4-3 Trigonometric Functions on the Unit Circle

Transcription

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.
3. (−4, −3)
Use x =
,y =
, and r = 5 to write the six trigonometric ratios.
5. (1, −8)
Use x = 1, y =
, and r =
to write the six trigonometric ratios.
7. (−8, 15)
eSolutions Manual - Powered by Cognero
Page 1
7. (−8, 15)
Use x =
, y = 15, and r = 17 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.
11. cot (–180°)
SOLUTION: The terminal side of
in standard position lies on the negative x-axis. Choose a point P(
side of the angle because r = 1.
, 0) on the terminal
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
15. tan π
SOLUTION: eSolutions Manual - Powered by Cognero
The terminal side of π in standard position lies on the negative x-axis. Choose a point P(
Page 2
, 0) on the terminal side
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
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.
19. SOLUTION: The terminal side of
lies in Quadrant II. Therefore, its reference angle is θ ' =
.
21. −405°
SOLUTION: A coterminal angle is −405° + 360(2)° or 315°. The terminal side of 315° lies in Quadrant IV, so its reference angle
is 360º – 315º or 45º. eSolutions Manual - Powered by Cognero
Page 3
21. −405°
SOLUTION: A coterminal angle is −405° + 360(2)° or 315°. The terminal side of 315° lies in Quadrant IV, so its reference angle
is 360º – 315º or 45º. 23. SOLUTION: The terminal side of lies in Quadrant II. Therefore, its reference angle is θ ' =
.
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.
27. sin
Because the terminal side of θ lies in Quadrant II, the reference angle θ ' is Page 4
or . In Quadrant II, sin θ is
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.
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.
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.
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
Page 5
sin θ and cos θ are positive, so θ must lie in Quadrant I. This means that both x and y are positive.
Find the exact values of the five remaining trigonometric functions of θ.
33. tan θ = 2, where sin θ > 0 and cos θ > 0
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.
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 =
Use x =
,y =
or , use the point (x,
, and r = 5 to write the five remaining trigonometric ratios.
37. secManual
θ = - Powered
, where
θ<0
eSolutions
by sin
Cognero
SOLUTION: ) and r= 5 to find x.
and cos θ > 0
Page 6
37. sec θ =
, where sin θ < 0 and cos θ > 0
negative.
Because sec θ = or
Use x = 1, y =
, use the point (1, y) and r =
, and r =
to find y.
39. tan θ = −1, where sin θ < 0
negative.
Because tan θ =
Use x = , y =
or
, use the point ( ,
, and r =
) to find r.
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.
Page 7
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° rotation
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 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).
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.
45. cos
SOLUTION: eSolutions
Manual - Powered by Cognero
Page 8
45. cos
SOLUTION: 47. csc 390°
SOLUTION: 49. csc 5400°
SOLUTION: There
csc 5400° is undefined.
51. SOLUTION: 53. tan
eSolutions
Manual - Powered by Cognero
SOLUTION: corresponds to the point (x, y) =
Page 9
on the unit circle.
53. tan
SOLUTION: corresponds to the point (x, y) =
on the unit circle.
55. SOLUTION: 57. tan
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
Page 10
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.
Page 11
The times and corresponding angle measures are shown in the table below.
Complete each trigonometric expression.
61. tan
= sin ___
SOLUTION: on the unit circle. So, tan corresponds to the point (x, y) =
= or 1.
On the unit circle, sin
63. cos
= 1. Therefore, tan
= sin .
= sin ___
SOLUTION: on the unit circle. So, cos
corresponds to the point (x, y) =
=
65. cos
= and sin
= . Therefore, cos
= sin or cos
= sin .
= sin ___
SOLUTION: on the unit circle. So, cos
=
= and sin
= Therefore, cos = sin = sin or cos
.
Use the given values to evaluate the trigonometric functions.
67. cos (−θ) = ; cos θ = ?; sec θ = ?
Because cos (–θ) =
and cos (–θ) = cos θ, cos θ =
Page 12
. So, sec θ =
or
.
on the unit circle. So, cos
=
4-3 Trigonometric
Functions
on the = Unit
Circle
= and sin
Therefore, cos = sin = sin or cos
.
Use the given values to evaluate the trigonometric functions.
67. cos (−θ) = ; cos θ = ?; sec θ = ?
SOLUTION: Because cos (–θ) =
69. sec θ = and cos (–θ) = cos θ, cos θ =
. So, sec θ =
or
.
; cos θ = ?; cos (−θ) = ?
SOLUTION: Because sec θ = and cos θ =
, cos θ =
. Because cos (–θ) = cos θ and cos θ =
, cos (–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.
73. Page 13
Find the coordinates of P for each circle with the given radius and angle measure.
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.
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 eSolutions Manual - Powered by Cognero
is − π or
Page 14
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 So, the coordinates of P are (−4, −4
is − π or
).
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
where x = 0, 1, 2, …, 24 corresponds to 12:00 midnight, 1:00
certain day, that function was
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
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
for x = 0, 1, 2, …, 24.
Depth (m)
5.4
5
5.4
6.5
8
9.5
10.6
11
10.6
9.5
8
6.5
5.4
Time
13
14
15
16
17
18
19
20
21
22
23
24
Depth (m)
5
5.4
6.5
8
9.5
10.6
11
10.6
9.5
8
6.5
5.4
Therefore, the maximum depth, or high tide, that day was 11 meters.
Page 15
So, the coordinates of P are (−4, −4
).
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
where x = 0, 1, 2, …, 24 corresponds to 12:00 midnight, 1:00
certain day, that function was
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
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
for x = 0, 1, 2, …, 24.
Depth (m)
5.4
5
5.4
6.5
8
9.5
10.6
11
10.6
9.5
8
6.5
5.4
Time
13
14
15
16
17
18
19
20
21
22
23
24
Depth (m)
5
5.4
6.5
8
9.5
10.6
11
10.6
9.5
8
6.5
5.4
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.
79. CHALLENGE For each statement, describe n.
a.
b.
SOLUTION: a. On the unit circle, cos θ = 0 when θ =
+ 2π or and θ =
+ 2π or
and
. Because the cosine function is periodic, cos θ = 0 when θ =
. So, in general,
when n is an odd integer.
b. Because csc θ =
So,
, csc θ is undefined when sin θ = 0. On the unit circle, sin θ = 0 when θ = 0, π, 2π, etc.
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.
81. Since
tan- (–t)
= –tan
t, the tangent of a negative angle is a negative number.
eSolutions
Manual
Powered
by Cognero
Page 16
SOLUTION: Sample answer: The expression tan (–t) = –tan t means that tangent is an odd function. The tangent of an angle
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.
4-3 Trigonometric
Functions
Circle
So,
when on
n = the
2, 4, Unit
6, etc. Therefore,
when n is an even integer.
REASONING Determine whether each statement is true or false . Explain your reasoning.
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.
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, on
the 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).
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 =
verify the relationship.
, but
. Now find –tan t to
which is equivalent to tan (−t). Thus, −tan t = tan (−t).
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.
Page 17
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″.
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°.
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.
Page 18
can be written
about
14.089°.
Therefore, 14° 5′ 20″
4-3 Trigonometric
Functions
onasthe
Unit
Circle
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.
93. log125 5 - Powered by Cognero
eSolutions Manual
SOLUTION: Page 19
4-3 Trigonometric
Functions on the Unit Circle
Therefore, the vertical rise is about 3.2 inches when set at an 8% incline.
Evaluate each logarithm.
93. log125 5
SOLUTION: 95. log4 128
SOLUTION: List all possible rational zeros of each function. Then determine which, if any, are zeros.
97. g(x) = x3 + 6x2 + 10x + 3
SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 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)
2
2
(x + 3x + 1). Because the factor (x + 3x + 1) yields no rational zeros, the rational zero of g is −3.
99. h(x) = 2x3 + 3x2 – 8x + 3
SOLUTION: The leading coefficient is 2 and the constant term is 3. The possible rational zeros are
By using synthetic division, it can be determined that x = −3 is a rational zero.
or Page 20
4-3
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)
2
2
Trigonometric
Functions
Circle
(x + 3x + 1). Because
the factor on
(x +the
3x +Unit
1) yields
no rational zeros, the rational zero of g is −3.
99. h(x) = 2x3 + 3x2 – 8x + 3
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.
101. g(x) = 4x3 + x2 + 8x + 2
or By using synthetic division, it can be determined that x = −
is a rational zero.
is a factor of g(x), we can use the final quotient to write a factored form of g(x) as g(x)
Because
=
2
2
(4x + 8). Because the factor (4x + 8) yields no real zeros, the rational zero of f is −
103. SAT/ACT In the figure,
and .
are tangents to circle C. What is the value of m?
Page 21
Because
is a factor of g(x), we can use the final quotient to write a factored form of g(x) as g(x)
2
2
4-3 Trigonometric
on the
Circle
=
(4x + 8).Functions
Because the factor
(4xUnit
+ 8) yields
no real zeros, the rational zero of f is −
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°.
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.
Page 22

4-3 Trigonometric Functions on the Unit Circle

Transcription

Similar documents

Worksheet 6 Chain rule and implicit differentiation 1 Chain rule 2

Higher Relationships and Calculus NAB Revision Pack

Calculus I Final, Sample

AS Entrance Examination Sample Paper Mathematics Instructions

square, circle

Document 6539622

Test 2 Review 1

Math 370 Sample Final Exam Questions 5 + 5i .

Sample Solutions to Quiz 3 for MATH3270A − y

MATH 307: Problem Set #6