4-3 Trigonometric Functions on the Unit Circle · CAROUSEL Zoe is on a carousel at the carnival. The diameter of the carousel is 80 feet. Find the position of her seat from the center

The given point lies on the terminal side of an angle θ in standard position. Find the values of the six trigonometric functions of θ.

1. (3, 4)

SOLUTION: Use the values of x and y to find r.

Use x = 3, y = 4, and r = 5 to write the six trigonometric ratios.

2. (−6, 6)


Use x = , y = 6, and r = to write the six trigonometric ratios.

3. (−4, −3)


Use x = , y = , and r = 5 to write the six trigonometric ratios.

4. (2, 0)


Use x = 2, y =0, and r = 2 to write the six trigonometric ratios.

5. (1, −8)


Use x = 1, y = , and r = to write the six trigonometric ratios.

6. (5, −3)



7. (−8, 15)


Use x = , y = 15, and r = 17 to write the six trigonometric ratios.

8. (−1, −2)


Use x = , y = , and r = to write the six trigonometric ratios.

Find the exact value of each trigonometric function, if defined. If not defined, write undefined.

9. sin

SOLUTION:

The terminal side of in standard position lies on the positive y-axis. Choose a point P(0, 1) on the terminal side

of the angle because r = 1.

10. tan 2π

SOLUTION:

The terminal side of in standard position lies on the positive x-axis. Choose a point P(1, 0) on the terminal side of the angle because r = 1.

11. cot (–180°)

SOLUTION:

The terminal side of in standard position lies on the negative x-axis. Choose a point P( , 0) on the terminalside of the angle because r = 1.

12. csc 270°

SOLUTION:

The terminal side of in standard position lies on the negative y-axis. Choose a point P(0, ) on the terminal side of the angle because r = 1.

13. cos (–270°)

SOLUTION:

The terminal side of in standard position lies on the positive y-axis. Choose a point P(0, 1) on the terminal side of the angle because r = 1.

14. sec 180°

SOLUTION:

The terminal side of in standard position lies on the negative x-axis. Choose a point P( , 0) on the terminal side of the angle because r = 1.

15. tan π

SOLUTION:

The terminal side of π in standard position lies on the negative x-axis. Choose a point P( , 0) on the terminal side of the angle because r = 1.

16.

SOLUTION:

The terminal side of − in standard position lies on the negative y-axis. Choose a point P(0, ) on the terminal

side of the angle because r = 1.

Sketch each angle. Then find its reference angle.17. 135°

SOLUTION:

The terminal side of 135º lies in Quadrant II. Therefore, its reference angle is θ ' = 180º – 135º or 45º.

18. 210°

SOLUTION:

The terminal side of 210º lies in Quadrant III. Therefore, its reference angle is θ ' = 210º – 180º or 30º.

19.

SOLUTION:

The terminal side of lies in Quadrant II. Therefore, its reference angle is θ ' = .

20.

SOLUTION:

A coterminal angle is − 2π or , which lies in Quadrant IV. So, the reference angle is θ ' is 2π − or

.

21. −405°

SOLUTION:

A coterminal angle is −405° + 360(2)° or 315°. The terminal side of 315° lies in Quadrant IV, so its reference angleis 360º – 315º or 45º.

22. −75°

SOLUTION:

A coterminal angle is −75° + 360° or 285°. The terminal side of 285° lies in Quadrant IV, so its reference angle is 360° − 285° or 75°.

23.

SOLUTION:


24.

SOLUTION:

A coterminal angle is + 2(−1)π or The terminal side of lies in Quadrant I, so the reference angle is

Find the exact value of each expression.

25. cos

SOLUTION:

Because the terminal side of θ lies in Quadrant III, the reference angle θ ' is – π or . In Quadrant III, cos

θ is negative.

26. tan

SOLUTION:

Because the terminal side of θ lies in Quadrant III, the reference angle θ ' is or . In Quadrant III, tan θ

is positive.

27. sin

SOLUTION:

Because the terminal side of θ lies in Quadrant II, the reference angle θ ' is or . In Quadrant II, sin θ is

positive.

28. cot (−45°)

SOLUTION:

A coterminal angle is −45° + 360° or 315°. Because the terminal side of 315° lies in Quadrant IV, the reference angle θ ' is 360° − 315° or 45°. Because tangent and cotangent are reciprocal functions and tan θ is negative in Quadrant IV, it follows that cot θ is also negative in Quadrant IV.

29. csc 390°

SOLUTION:

A coterminal angle is 390° + 360° or 30°, which lies in Quadrant I. So, the reference angle θ ' is 360° − 30° or 30°. Because sine and cosecant are reciprocal functions and sin θ is positive in Quadrant I, it follows that csc θ is also positive in Quadrant I.

30. sec (−150°)

SOLUTION:

A coterminal angle is −150° + 360° or 210°, which lies in Quadrant III. Because the terminal side of θ lies in Quadrant III. , the reference angle θ ' is 210º – 180º or 30º. Because secant and cosine are reciprocal functions and cos θ is negative in Quadrant III, it follows that sec θ is also negative in Quadrant III.

31. tan

SOLUTION:

Because the terminal side of θ lies in Quadrant IV, the reference angle θ ' is or . In Quadrant IV, tan

θ is negative.

32. sin 300°

SOLUTION:

Because the terminal side of θ lies in Quadrant IV, the reference angle θ ' is or . In Quadrant IV,sin θ is negative.

Find the exact values of the five remaining trigonometric functions of θ.33. tan θ = 2, where sin θ > 0 and cos θ > 0

SOLUTION:

To find the other function values, you must find the coordinates of a point on the terminal side of θ. You know that sin θ and cos θ are positive, so θ must lie in Quadrant I. This means that both x and y are positive.

Because tan θ = or , use the point (1, 2) to find r.

Use x = 1, y = 2, and r = to write the five remaining trigonometric ratios.

34. csc θ = 2, where sin θ > 0 and cos θ < 0

SOLUTION:

To find the other function values, you must find the coordinates of a point on the terminal side of θ. You know that sin θ is positive cos θ are negative, so θ must lie in Quadrant II. This means that x is negative and y is positive.

Because csc θ = or , use the point (x, 1) and r = 2 to find x.

Use x = 1, y = 1, and r = 2to write the five remaining trigonometric ratios.

35.

SOLUTION:

To find the other function values, you must find the coordinates of a point on the terminal side of θ. You know that sin θ is negative and cos θ is positive, so θ must lie in Quadrant IV. This means that x is positive and y is negative.

Because sin θ = or , use the point (x, ) and r= 5 to find x.

Use x = , y = , and r = 5 to write the five remaining trigonometric ratios.

36.

SOLUTION:

To find the other function values, you must find the coordinates of a point on the terminal side of θ. You know that sin θ and cos θ are negative, so θ must lie in Quadrant III. This means that both x and y are negative.

Because cos θ = or , use the point ( , y) and r = 13 to find y .


37. sec θ = , where sin θ < 0 and cos θ > 0

SOLUTION:


Because sec θ = or , use the point (1, y) and r = to find y .

Use x = 1, y = , and r = to write the five remaining trigonometric ratios.

38. cot θ = 1, where sin θ < 0 and cos θ < 0

SOLUTION:


Because cot θ = or , use the point ( , ) to find r.

Use x = , y = , and r = to write the five remaining trigonometric ratios.

39. tan θ = −1, where sin θ < 0

SOLUTION:


Because tan θ = or , use the point ( , ) to find r.


40.

SOLUTION:

To find the other function values, you must find the coordinates of a point on the terminal side of θ. You know that sin θ is positive and cos θ is negative, so θ must lie in Quadrant II. This means that x is negative and y is positive.

Because cos θ = or , use the point ( , y) and r = 2 to find y. .


41. CAROUSEL Zoe is on a carousel at the carnival. The diameter of the carousel is 80 feet. Find the position of her seat from the center of the carousel after a rotation of 210º.

SOLUTION:

Let the center of the carousel represent the origin on the coordinate plane and Zoe’s position after the 210° rotationhave coordinates (x, y). The definitions of sine and cosine can then be used to find the values of x and y . The value

of r is 80 ÷ 2 or 40. The seat rotates 210º, so the reference angle is 210º – 180º or 30º. Because the final position of the seat corresponds to Quadrant III, the sine and cosine of 210° are negative.

Therefore, the position of her seat relative to the center of the carousel is or (–34.6, –20).

42. COIN FUNNEL A coin is dropped into a funnel where it spins in smaller circles until it drops into the bottom of the bank. The diameter of the first circle the coin makes is 24 centimeters. Before completing one full circle, the

coin travels 150° and falls over. What is the new position of the coin relative to the center of the funnel?

SOLUTION: Let the center of the funnel represent the origin on the coordinate plane and the final position of the coin have coordinates (x, y). The definitions of sine and cosine can then be used to find the values of x and y . The value of r, 12 cm, is the length of the radius of the first circle. The coin travels through an angle of 150º, so the reference angle is 180º – 150º or 30º. Since the final position of the coin corresponds to Quadrant II, the cosine of 150º is negative and the sine of 150º is positive.

Therefore, the coordinates of the final position of the coin are or about (–10.4, 6).

Find the exact value of each expression. If undefined, write undefined.

43. sec 120°

SOLUTION:

120º corresponds to the point (x, y) = on the unit circle.

44. sin 315°

SOLUTION:


45. cos

SOLUTION:

46.

SOLUTION:

47. csc 390°

SOLUTION:

48. cot 510°

SOLUTION:

49. csc 5400°

SOLUTION:

There

csc 5400° is undefined.

50. sec

SOLUTION:

corresponds to the point (x, y) = on the unit circle.

Therefore, sec is undefined.

51.

SOLUTION:

52. csc

SOLUTION:

53. tan

SOLUTION:


54. sec

SOLUTION:


55.

SOLUTION:

56. cos

SOLUTION:


57. tan

SOLUTION:

58.

SOLUTION:

59. RIDES Jae and Anya are on a ride at an amusement park. After the first several swings, the angle the ride makes with the vertical is modeled by θ = 22 cos t, with θ measured in radians and t measured in seconds. Determine the measure of the angle in radians for t = 0, 0.5, 1, 1.5, 2, and 2.5.

SOLUTION: Use the unit circle to find each angle measure. t = 0

t = 0.5

t = 1

t = 1.5

t = 2

t = 2.5

The times and corresponding angle measures are shown in the table below.

Complete each trigonometric expression.60. cos 60° = sin ___

SOLUTION:

60° corresponds to the point (x, y) = on the unit circle. So, cos 60° = .

On the unit circle, sin 30° = and sin 150° = . Therefore, cos 60° = sin 30° or cos 60° = sin 150° .

61. tan = sin ___

SOLUTION:

corresponds to the point (x, y) = on the unit circle. So, tan = or 1.

On the unit circle, sin = 1. Therefore, tan = sin .

62. sin = cos ___

SOLUTION:

corresponds to the point (x, y) = on the unit circle. So, sin =

On the unit circle, cos = and cos = . Therefore, sin = cos or sin = cos .

63. cos = sin ___

SOLUTION:

corresponds to the point (x, y) = on the unit circle. So, cos =

On the unit circle, sin = and sin = . Therefore, cos = sin or cos = sin .

64. sin (−45°) = cos ___

SOLUTION:

Rewrite −45° as the sum of a number and an integer multiple of 360º. –45º + 360(1)º = 315º

315º corresponds to the point (x, y) = on the unit circle. So, sin (–45º) =

On the unit circle, cos 135º = and cos 225º = Therefore, sin (−45°) = cos 135º or sin (−45°) = cos

225º.

65. cos = sin ___

SOLUTION:


On the unit circle, sin = and sin = Therefore, cos = sin or cos = sin .

66. ICE CREAM The monthly sales in thousands of dollars for Fiona’s Fine Ice Cream shop can be modeled by

, where t = 1 represents January, t = 2 represents February, and so on.

a. Estimate the sales for January, March, July, and October. b. Describe why the ice cream shop’s sales can be represented by a trigonometric function.

SOLUTION: a. January corresponds to t = 1.

March corresponds to t = 3.

July corresponds to t = 7.

October corresponds to t = 10.

b. Sample answer: The ice cream shop’s sales can be represented by a trigonometric function because people eat more ice cream in the summer and less in the winter.

Use the given values to evaluate the trigonometric functions.

67. cos (−θ) = ; cos θ = ?; sec θ = ?

SOLUTION:

Because cos (–θ) = and cos (–θ) = cos θ, cos θ = . So, sec θ = or .

68. sin (−θ) = ; sin θ = ?; csc θ = ?

SOLUTION:

Because sin(–θ) = and sin(–θ) = −sin θ, −sin θ = or sin θ = − . So, csc θ = or – .

69. sec θ = ; cos θ = ?; cos (−θ) = ?

SOLUTION:

Because sec θ = and cos θ = , cos θ = . Because cos (–θ) = cos θ and cos θ = , cos (–θ) = .

70. csc θ = ; sin θ = ?; sin (−θ) = ?

SOLUTION:

Because csc θ = and sin θ = , sin θ = . Because sin (–θ) = −sin θ and sin θ = , sin (–θ) = − .

71. GRAPHS Suppose the terminal side of an angle θ in standard position coincides with the graph of y = 2x in Quadrant III. Find the six trigonometric functions of θ.

SOLUTION: Graph y = 2x.

One point that lies on the line in Quadrant III is (−2, −4). So, x = −2 and y = −4. Find r.

Use x = −2, y = −4, and r = to write the six trigonometric ratios.

Find the coordinates of P for each circle with the given radius and angle measure.

72.

SOLUTION: Use the definitions of the cosine and sine functions to find the values of x and y . Because P is in Quadrant IV, the

cosine of is positive and the sine of is negative. The reference angle for is 2π − or and the

radius r is 3.

So, the coordinates of P are .

73.

SOLUTION: Use the definitions of the cosine and sine functions to find the values of x and y . Because P is in Quadrant II, the

cosine of is negative and the sine of is positive. The reference angle for is π − or and the

radius r is 5.


74.

SOLUTION: Use the definitions of the cosine and sine functions to find the values of x and y . Because P is in Quadrant I, the

cosine and sine of are positive. The radius r is 6.

So, the coordinates of P are

75.

SOLUTION: Use the definitions of the cosine and sine functions to find the values of x and y . Because P is in Quadrant III, the

cosine and sine of are negative. The reference angle for is − π or and the radius r is 8.

So, the coordinates of P are (−4, −4 ).

76. COMPARISON Suppose the terminal side of an angle θ1 in standard position contains the point (7, −8), and the

terminal side of a second angle θ2 in standard position contains the point (−7, 8). Compare the sines of θ1 and θ2.

SOLUTION:

An angle in standard position that contains the point (7, −8) lies in Quadrant IV. Use x = 7 and y = −8 to find r.

So,

An angle in standard position that contains the point (−7, 8) lies in Quadrant II. Use x = −7 and y = 8 to find r.

So,

Therefore, sin θ1 = –sinθ2.

77. TIDES The depth y in meters of the tide on a beach varies as a sine function of x, the hour of the day. On a

certain day, that function was where x = 0, 1, 2, …, 24 corresponds to 12:00 midnight, 1:00

A.M., 2:00 A.M., …, 12:00 midnight the next night. a. What is the maximum depth, or high tide, that day? b. At what time(s) does the high tide occur?

SOLUTION:

a. Evaluate for x = 0, 1, 2, …, 24.

Therefore, the maximum depth, or high tide, that day was 11 meters. b. The high tides occurred when x = 7 and x = 19. Because x = 0 corresponds to midnight, x = 7 corresponds to 7 A.M. and x = 19 corresponds to 7:00 P.M.

Time Depth (m) Time Depth (m) 0 5.4 13 5 1 5 14 5.4 2 5.4 15 6.5 3 6.5 16 8 4 8 17 9.5 5 9.5 18 10.6 6 10.6 19 11 7 11 20 10.6 8 10.6 21 9.5 9 9.5 22 8

10 8 23 6.5 11 6.5 24 5.4 12 5.4

78. MULTIPLE REPRESENTATIONS In this problem, you will investigate the period of the sine function. a. TABULAR Copy and complete a table similar to the one below that includes all 16 angle measures from the unit circle.

b. VERBAL After what values of θ do sin θ, sin 2θ, and sin 4θ repeat their range values? In other words, what are the periods of these functions? c. VERBAL Make a conjecture as to how the period of y = sin nθ is affected for different values of n.

SOLUTION: a.

b. The period of sin θ is 2π. sin 2θ repeats its values after π. Therefore, the period of sin 2θ is π. sin 4 θ repeats its

values after . Therefore, the period of sin 4θ is .

c. Sample answer: The period decreases as the value of n increases.

79. CHALLENGE For each statement, describe n.

a.

b.

SOLUTION:

a. On the unit circle, cos θ = 0 when θ = and . Because the cosine function is periodic, cos θ = 0 when θ =

+ 2π or and θ = + 2π or . So, in general, when n is an odd integer.

b. Because csc θ = , csc θ is undefined when sin θ = 0. On the unit circle, sin θ = 0 when θ = 0, π, 2π, etc.

So, when n = 2, 4, 6, etc. Therefore, when n is an even integer.

REASONING Determine whether each statement is true or false . Explain your reasoning.80. If cos θ = 0.8, sec θ – cos (–θ ) = 0.45.

SOLUTION:

Sample answer: The cosine function is an even function, so cos (−θ) = cos θ.

Therefore, the statement is true.

81. Since tan (–t) = –tan t, the tangent of a negative angle is a negative number.

SOLUTION: Sample answer: The expression tan (–t) = –tan t means that tangent is an odd function. The tangent of an angle depends on what quadrant the terminal side of the angle lies in. Therefore, the statement is false.

82. Writing in Math Explain why the attendance at a year-round theme park could be modeled by a periodic function. What issues or events could occur over time to alter this periodic depiction?

SOLUTION: Sample answer: Theme park attendance is much higher in the spring and summer because students are out of school and people take more vacations. During the winter, attendance is lower because fewer people take vacations. Attendance fluctuates every year; most likely, the period of this function would be one year. This depiction would change if theme parks hosted events in the winter that attracted more people or if people vacationed more in the winter.

REASONING Use the unit circle to verify each relationship.83. sin (–t) = –sin t

SOLUTION: Sample answer: The sine function is represented by the y-coordinate on the unit circle. Comparing sin t and sin (−t)for different values of t, notice that the y-coordinate is positive for sin t and is negative for sin (−t). For instance, onthe first unit circle, sin t = b and sin (−t) = −b. Now find –(sin t) to verify the relationship. –(sin t) = –(b) or –b, which is equivalent to sin (−t). Thus, −sin t = sin (−t).

84. cos (–t) = cos t

SOLUTION: Sample answer: The cosine function is represented by the x-coordinate on the unit circle. Comparing cos t and cos (−t) for different values of t, notice that the value of cosine, the x-coordinate, will be the same regardless of the sign of t. Thus, cos t = cos (−t).

85. tan (–t) = –tan t

SOLUTION:

Sample answer: Since tan t = , we can analyze –tan t and tan (−t) by first looking at tan t and tan (−t) on the

unit circle for a given value of t. Regardless of the sign of t, the value of cosine remains the same. However, the

value of sine is positive for t but negative for –t. This results in tan t = , but . Now find –tan t to

verify the relationship. which is equivalent to tan (−t). Thus, −tan t = tan (−t).

86. Writing in Math Make a conjecture as to the periods of the secant, cosecant, and cotangent functions. Explain.

SOLUTION:

Sample answer: The period of the secant function will be 2π because it is the reciprocal of the cosine function and the period of the cosine function is 2π. The period of the cosecant function will be 2 because it is the reciprocal of the sine function and the period of the sine function is 2 . The period of the cotangent function will be π because it is the reciprocal of the tangent function and the period of the tangent function is .

Write each decimal degree measure in DMS form and each DMS measure in decimal degree form to the nearest thousandth.

87. 168.35°

SOLUTION:

Convert 0. 35° into minutes and seconds.

Therefore, 168.35° can be written as 168° 21′ 23″.

88. 27.465°

SOLUTION: First, convert 0. 465º into minutes and seconds.

Next, convert 0.9' into seconds.


89. 14° 5′ 20″

SOLUTION:

Each minute is of a degree and each second is of a minute, so each second is of a degree.

Therefore, 14° 5′ 20″ can be written as about 14.089°.

90. 173° 24′ 35″

SOLUTION:



91. EXERCISE A preprogrammed workout on a treadmill consists of intervals walking at various rates and angles of incline. A 1% incline means 1 unit of vertical rise for every 100 units of horizontal run.

a. At what angle, with respect to the horizontal, is the treadmill bed when set at a 10% incline? Round to the nearest degree. b. If the treadmill bed is 40 inches long, what is the vertical rise when set at an 8% incline?

SOLUTION: a. A 1% incline is equivalent to 1 unit of vertical rise for every 100 units of horizontal run. So, a 10% incline is equivalent to 10 units of vertical rise for every 100 units of horizontal run. Draw a diagram to model the situation.

Use the tangent function to find θ.

So, when set at a 10% incline, the treadmill bed would be at an angle of about 5.7° to the horizontal.

b. When set at an 8% incline, the treadmill bed would be at an angle of θ = or about 4.57º.

Draw a diagram to model the situation.

Use the tangent function to find x.

Therefore, the vertical rise is about 3.2 inches when set at an 8% incline.

Evaluate each logarithm.92. log8 64

SOLUTION:

93. log125 5

SOLUTION:

94. log2 32

SOLUTION:

95. log4 128

SOLUTION:

List all possible rational zeros of each function. Then determine which, if any, are zeros.

96. f (x) = x3 – 4x2 + x + 2

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 2.

Therefore, the possible rational zeros of f are By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x – 1) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x − 1)

(x2 −3 − 2). Because the factor (x

2 −3 − 2) yields no rational zeros, the rational zero of f is 1.

97. g(x) = x3 + 6x2 + 10x + 3


Therefore, the possible rational zeros of g are By using synthetic division, it can be determined that x = −3 is a rational zero.

Because (x + 3) is a factor of g(x), we can use the final quotient to write a factored form of g(x) as g(x) = (x + 3)

(x2 + 3x + 1). Because the factor (x

2 + 3x + 1) yields no rational zeros, the rational zero of g is −3.

98. h(x) = x4 – x2 + x – 1

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −1. Therefore, the possible rational zeros of h are By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x – 1) is a factor of h(x), we can use the final quotient to write a factored form of h(x) as h(x) = (x − 1)

(x3 + x

2 + 1). Because the factor (x

3 + x

2 + 1) yields no rational zeros, the rational zero of h is 1.

99. h(x) = 2x3 + 3x2 – 8x + 3

SOLUTION:

The leading coefficient is 2 and the constant term is 3. The possible rational zeros are or

By using synthetic division, it can be determined that x = −3 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of h(x), we can use the final quotient to write a factored form of h(x) as h(x) = (x + 3)

(x − 1)(2x − 1). Therefore, the rational zeros of h are –3, , and 1.

100. f (x) = 2x4 + 3x3 – 6x2 – 11x – 3

SOLUTION:

The leading coefficient is 2 and the constant term is −3. The possible rational zeros are or

By using synthetic division, it can be determined that x = − is a rational zero.

Because is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x3 − 6x − 2). Because the factor (x

3 − 6x − 2) yields no rational zeros, the rational zero of f is − .

101. g(x) = 4x3 + x2 + 8x + 2

SOLUTION:



Because is a factor of g(x), we can use the final quotient to write a factored form of g(x) as g(x)

= (4x2 + 8). Because the factor (4x

2 + 8) yields no real zeros, the rational zero of f is − .

102. NAVIGATION A global positioning system (GPS) uses satellites to allow a user to determine his or her position on Earth. The system depends on satellite signals that are reflected to and from a hand-held transmitter. The time that the signal takes to reflect is used to determine the transmitter’s position. Radio waves travel through air at a speed of 299,792,458 meters per second. Thus, d(t) = 299,792,458t relates the time t in seconds to the distance traveled d(t) in meters. a. Find the distance a radio wave will travel in 0.05, 0.2, 1.4, and 5.9 seconds. b. If a signal from a GPS satellite is received at a transmitter in 0.08 second, how far from the transmitter is the satellite?

SOLUTION: a. Use d(t) = 299,792,458t to find d(0.05).

Find d(0.2).

Find d(1.4).

Find d(5.9).

b. Find d(0.08).

103. SAT/ACT In the figure, and are tangents to circle C. What is the value of m?

SOLUTION: From geometry, any line drawn tangent to a circle is perpendicular to a radius drawn to the point of tangency. Therefore, the quadrilateral formed by the tangent lines and radii has two 90º angles.

Therefore, m = 45°.

104. Suppose θ is an angle in standard position with sin θ > 0. In which quadrant(s) does the terminal side of θ lie?A I only B I and II C I and III D I and IV

SOLUTION:

Because sin θ is greater than zero in Quadrants I and II, the correct answer is B.

105. REVIEW Find the angular speed in radians per second of a point on a bicycle tire if it completes 2 revolutions in 3 seconds.

F

G

H

J

SOLUTION:

Because each revolution measures 2π radians, 2 revolutions correspond to an angle of rotation of 2 × 2π or 4π radians.

Therefore, the correct answer is J.

106. REVIEW Which angle has a tangent and cosine that are both negative? A 110° B 180° C 210° D 340°

SOLUTION:

Tangent and cosine are both negative in Quadrant II. Because 110° is the only angle that is in Quadrant II, the

eSolutions Manual - Powered by Cognero Page 1

4-3 Trigonometric Functions on the Unit Circle


1. (3, 4)



2. (−6, 6)



3. (−4, −3)



4. (2, 0)



5. (1, −8)



6. (5, −3)



7. (−8, 15)



8. (−1, −2)




9. sin

SOLUTION:



10. tan 2π

SOLUTION:


11. cot (–180°)

SOLUTION:


12. csc 270°

SOLUTION:


13. cos (–270°)

SOLUTION:


14. sec 180°

SOLUTION:


15. tan π

SOLUTION:


16.

SOLUTION:




SOLUTION:


18. 210°

SOLUTION:


19.

SOLUTION:


20.

SOLUTION:


.

21. −405°

SOLUTION:


22. −75°

SOLUTION:


23.

SOLUTION:


24.

SOLUTION:



25. cos

SOLUTION:


θ is negative.

26. tan

SOLUTION:


is positive.

27. sin

SOLUTION:


positive.

28. cot (−45°)

SOLUTION:


29. csc 390°

SOLUTION:


30. sec (−150°)

SOLUTION:


31. tan

SOLUTION:


θ is negative.

32. sin 300°

SOLUTION:



SOLUTION:





SOLUTION:




35.

SOLUTION:




36.

SOLUTION:





SOLUTION:





SOLUTION:





SOLUTION:




40.

SOLUTION:





SOLUTION:









43. sec 120°

SOLUTION:


44. sin 315°

SOLUTION:


45. cos

SOLUTION:

46.

SOLUTION:

47. csc 390°

SOLUTION:

48. cot 510°

SOLUTION:

49. csc 5400°

SOLUTION:

There


50. sec

SOLUTION:



51.

SOLUTION:

52. csc

SOLUTION:

53. tan

SOLUTION:


54. sec

SOLUTION:


55.

SOLUTION:

56. cos

SOLUTION:


57. tan

SOLUTION:

58.

SOLUTION:



t = 0.5

t = 1

t = 1.5

t = 2

t = 2.5



SOLUTION:



61. tan = sin ___

SOLUTION:



62. sin = cos ___

SOLUTION:



63. cos = sin ___

SOLUTION:



64. sin (−45°) = cos ___

SOLUTION:




225º.

65. cos = sin ___

SOLUTION:












67. cos (−θ) = ; cos θ = ?; sec θ = ?

SOLUTION:


68. sin (−θ) = ; sin θ = ?; csc θ = ?

SOLUTION:


69. sec θ = ; cos θ = ?; cos (−θ) = ?

SOLUTION:


70. csc θ = ; sin θ = ?; sin (−θ) = ?

SOLUTION:







72.



radius r is 3.


73.



radius r is 5.


74.




75.






SOLUTION:


So,


So,





SOLUTION:

a. Evaluate for x = 0, 1, 2, …, 24.



10 8 23 6.5 11 6.5 24 5.4 12 5.4



SOLUTION: a.





a.

b.

SOLUTION:






SOLUTION:








SOLUTION: Sample answer: The sine function is represented by the y-coordinate on the unit circle. Comparing sin t and sin (−t)for different values of t, notice that the y-coordinate is positive for sin t and is negative for sin (−t). For instance, onthe first unit circle, sin t = b and sin (−t) = −b. Now find –(sin t) to verify the relationship. –(sin t) = –(b) or –b, which is equivalent to sin (−t). Thus, −sin t = sin (−t).

84. cos (–t) = cos t

SOLUTION: Sample answer: The cosine function is represented by the x-coordinate on the unit circle. Comparing cos t and cos (−t) for different values of t, notice that the value of cosine, the x-coordinate, will be the same regardless of the sign of t. Thus, cos t = cos (−t).

85. tan (–t) = –tan t

SOLUTION:

Sample answer: Since tan t = , we can analyze –tan t and tan (−t) by first looking at tan t and tan (−t) on the

unit circle for a given value of t. Regardless of the sign of t, the value of cosine remains the same. However, the

value of sine is positive for t but negative for –t. This results in tan t = , but . Now find –tan t to

verify the relationship. which is equivalent to tan (−t). Thus, −tan t = tan (−t).

86. Writing in Math Make a conjecture as to the periods of the secant, cosecant, and cotangent functions. Explain.

SOLUTION:

Sample answer: The period of the secant function will be 2π because it is the reciprocal of the cosine function and the period of the cosine function is 2π. The period of the cosecant function will be 2 because it is the reciprocal of the sine function and the period of the sine function is 2 . The period of the cotangent function will be π because it is the reciprocal of the tangent function and the period of the tangent function is .

Write each decimal degree measure in DMS form and each DMS measure in decimal degree form to the nearest thousandth.

87. 168.35°

SOLUTION:

Convert 0. 35° into minutes and seconds.


88. 27.465°

SOLUTION: First, convert 0. 465º into minutes and seconds.

Next, convert 0.9' into seconds.


89. 14° 5′ 20″

SOLUTION:



90. 173° 24′ 35″

SOLUTION:



91. EXERCISE A preprogrammed workout on a treadmill consists of intervals walking at various rates and angles of incline. A 1% incline means 1 unit of vertical rise for every 100 units of horizontal run.

a. At what angle, with respect to the horizontal, is the treadmill bed when set at a 10% incline? Round to the nearest degree. b. If the treadmill bed is 40 inches long, what is the vertical rise when set at an 8% incline?

SOLUTION: a. A 1% incline is equivalent to 1 unit of vertical rise for every 100 units of horizontal run. So, a 10% incline is equivalent to 10 units of vertical rise for every 100 units of horizontal run. Draw a diagram to model the situation.

Use the tangent function to find θ.

So, when set at a 10% incline, the treadmill bed would be at an angle of about 5.7° to the horizontal.

b. When set at an 8% incline, the treadmill bed would be at an angle of θ = or about 4.57º.

Draw a diagram to model the situation.

Use the tangent function to find x.

Therefore, the vertical rise is about 3.2 inches when set at an 8% incline.

Evaluate each logarithm.92. log8 64

SOLUTION:

93. log125 5

SOLUTION:

94. log2 32

SOLUTION:

95. log4 128

SOLUTION:

List all possible rational zeros of each function. Then determine which, if any, are zeros.

96. f (x) = x3 – 4x2 + x + 2


Therefore, the possible rational zeros of f are By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x – 1) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x − 1)

(x2 −3 − 2). Because the factor (x

2 −3 − 2) yields no rational zeros, the rational zero of f is 1.

97. g(x) = x3 + 6x2 + 10x + 3


Therefore, the possible rational zeros of g are By using synthetic division, it can be determined that x = −3 is a rational zero.

Because (x + 3) is a factor of g(x), we can use the final quotient to write a factored form of g(x) as g(x) = (x + 3)

(x2 + 3x + 1). Because the factor (x

2 + 3x + 1) yields no rational zeros, the rational zero of g is −3.

98. h(x) = x4 – x2 + x – 1

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −1. Therefore, the possible rational zeros of h are By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x – 1) is a factor of h(x), we can use the final quotient to write a factored form of h(x) as h(x) = (x − 1)

(x3 + x

2 + 1). Because the factor (x

3 + x

2 + 1) yields no rational zeros, the rational zero of h is 1.

99. h(x) = 2x3 + 3x2 – 8x + 3

SOLUTION:


By using synthetic division, it can be determined that x = −3 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of h(x), we can use the final quotient to write a factored form of h(x) as h(x) = (x + 3)

(x − 1)(2x − 1). Therefore, the rational zeros of h are –3, , and 1.

100. f (x) = 2x4 + 3x3 – 6x2 – 11x – 3

SOLUTION:

The leading coefficient is 2 and the constant term is −3. The possible rational zeros are or


Because is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x3 − 6x − 2). Because the factor (x

3 − 6x − 2) yields no rational zeros, the rational zero of f is − .

101. g(x) = 4x3 + x2 + 8x + 2

SOLUTION:



Because is a factor of g(x), we can use the final quotient to write a factored form of g(x) as g(x)

= (4x2 + 8). Because the factor (4x

2 + 8) yields no real zeros, the rational zero of f is − .

102. NAVIGATION A global positioning system (GPS) uses satellites to allow a user to determine his or her position on Earth. The system depends on satellite signals that are reflected to and from a hand-held transmitter. The time that the signal takes to reflect is used to determine the transmitter’s position. Radio waves travel through air at a speed of 299,792,458 meters per second. Thus, d(t) = 299,792,458t relates the time t in seconds to the distance traveled d(t) in meters. a. Find the distance a radio wave will travel in 0.05, 0.2, 1.4, and 5.9 seconds. b. If a signal from a GPS satellite is received at a transmitter in 0.08 second, how far from the transmitter is the satellite?

SOLUTION: a. Use d(t) = 299,792,458t to find d(0.05).

Find d(0.2).

Find d(1.4).

Find d(5.9).

b. Find d(0.08).

103. SAT/ACT In the figure, and are tangents to circle C. What is the value of m?

SOLUTION: From geometry, any line drawn tangent to a circle is perpendicular to a radius drawn to the point of tangency. Therefore, the quadrilateral formed by the tangent lines and radii has two 90º angles.

Therefore, m = 45°.

104. Suppose θ is an angle in standard position with sin θ > 0. In which quadrant(s) does the terminal side of θ lie?A I only B I and II C I and III D I and IV

SOLUTION:

Because sin θ is greater than zero in Quadrants I and II, the correct answer is B.

105. REVIEW Find the angular speed in radians per second of a point on a bicycle tire if it completes 2 revolutions in 3 seconds.

F

G

H

J

SOLUTION:

Because each revolution measures 2π radians, 2 revolutions correspond to an angle of rotation of 2 × 2π or 4π radians.

Therefore, the correct answer is J.

106. REVIEW Which angle has a tangent and cosine that are both negative? A 110° B 180° C 210° D 340°

SOLUTION:

Tangent and cosine are both negative in Quadrant II. Because 110° is the only angle that is in Quadrant II, the

eSolutions Manual - Powered by Cognero Page 2

4-3 Trigonometric Functions on the Unit Circle


1. (3, 4)



2. (−6, 6)



3. (−4, −3)



4. (2, 0)



5. (1, −8)



6. (5, −3)



7. (−8, 15)



8. (−1, −2)




9. sin

SOLUTION:



10. tan 2π

SOLUTION:


11. cot (–180°)

SOLUTION:


12. csc 270°

SOLUTION:


13. cos (–270°)

SOLUTION:


14. sec 180°

SOLUTION:


15. tan π

SOLUTION:


16.

SOLUTION:




SOLUTION:


18. 210°

SOLUTION:


19.

SOLUTION:


20.

SOLUTION:


.

21. −405°

SOLUTION:


22. −75°

SOLUTION:


23.

SOLUTION:


24.

SOLUTION:



25. cos

SOLUTION:


θ is negative.

26. tan

SOLUTION:


is positive.

27. sin

SOLUTION:


positive.

28. cot (−45°)

SOLUTION:


29. csc 390°

SOLUTION:


30. sec (−150°)

SOLUTION:


31. tan

SOLUTION:


θ is negative.

32. sin 300°

SOLUTION:



SOLUTION:





SOLUTION:




35.

SOLUTION:




36.

SOLUTION:





SOLUTION:





SOLUTION:





SOLUTION:




40.

SOLUTION:





SOLUTION:









43. sec 120°

SOLUTION:


44. sin 315°

SOLUTION:


45. cos

SOLUTION:

46.

SOLUTION:

47. csc 390°

SOLUTION:

48. cot 510°

SOLUTION:

49. csc 5400°

SOLUTION:

There


50. sec

SOLUTION:



51.

SOLUTION:

52. csc

SOLUTION:

53. tan

SOLUTION:


54. sec

SOLUTION:


55.

SOLUTION:

56. cos

SOLUTION:


57. tan

SOLUTION:

58.

SOLUTION:



t = 0.5

t = 1

t = 1.5

t = 2

t = 2.5



SOLUTION:



61. tan = sin ___

SOLUTION:



62. sin = cos ___

SOLUTION:



63. cos = sin ___

SOLUTION:



64. sin (−45°) = cos ___

SOLUTION:




225º.

65. cos = sin ___

SOLUTION:












67. cos (−θ) = ; cos θ = ?; sec θ = ?

SOLUTION:


68. sin (−θ) = ; sin θ = ?; csc θ = ?

SOLUTION:


69. sec θ = ; cos θ = ?; cos (−θ) = ?

SOLUTION:


70. csc θ = ; sin θ = ?; sin (−θ) = ?

SOLUTION:







72.



radius r is 3.


73.



radius r is 5.


74.




75.






SOLUTION:


So,


So,





SOLUTION:

a. Evaluate for x = 0, 1, 2, …, 24.



10 8 23 6.5 11 6.5 24 5.4 12 5.4



SOLUTION: a.





a.

b.

SOLUTION:






SOLUTION:








SOLUTION: Sample answer: The sine function is represented by the y-coordinate on the unit circle. Comparing sin t and sin (−t)for different values of t, notice that the y-coordinate is positive for sin t and is negative for sin (−t). For instance, onthe first unit circle,

Documents

4-3 Trigonometric Functions on the Unit Circle · CAROUSEL Zoe is on a carousel at the carnival. The diameter of the carousel is 80 feet. Find the position of her seat from the center