4-3 Trigonometric Functions on the Unit Circle · Find the exact values of the five remaining trigonometric functions of . tan = 2, ... 12 cm, is the length of the radius of the first

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

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 = towritethesixtrigonometricratios.

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 = towritethesixtrigonometricratios.

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 = towritethesixtrigonometricratios.

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

9.sin

SOLUTION:

Theterminalsideof instandardpositionliesonthepositivey-axis. Choose a point P(0, 1) on the terminal side

of the angle because r = 1.

10.tan 2

SOLUTION:

The terminal side of instandardpositionliesonthepositivex-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 instandardpositionliesonthenegativex-axis. Choose a point P( , 0) on the terminalside of the angle because r = 1.

12.csc 270

SOLUTION:

The terminal side of instandardpositionliesonthenegativey-axis. Choose a point P(0, ) on the terminal side of the angle because r = 1.

13.cos (270)

SOLUTION:

The terminal side of instandardpositionliesonthepositivey-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 instandardpositionliesonthenegativex-axis. Choose a point P( , 0) on the terminal side of the angle because r = 1.

15.tan

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.

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 135 lies in Quadrant II. Therefore, its reference angle is ' = 180 135 or 45.

18.210

SOLUTION:

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

19.

SOLUTION:

The terminal side of liesinQuadrantII.Therefore,itsreferenceangleis ' = .

20.

SOLUTION:

A coterminal angle is 2 or , which lies in Quadrant IV. So, the reference angle is ' is 2 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 360 315 or 45.

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:

Theterminalsideof liesinQuadrantII.Therefore,itsreferenceangleis ' = .

24.

SOLUTION:

A coterminal angle is + 2(1) or Theterminalsideof lies in Quadrant I, so the reference angle is

Find the exact value of each expression.

25.cos

SOLUTION:

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

is negative.

26.tan

SOLUTION:

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

is positive.

27.sin

SOLUTION:

Because the terminal side of lies in Quadrant II, the reference angle ' is or . In Quadrant II, sin 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 ' is 360 315 or 45. Because tangent and cotangent are reciprocal functions and tan is negative in Quadrant IV, it follows that cot 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 ' is 360 30 or 30. Because sine and cosecant are reciprocal functions and sin is positive in Quadrant I, it follows that csc 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 lies in Quadrant III. , the reference angle ' is 210 180 or 30. Because secant and cosine are reciprocal functions and cos is negative in Quadrant III, it follows that sec is also negative in Quadrant III.

31.tan

SOLUTION:

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

is negative.

32.sin 300

SOLUTION:

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

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

SOLUTION:

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

Because tan = 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 = 2, where sin > 0 and cos < 0

SOLUTION:

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

Because csc = or , use the point (x,1)andr = 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 . You know that sin is negative and cos is positive, so must lie in Quadrant IV. This means that x is positive and y is negative.

Because sin = 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 . You know that sin and cos are negative, so must lie in Quadrant III. This means that both x and y are negative.

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


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

SOLUTION:

To find the other function values, you must find the coordinates of a point on the terminal side of . You know that sin is negative and cos is positive, so mustlieinQuadrantIV.Thismeansthatx is positive and y is negative.

Because sec = or , use the point (1, y) and r = tofindy .

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

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

SOLUTION:


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

Use x = , y = , and r = towritethefiveremainingtrigonometricratios.

39.tan = 1, where sin < 0

SOLUTION:


Because tan = 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 . You know that sin is positive and cos is negative, so must lie in Quadrant II. This means that x is negative and y is positive.

Because cos = or , use the point ( ,y)andr = 2 to find y. .


41.CAROUSELZoe 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 210.

SOLUTION:

Let the center of the carousel represent the origin on the coordinate plane and Zoes 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 210, so the reference angle is 210 180 or 30. 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 FUNNELAcoinisdroppedintoafunnelwhereitspinsinsmallercirclesuntilitdropsintothebottomofthe 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 150, so the reference angle is 180 150 or 30. Since the final position of the coin corresponds to Quadrant II, the cosine of 150 is negative and the sine of 150 is positive.

Therefore, the coordinates of the final position of the coin are orabout(10.4, 6).

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

43.sec 120

SOLUTION:

120 corresponds to the point (x, y) = ontheunitcircle.

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) = ontheunitcircle.

Therefore, sec isundefined.

51.

SOLUTION:

52.csc

SOLUTION:

53.tan

SOLUTION:


54.sec

SOLUTION:


55.

SOLUTION:

56.cos

SOLUTION:


57.tan

SOLUTION:

58.

SOLUTION:

59.RIDESJaeandAnyaareonarideatanamusementpark.Afterthefirstseveralswings,theangletheridemakes with the vertical is modeled by = 22 cos t, with 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.cos60=sin___

SOLUTION:

60 corresponds to the point (x, y) = ontheunitcircle.So,cos60 = .

On the unit circle, sin 30 = and sin 150 = .Therefore,cos60=sin30 orcos60=sin150 .

61.tan =sin___

SOLUTION:

corresponds to the point (x, y) = ontheunitcircle.So,tan = or 1.

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

62.sin =cos___

SOLUTION:

corresponds to the point (x, y) = ontheunitcircle.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) = ontheunitcircle.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 360. 45 + 360(1) = 315

315 corresponds to the point (x, y) = ontheunitcircle.So,sin(45) =

On the unit circle, cos 135 = and cos 225 = Therefore, sin (45)=cos135 or sin (45)=cos

225.

65.cos =sin___

SOLUTION:


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

66.ICE CREAMThemonthlysalesinthousandsofdollarsforFionas 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 shops 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 shops 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 ()= ; cos =?sec = ?

SOLUTION:

Because cos () = andcos() = cos , cos = . So, sec = or .

68.sin ()= ; sin =?csc = ?

SOLUTION:

Because sin() = andsin() = sin , sin = or sin = . So, csc = or .

69.sec = ; cos =?cos () = ?

SOLUTION:

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

70.csc = ; sin =?sin () = ?

SOLUTION:

Because csc = and sin = , sin = . Because sin () = sin and sin = , sin () = .

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

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 ispositiveandthesineof isnegative.Thereferenceanglefor is2 or andthe

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 isnegativeandthesineof ispositive.Thereferenceanglefor is or andthe

radius ris5.


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 arepositive.Theradiusr 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 arenegative.Thereferenceanglefor is or andtheradiusris8.

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

76.COMPARISON Supposetheterminalsideofanangle 1 in standard position contains the point (7, 8), and the

terminal side of a second angle 2 in standard position contains the point (7, 8). Compare the sines of 1 and 2.

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 1 = sin2.

77.TIDES Thedepthy 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 wherex = 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 forx = 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 REPRESENTATIONSIn this problem, you will investigate the period of the sine function. a. TABULARCopyandcompleteatablesimilartotheonebelowthatincludesall16anglemeasuresfromtheunit circle.

b. VERBALAfterwhatvaluesof do sin , sin 2, and sin 4 repeat their range values? In other words, what are the periods of these functions? c. VERBALMakeaconjectureastohowtheperiodofy = sin n is affected for different values of n.

SOLUTION:a.

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

values after . Therefore, the period of sin 4 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 = 0 when = and . Because the cosine function is periodic, cos = 0 when =

+2 or and = +2 or . So, in general, whenn is an odd integer.

b. Because csc = , csc is undefined when sin = 0. On the unit circle, sin = 0 when = 0, , 2, 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 = 0.8, sec cos ( ) = 0.45.

SOLUTION:

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

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 = ,wecananalyzetan 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. whichisequivalenttotan(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 2 because it is the reciprocal of the cosine function and the period of the cosine function is 2. 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 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,168.35canbewrittenas168 21 23.

88.27.465

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

Next, convert 0.9' into seconds.


89.145 20

SOLUTION:

Each minute is ofadegreeandeachsecondis ofaminute,soeachsecondis ofadegree.

Therefore,145 20 can be written as about 14.089.

90.17324 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 .

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 = or about 4.57.

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:


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

Because isafactoroff (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 isafactorofg(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.NAVIGATIONAglobalpositioningsystem(GPS)usessatellitestoallowausertodeterminehisorherpositionon 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 transmitters 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. Findthedistancearadiowavewilltravelin0.05,0.2,1.4,and5.9seconds. b. IfasignalfromaGPSsatelliteisreceivedatatransmitterin0.08second,howfarfromthetransmitteristhesatellite?

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 aretangentstocircleC. 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 90 angles.

Therefore, m = 45.

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

SOLUTION:

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

105.REVIEWFind 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 2 radians, 2 revolutions correspond to an angle of rotation of 2 2 or 4 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 2

SOLUTION:


11.cot (180)

SOLUTION:


12.csc 270

SOLUTION:


13.cos (270)

SOLUTION:


14.sec 180

SOLUTION:


15.tan

SOLUTION:


16.

SOLUTION:




SOLUTION:


18.210

SOLUTION:


19.

SOLUTION:


20.

SOLUTION:


.

21.405

SOLUTION:


22.75

SOLUTION:


23.

SOLUTION:


24.

SOLUTION:



25.cos

SOLUTION:


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:


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:




225.

65.cos =sin___

SOLUTION:












67.cos ()= ; cos =?sec = ?

SOLUTION:


68.sin ()= ; sin =?csc = ?

SOLUTION:


69.sec = ; cos =?cos () = ?

SOLUTION:


70.csc = ; sin =?sin () = ?

SOLUTION:







72.



radius r is 3.


73.



radius ris5.


74.




75.






SOLUTION:


So,


So,





SOLUTION:

a. Evaluate forx = 0, 1, 2, , 24.



10 8 23 6.5 11 6.5 24 5.4 12 5.4



SOLUTION:a.





a.

b.

SOLUTION:






SOLUTION:









84.cos (t) = cos t


85.tan (t) = tan t

SOLUTION:






SOLUTION:



87.168.35

SOLUTION:



88.27.465




89.145 20

SOLUTION:



90.17324 35

SOLUTION:













SOLUTION:

93.log125 5

SOLUTION:

94.log2 32

SOLUTION:

95.log4 128

SOLUTION:


96.f (x) = x3 4x2 + x + 2






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






98.h(x) = x4 x2 + x 1

SOLUTION:



(x3 + x


3 + x


99.h(x) = 2x3 + 3x2 8x + 3

SOLUTION:






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

SOLUTION:






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

SOLUTION:








Find d(0.2).

Find d(1.4).

Find d(5.9).

b. Find d(0.08).



Therefore, m = 45.


SOLUTION:



F

G

H

J

SOLUTION:




SOLUTION:





1.(3, 4)



2.(6, 6)



3.(4, 3)



4.(2, 0)



5.(1, 8)



6.(5, 3)



7.(8, 15)



8.(1, 2)


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4-3 Trigonometric Functions on the Unit Circle · Find the exact values of the five remaining trigonometric functions of . tan = 2, ... 12 cm, is the length of the radius of the first