketch a reference triangle and label what we know.

Since sin θ =

!

b

r = –

!

4

5, we know that b = –4 and r = 5.

Use the Pythagorean theorem to find a: a2 + (–4)2 = 52 a2 = 25 – 16 = 9 a = –3 a is negative since P(a, b) is in quadrant III.

We can now find the other five functions using Definition 1:

cos θ = –

!

3

5, tan θ =

!

4

3, cot θ =

!

3

4, sec θ = –

!

5

3, csc θ = –

!

5

4.

terminal side

of !

b

a

–5 5

5

–5P(a, b)

18. cot x = 4 and x is a Quadrant I angle: a = 4, b = 1 and r =

!

(4)2

+ (1)2 =

!

17 , so

sin x =

!

b

r =

!

1

17

cos x =

!

a

r =

!

4

17

tan x =

!

b

a =

!

1

4

sec x =

!

r

a =

!

17

4

csc x =

!

r

b =

!

17

20. tan x = –

!

1

3 and x is a Quadrant IV angle: a = 3, b = –1 and r =

!

10 , so

sin x =

!

b

r = –

!

1

10

cos x =

!

a

r =

!

3

10

cot x =

!

a

b =

!

3

–1 = –3

sec x =

!

r

a =

!

10

3

csc x =

!

r

b =

!

10

"1 = –

!

10


27

22. csc x = –

!

25

7 and x is a Quadrant III angle: b = –7, r = 25, so a = –

!

(25)2" (7)

2 = –24, and

sin x =

!

b

r = –

!

"7

25

cos x =

!

a

r =

!

"24

25

tan x =

!

b

a =

!

"7

–24 =

!

7

24

cot x =

!

a

b =

!

"24

"7 =

!

24

7

sec x =

!

r

a =

!

25

"24 = –

!

25

24

24. sec x = –

!

2 and x is a Quadrant II angle: r =

!

2 , a = –1, so b =

!

2"1 = 1, and

sin x =

!

b

r =

!

1

2

cos x =

!

a

r =

!

"1

2

tan x =

!

b

a =

!

"1

1 = –1

cot x =

!

a

b =

!

1

"1 = –1

csc x =

!

r

b =

!

2

1 =

!

2

26. No. For all those values of θ for which both are defined, tan θ = 1/(cot θ), hence both are either positive or both are negative.

28. Radian mode: cos(7 rad) = 0.7539 30. Radian mode: cot (5 rad) = –0.2958

32. Degree mode: csc 129° =

!

1

sin129° = 1.28 34. Degree mode: sin 198° = –0.3090

36. Degree mode: tan 483° = –1.540 38. Radian mode: sec

!

39"

5 =

!

1

cos39"5

= 1.236

40. Radian mode: csc (–9) =

!

1

sin(–9) = –2.426 42. Degree mode: cos (–55°) = 0.5736

44. Degree mode: tan(–138°) = 0.9004

52. (a, b) = (1, 1), r =

!

a2

+ b2 =

!

12

+12 =

!

2

sin θ = br =

!

1

2

cos θ = ar =

!

1

2

tan θ = ba = 1

csc θ = rb =

!

2 sec θ = ra =

!

2 cot θ = ab = 1


28

54. (a, b) = (–1,

!

3 ), r =

!

a2

+ b2 =

!

(–1)2

+ ( 3)2 = 2

sin θ = br =

!

3

2 cos θ =

ar = –

12 tan θ =

ba = –

!

3

csc θ = rb =

!

2

3

sec θ = ra = –2 cot θ =

ab = –

!

1

3

56. I, II 58. I, III 60. I, II 62. II, III 64. II, IV 66. II and III

68. Given cos θ = –

!

1

2, cot θ > 0, then the terminal side of θ is in quadrant

III. We sketch the reference triangle and label what we know.

Since cos θ =

!

a

r =

!

"1

2, we know that a = –1 and r = 2.

Use the Pythagorean theorem to find b: (–1)2 + b2 = (2)2 b2 = 4 – 1 = 3 b = –

!

3 b is negative since P(a, b) is in quadrant III.

The other five functions are:

sin θ = –

!

3

2, tan θ =

!

3 , cot θ =

!

1

3

, sec θ = –2,

csc θ = –

!

2

3

.

3–

terminal sideof !

b

a

-1

P(a, b)

70. Given csc θ =

!

5 (or sin θ =

!

1

5

), cos θ < 0, then the terminal side of

θ is in quadrant II. We sketch the reference triangle and label what we know.

Since sin θ =

!

b

r =

!

1

5

, we know that b = 1 and r =

!

5 .

Use the Pythagorean theorem to find a: a2 + (1)2 = (

!

5 )2 a2 = 5 – 1 = 4 a = –2 a is negative since P(a, b) is in quadrant II.

terminal sideof !

b

a

–1

P(a, b)

–2 1 2

1

Now we have: cos θ = –

!

2

5

, tan θ = –

!

1

2, cot θ = –2, sec θ = –

!

5

2.


29

72. tan x = –

!

1

2 and cos x > 0, so x is a Quadrant IV angle: a = 2, b = –1 and r =

!

5 , so

sin x =

!

b

r =

!

–1

5

cos x =

!

a

r =

!

2

5

cot x =

!

a

b =

!

2

"1 = –2

sec x =

!

r

a =

!

5

2

csc x =

!

r

b =

!

5

"1 = –

!

5

74. No. For example cos

!

"

3

#

$ %

&

' ( =

1

2 = cos

!

–"

3

#

$ %

&

' ( , but α =

!

3 and β = –

!

3 are not coterminal.

76. Use the reciprocal identity sec x =

!

1

cos x. sec x =

!

1

–0.41615 = –2.40298

78. Degree mode: sin 37.85° = 0.6136

80. Use the reciprocal relation sec θ =

!

1

cos". Degree mode: sec 107.53° =

!

1

cos 107.53° = –3.320

82. Radian mode: tan 4.738 = –39.04

84. Use the reciprocal relation csc θ =

!

1

sin". Radian mode: csc(–0.408) =

!

1

sin (–0.408) = –2.520

86. Degree mode: cot 352°5'55" = cot

!

352+5

60+

55

3,600

"

# $

%

& ' ° = –7.205

88. Degree mode: cos(–432.18°) = 0.3060

90. Degree mode: sin 605°42'75" = sin

!

605+42

60+

75

3,600

"

# $

%

& ' ° = –0.9116

92. Use the reciprocal relation sec θ =

!

1

cos". Radian mode: sec 3.55 =

!

1

cos 3.55 = –1.090

94. Degree mode: cot(–567.43°) = –1.927 96. Cotangent and cosecant. Since cot x = a/b and csc x = r/b, neither is defined when the terminal side

of the angle lies along the positive or negative horizontal axis, because b will be 0, and division by zero is not defined.

98. (A) Since θ = s

r, and s = 10, and r, the measure of CA, is 4, we have θ =

10

4 = 2.5 rad.

(B) Since cos θ = a

r and sin θ =

b

r, we have

a = r cos θ = 4 cos 2.5 b = r sin θ = 4 sin 2.5 Thus, (a, b) = (4 cos 2.5, 4 sin 2.5) = (–3.20, 2.39)


30

100. (A) Since θ = s

r, and s = 4, and r, the measure of CA, is 1, we have θ =

4

1 = 4 rad.

(B) Since cos θ = a

r and sin θ =

b

r, we have

a = r cos θ = 1 cos 4 b = r sin θ = 1 sin 4 Thus, (a, b) = (1 cos 4, 1 sin 4) = (–0.654, –0.757)

102. From the figure, we note: s = rθ

r =

!

a2

+ b2 =

!

32

+ 42 =

!

25 = 5

tan θ = b

a = 4

3

θ = tan–1 4

3

Thus, s = 5 tan–1 4

3 ≈ 4.64 units (calculator in radian mode).

b

a

5

4(3,4)

s

3

104. We sketch the reference triangle and label what we know.

Since cot θ =

!

a

b =

!

1

2 and since the terminal side of θ is in

quadrant III, we know that a = –1 and b = –2. Use the Pythagorean theorem to find r: r2 = a2 + b2 = (–1)2 + (–2)2 = 1 + 4 = 5 or r =

!

5

terminal sideof !

b

a

–1

P(a, b)

–2

Now we have:

sin θ = –

!

2

5

, cos θ = –

!

1

5

, tan θ = 2, sec θ = –

!

5 , csc θ = –

!

5

2.

106. If θ = 60°, I = k cos 60° = k(0.50) = 0.50k. Thus, at the angle θ = 60° the light intensity I will be 50% of the vertical intensity.

108. If θ = 8° (summer solstice), E = k cos θ = k cos 8° = 0.99k. If θ = 55° (winter solstice), E = k cos θ = k cos 55° = 0.57 k.

110. (A) If n = 6, A = n tan180

n

! " # $

% &

° = 6 tan

180

6

! " # $

% &

° = 6 tan 30° = 3.46410.

If n = 10, A = 10 tan180

10

! " # $

% &

° = 10 tan 18° = 3.24920.

If n = 100, A = 100 tan180

100

! " # $

% &

° = 100 tan 1.8° = 3.14263.

If n = 1,000, A = 1,000 tan180

1, 000

!

" #

$

% &

° = 1,000 tan 0.18° = 3.14160.

If n = 10,000, A = 10,000 tan180

10, 000

!

" #

$

% &

° = 10,000 tan 0.018° = 3.14159.

n 6 10 100 1000 10,000 An 3.46410 3.24920 3.14263 3.14160 3.14159


31

(B) The area of the circle is A = πr2 = π(1)2 = π, and An seem to approach π, the area of the circle, as n increases.

(C) No. An n sided polygon is always a polygon, no matter the size of n, but circumscribed polygon can be made as close to the circle as you like by taking n sufficiently large.

112. From Problem 111, x = cos 20πt =

!

25– (sin20"t)2

Now for t = 0, we have

x = cos 0 +

!

25– (sin0)2 = 1 + 5 = 6

For t = 0.01 sec, we have

x = cos 0.2π +

!

25– (sin0.2")2

= 0.81 + 4.96 = 5.77

114. From Problem 113, I = 35 sin(48πt – 12π) Now for t = 0.310 sec, we have I = 35 sin(14.88π – 12π) = 35 sin(2.88π) ≈ 13 amp

116. (A) If the angle of inclination θ = 89.2°, then m = tan θ = tan 89.2° = 71.62 If the angle of inclination θ = 179°, then m = tan θ = tan 179° = –0.02

(B) If the angle of inclination θ = 101°, then m = tan θ = tan 101° Then the equation of the line is given by y + 4 = tan 101° (x – 7) y = tan 101° (x – 7) – 4 y = x tan 101° – 7 tan 101° – 4 = –5.14x + 32.01

EXERCISE 2.4 Additional Applications

10. Use

!

n2

n1

=

!

sin"

sin#, 12. Use

!

n2

n1

=

!

sin"

sin#,

where n2 = 1.33, n1 = 1.00, and α = 34.2° where n2 = 1.66, n1 = 1.33, and α = 45.0°

Solve for β:

!

1.33

1.00 =

sin34.2°

sin ! Solve for β:

!

1.66

1.33 =

sin 45.0°

sin!

sin β = sin34.2°

1.33 sin β =

1.33 sin 45.0°

1.66

β = sin–1 sin 34.2°

1.33

!

" # #

$

% & & β = sin–1 1.33 sin 45.0°

1.66

!

" # $

% &

= 25° = 34.5° 14. The index of refraction for flint glass is n1 = 1.66 and that for air is 1.00. Find the angle of incidence

α such that the angle of refraction β is 90°.

!

sin"

sin# =

!

n2

n1

; sin α = 1.00

1.66sin 90°; α = sin–1 1.00

1.66(1)

! " # $

% & = 37°

Exercise 2.4 Additional Applications

32

16. A diver looking straight up towards the surface receives a cone of light rays, since all of the light rays above the surface are refracted downward. Outside of the cone will be dark. Within the cone the driver will see what is above the surface in all directions, including boats and buoys on the water, people standing on the shore or dock, and so on. The cone angle θ is about 49°.

18. Use n2n1

= sin αsin β ,

where n1 = 1.00. α = 90° – 43° = 47°

sin β =

!

7.2

r =

!

7.2

7.22

+122

!

"r

43°

12 in

7.2 in

Solve for n2:

!

n2

1.00= sin 47° ÷

!

7.2

7.22

+122

≈ 1.4

22. We use sin

!

"

2 =

!

Sw

Sb

, where θ = 54° and Sb = 55 km/hr. Then we solve for Sw:

sin

!

54°

2 =

!

Sw

55; Sw = 55 sin 27° = 25 km/hr

24. We use sin

!

"

2 =

!

Ss

Sa

, where Sa = Mach 3.2 = 3.2Ss . Then we solve for θ:

sin

!

"

2 =

!

Ss

3.2Ss

; sin

!

"

2 =

!

1

3.2. Thus

!

"

2 = 18°, θ = 36°.

26. We use sin

!

"

2 =

!

S1

Sp, where θ = 120° and S1 = 2 × 1010 cm/sec. Then we solve for Sp:

sin

!

120°

2 =

!

2"1010

Sp; Sp =

!

2"1010

sin 63° = 2.24 × 1010 cm/sec

28. d = a + b sin 4θ = –1.8 + (–4.2) sin(4 · 40°) = –1.8 – 4.2 sin 160° = –3°

30. Let x =

!

OA , then r2 = (x + 42)2 + (48)2 r = 192' = 2,304" and (x + 42) =

!

(2,304)2– (48)

2 = 2,303.50" or x = 2,303.50 – 42 = 2,261.50"

We observe that:

!

2,261.50

2,303.50 =

!

40.25+t

2

48 =

!

90.5+ t

96

Solving for t we obtain:

t =

!

(2,261.50)(96) – (2,303.50)(90.5)

2,303.50 ≈ 3.75 in.

48" 96"

40.25"

t

2

48"

40.25"

t

2

42"

x

r

A

O


33

EXERCISE 2.5 Exact Value for Special Angles and Real Numbers

Note for Problems 6—16: The reference angle α is the angle (always taken positive) between the terminal side of θ and the horizontal axis.

6. α = θ = 45°

45°

!

P

8. α = |–45°| = 45°

-45°

!

P

10. α = |–π/4| = π/4

!

P

-!4

12. α = π –

!

5"

6 =

!

"

6

!

P

5!6

14. α = 180° – 150° = 30°

-150°

!

P

16. α = 2π –

!

5"

3 =

!

"

3

!

P

-5!3

18. (a, b) = (1, 0), r = 1

cos 0° =

!

a

r = 1

1 = 1

a

b

20. (a, b) = (1, 0), r = 1

cot 0 =

!

a

b which is not defined

since b = 0

a

b

22. Use the special 30°–60° triangle as the reference triangle. Use the sides of the reference triangle to determine P(a, b) and r. Then use Definition 1.

a

b

30°

2

( , 1)

3

3

(a, b) = (

!

3 , 1), r = 2

sin 30° =

!

b

r =

!

1

2

Exercise 2.5 Exact Value for Special Angles and Real Numbers

34

24. Use the special 30°–60° triangle as the reference triangle. Use the sides of the reference triangle to determine P(a, b). Then use Definition 1.

a

b

30°

2

( , 1)

3

3

1

(a, b) = (

!

3 , 1), r = 2

tan 30° =

!

b

a =

!

1

3

or

!

3

3

26. Locate the 30°–60° reference triangle, determine P(a, b) and r, then evaluate.

3

a

b

1

2

3(1, )

!/3

(a, b) = (1,

!

3 ), r = 2

csc

!

"

3 =

!

r

b =

!

2

3

or

!

2 3

3

28. (a, b) = (0, 1), r = 1

cos

!

–3"

2

#

$ %

&

' ( = cos

!

"

2

#

$ %

&

' (

=

!

a

r =

!

0

1 = 0

a

b

!/2

–3!/2

30. In view of Problem 28, tan

!

"

2 =

!

b

a =

!

1

0 which is not defined.

32. Locate the 30°–60° reference triangle, determine (a, b) and r, then evaluate.

3

a

b

1

3 (1, – )

–2

sin

!

–"

3

#

$ %

&

' ( =

!

b

r = –

!

3

2


a

b

2–1

3

-30°

cot(–30°) =

!

a

b =

!

3

–1 = –

!

3

36. Locate the 45° reference triangle, determine (a, b) and r, then evaluate.

a

b

– 1

45°

1

2

(1, –1)

7!/4

cos

!

7"

4 =

!

a

r =

!

1

2

or

!

2

2


a

b

2– 1

3

-30°

3( , –1)

cot

!

11"

6 =

!

a

b =

!

3

–1 = –

!

3

40. (a, b) = (1, 0), r = 1

sin(–2π) = b

r = 0

1 = 0

csc(–2π) = r

b = 1

0

which is not defined.


a

b

2

1

60°

30°

3

3(1, )

–300°

sin(–300°) = sin 60° =

!

b

r =

!

3

2


35


360°

a

b

-1

2

480°

3(–1, )

tan 480° =

!

3

–1 = –

!

3


30°a

b

-1

–510°

3–

2

sec(–510°) =

!

r

a =

!

2

– 3

= –

!

2

3

or –

!

2 3

3

48. The cotangent function is not defined at θ = 0, π, and 2π, because cot θ = a/b and b = 0 for any point on the horizontal axis.

50. The secant function is not defined at θ = π/2 and 3π/2, because sec θ = r/a and a = 0 for any point on the vertical axis.

52. It is shown that cos(90°) = 0 and sin

!

"

6

#

$ %

&

' ( =

!

1

2 = 0.5000, so that the

calculator displays exact values of these. Thus sin

!

2"

3

#

$ %

&

' ( is not

given exactly. To find sin

!

2"

3

#

$ %

&

' ( , locate the 30–60° reference

triangle, determine (a, b) and r, then evaluate.

b

a–1

3 2!/3

2

60°

30°

sin

!

2"

3

#

$ %

&

' ( =

!

3

2

54. It is shown that sin(–150°) = –

!

1

2 = –0.5000 and

tan

!

5"

4

#

$ %

&

' ( = 1.000, so that the calculator displays exact

values for these. Thus tan(–150°) is not given exactly. To find tan(–150°), locate the 30°–60° reference triangle, determine (a, b) and r, then evaluate.

30°a

b

-1

–150°

3–

2

tan(–150°) =

!

a

b =

!

–1

– 3

=

!

1

3

or

!

3

3

56. Draw a reference triangle in the first quadrant with side adjacent reference angle 1 and hypotenuse

!

2 . Observe that this is a special 45° triangle.

a

b

12

1

!

(A) θ = 45° (B) θ =

!

"

4

58. Draw a reference triangle in the third quadrant with side opposite reference angle –1 and hypotenuse 2. Observe that this is a special 30°–60° triangle.

30°a

b

-1

3–

2

!

(A) θ = 210° (B) θ =

!

7"

6

Exercise 2.5 Exact Value for Special Angles and Real Numbers

36

60. Draw a reference triangle in the second quadrant with side adjacent reference angle –1 and side opposite 1. Observe that this is a special 45° triangle.

(A) θ = 135° (B) θ =

!

3"

4

a

b

1

2

–1

!

64. All should equal –0.0379, because the cosine function is periodic with period 2π.

66. Calculator in radian mode: (A) cot(–1) = –0.64 (B) cot 8.7 = –1.1 (C) cot(–12.64) = 14

!

cos(–1)

sin(–1) = –0.64

!

cos 8.7

sin 8.7 = –1.1

!

cos(–12.64)

sin(–12.64) = 14

68. (A) cos 5 = 0.28 (B) cos(–13.4) = 0.67 (C) cos(–1,003) = –0.67 cos(–5) = 0.28 cos(13.4) = 0.67 cos(1,003) = –0.67

70. (A) 1 – sin2 14 = 0.02 (B) 1 – sin2 (–16.3) = 0.69 (C) 1 – sin2 766 = 0.73 cos2 14 = 0.02 cos2 (–16.3) = 0.69 cos2 766 = 0.73

72. cos x sec x = cos x

!

1

cos x

"

# $

%

& ' = 1 Use Identity (2)

74. tan x csc x =

!

sin x

cos x ·

!

1

sin x =

!

1

cos x = sec x Use Identities (4) and (1)

76.

!

cos x

1 – sin2x

=

!

cos x

cos2x

=

!

1

cos x = sec x Use Identity (9)

78. tan(–x) cos(–x) =

!

sin(–x)

cos(–x) · cos(–x) = sin(–x) = –sin x Use Identity (4)

80. We can draw a reference triangle in the second quadrant with side opposite reference angle

!

3 and side adjacent –1. We can also draw a reference triangle in the fourth quadrant with side opposite reference angle –

!

3 and side adjacent 1. Each triangle is a special 30°–60° triangle.

a

b

3

3–

1

60°

60°

θ = 120° or θ = 300°

82. We can draw a reference triangle in the second quadrant with side adjacent reference angle –

!

3 and hypotenuse 2. We can also draw a reference triangle in the third quadrant with side adjacent –

!

3 and hypotenuse 2. Each triangle is a special 30°–60° triangle.

1

30°a

b

–1

3–

2

2

θ =

!

5"

6 or θ =

!

7"

6


37

84. For x =

!

"

4, sin x =

!

1

2

, which is the least positive x in radon measure for which sin x =

!

1

2

.

86. For x =

!

2"

3, tan x = –

!

3 ; x =

!

2"

3 is the least positive x.

88. (A) Since

!

4

x = cos 30° and cos 30° =

!

3

2;

!

4

x =

!

3

2; x =

!

8

3

Since

!

y

4 = tan 30° and tan 30° =

!

3

3;

!

y

4 =

!

3

3; y =

!

4

3

(B) Since

!

5

y = tan 45° and tan 45° = 1;

!

5

y = 1; y = 5

Since

!

5

x = sin 45° and sin 45° =

!

2

2;

!

5

x =

!

2

2; x = 5

!

2

(C) Since

!

x

3 = cos 60° and cos 60° =

!

1

2;

!

x

3 =

!

1

2; x =

!

3

2

Since

!

y

3 = sin 60° and sin 60° =

!

3

2;

!

y

3 =

!

3

2; y =

!

3 3

2

90. Since (a, b) is on a unit circle with (a, b) = (0.87096774, 0.4913402), we can solve cos s = 0.87096774 or sin s = 0.4913402. Then s = cos–1(0.87096774) = 0.514 or s = sin–1(0.4913402) = 0.514.

92. (A) Identity (5) (B) Identity (9) (C) Identity (1)

94. 2π or 360°

96. g(x) = 4 cos

!

x

3

"

# $ %

& ' is periodic because cosine function is periodic. If T is the period of g(x), then we

should have: g(x) = g(x + T)

g(x + T) = 4 cos

!

x +T

3

"

# $

%

& ' = 4 cos

!

x

3+T

3

"

# $

%

& '

= 4

!

cosx

3

"

# $

%

& ' cos

T

3

"

# $

%

& ' ( sin

x

3

"

# $

%

& ' sin

T

3

"

# $

%

& '

) * +

, - .

The right-hand-side will be equal to g(x) = 4 cos

!

x

3

"

# $ %

& ' if cos

!

T

3

"

# $

%

& ' = 0 and sin

!

T

3

"

# $

%

& ' = 0 .

Thus, T = 6π is the period.

98. k(x) = 2x cos x is not periodic since cos x is periodic but x is not. The product of two terms (cos x and x) is periodic if both factors are periodic and with the same period.

100. S1 = 0.5 S2 = S1 + cos S1 = 0.5 + cos 0.5 = 1.377583 S3 = S2 + cos S2 = 1.377583 + cos 1.377583 = 1.569596 S4 = S3 + cos S3 = 1.569596 + cos 1.569596 = 1.570796 S5 = S4 + cos S4 = 1.570796 + cos 1.570796 = 1.570796

!

"

2 = 1.570796

Chapter 2 Trigonometric Functions40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com/...Chapter 2...

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Transcript of Chapter 2 Trigonometric Functions40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com/...Chapter 2...