Download The Other Trigonometric Functions

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The Other Trigonometric Functions
∗
OpenStax College
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 4.0
†
Abstract
In this section, you will:
•
•
Find exact values of the trigonometric functions secant, cosecant, tangent, and cotangent of
π π
, , and π6 .
3 4
Use reference angles to evaluate the trigonometric functions secant, cosecant, tangent, and cotangent.
•
•
•
Use properties of even and odd trigonometric functions.
Recognize and use fundamental identities.
Evaluate trigonometric functions with a calculator.
A wheelchair ramp that meets the standards of the Americans with Disabilities Act must make an angle
1
12 or less, regardless of its length. A tangent represents a ratio, so this
means that for every 1 inch of rise, the ramp must have 12 inches of run. Trigonometric functions allow us
with the ground whose tangent is
to specify the shapes and proportions of objects independent of exact dimensions. We have already dened
the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often
used, there are four others. Together they make up the set of six trigonometric functions. In this section,
we will investigate the remaining functions.
1 Finding Exact Values of the Trigonometric Functions Secant, Cosecant, Tangent, and Cotangent
To dene the remaining functions, we will once again draw a unit circle with a point
(x, y)
an angle of t,as shown in Figure 1. As with the sine and cosine, we can use the
coordinates to nd the
other functions.
∗ Version
1.7: Feb 24, 2015 11:46 am -0600
† http://creativecommons.org/licenses/by/4.0/
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(x, y)
corresponding to
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Figure 1
The rst function we will dene is the tangent.
The
tangent
to the x -value of the corresponding point on the unit circle.
of an angle is the ratio of the y -value
In Figure 1, the tangent of angle t is equal
y
x , x 6= 0. Because the y -value is equal to the sine of t, and the x -value is equal to the cosine of t, the
sin t
tangent of angle t can also be dened as
cos t , cos t 6= 0.The tangent function is abbreviated as tan. The
remaining three functions can all be expressed as reciprocals of functions we have already dened.
to
•
The
•
secant function is the reciprocal of the cosine function.
1
cos t
The
to
= x1 , x 6= 0. The
cotangent function is the reciprocal of the tangent function.
equal to
•
The
cos t
sin t
= xy , y 6= 0. The
1
sin t
= y1 , y 6= 0. The
cosecant
equal to
t
If
t
In Figure 1, the cotangent of angle t is
cotangent function is abbreviated as cot.
function is the reciprocal of the sine function.
A General Note:
of
In Figure 1, the secant of angle t is equal
secant function is abbreviated as sec.
In Figure 1, the cosecant of angle t is
cosecant function is abbreviated as csc.
is a real number and
(x, y)
is a point where the terminal side of an angle
radians intercepts the unit circle, then
tan t = xy , x 6= 0
sec t = x1 , x 6= 0
csc t = y1 , y 6= 0
(1)
cot t = xy , y 6= 0
Example 1
Finding Trigonometric
Functions from a Point on the Unit Circle
√
The point
−
3 1
2 , 2
is on the unit circle, as shown in Figure 2. Find sin t, cos t, tan t, sec t, csc t,
and cot t.
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Figure 2
Solution
Because we know the
(x, y)
coordinates of the point on the unit circle indicated by angle t, we can
use those coordinates to nd the six functions:
sin t = y =
1
2
cos t = x = −
tan t =
y
x
=
sec t =
1
x
=
csc t =
1
y
=
cot t =
x
y
=
√
3
2
1
2
√
− 23
1
√
− 3
2
1
1
2
√
− √23 = − √13 = − 33
√
(2)
= − √23 = − 2 3 3
=2
1
2 √
3
2
1
2
−
=
√
=−
3
2
2
1
√
=− 3
Try It:
Exercise 2
The point
(Solution on p. 28.)
√ 2
2
,
−
is on the unit circle, as shown in Figure 3. Find sin t, cos t, tan t, sec t, csc t,
2
2
√
and cot t.
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Figure 3
Example 2
Finding the Trigonometric Functions of an Angle
Find sin t, cos t, tan t, sec t, csc t, and cot t when t
=
π
6.
Solution
We have previously used the properties of equilateral triangles to demonstrate that sin
√
π
6
=
3
1
π
2 and cos 6 = 2 . We can use these values and the denitions of tangent, secant, cosecant, and
cotangent as functions of sine and cosine to nd the remaining function values.
tan
π
6
=
sec π6 =
=
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=
sin
cos
1
2
√
3
2
=
1
cos π
6
1
√
3
2
π
6
π
6
=
√1
3
√2
3
√
=
=
(3)
3
3
√
2 3
3
(4)
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csc
π
1
1
=
= 1 =2
6
sin π6
2
cot π6 =
=
cos π
6
π
sin
√ 6
3
2
1
2
=
√
(5)
(6)
3
Try It:
Exercise 4
(Solution on p. 28.)
Find sin t, cos t, tan t, sec t, csc t, and cot t when t
=
π
3.
Because we know the sine and cosine values for the common rst-quadrant angles, we can nd the other
function values for those angles as well by setting x equal to the cosine and y equal to the sine and then using
the denitions of tangent, secant, cosecant, and cotangent. The results are shown in Table 1.
Angle
Cosine
Sine
Tangent
Secant
Cosecant
Cotangent
1
π
6,
√
3
2
1
2
√
3
3
√
2 3
3
Undened
2
0
1
0
0
√
Undened
or 30 ◦
π
4,
√
2
2
√
2
2
1
√
√
3
2
2
1
or 45 ◦
π
3,
1
2
√
3
2
√
or 60 ◦
π
2,
or 90 ◦
0
1
3
Undened
2
Undened
√
2 3
3
√
3
3
1
0
Table 1
2 Using Reference Angles to Evaluate Tangent, Secant, Cosecant, and Cotangent
We can evaluate trigonometric functions of angles outside the rst quadrant using reference angles as we
have already done with the sine and cosine functions. The procedure is the same: Find the
reference angle
formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values
for the original angle will be the same as those for the reference angle, except for the positive or negative
sign, which is determined by x - and y -values in the original quadrant. Figure 4 shows which functions are
positive in which quadrant.
To help us remember which of the six trigonometric functions are positive in each quadrant, we can use
the mnemonic phrase A Smart Trig Class.
Each of the four words in the phrase corresponds to one of
the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is all
of the six trigonometric functions are positive.
function, cosecant, are positive. In quadrant III, are positive. Finally, in quadrant IV, Class,
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In quadrant II, Smart,
only
sine
A,
and its reciprocal
Trig, only tangent and its reciprocal function, cotangent,
cosine and its reciprocal function, secant, are positive.
only
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Figure 4
Given an angle not in the rst quadrant, use reference angles to nd all six
trigonometric functions.
How To:
1.Measure the angle formed by the terminal side of the given angle and the horizontal axis. This
is the reference angle.
2.Evaluate the function at the reference angle.
3.Observe the quadrant where the terminal side of the original angle is located. Based on the
quadrant, determine whether the output is positive or negative.
Example 3
Using Reference Angles to Find Trigonometric Functions
Use reference angles to nd all six trigonometric functions of
−
Solution
5π
6 .
π
6 , so that is the reference angle.
5π
Since −
is in the third quadrant, where both x and y are negative, cosine, sine, secant, and cosecant
6
will be negative, while tangent and cotangent will be positive.
The angle between this angle's terminal side and the x -axis is
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√
√3
cos − 5π
= − 23 , sin − 5π
= − 21 , tan − 5π
= 3
6
6
6
√
√
5π
sec −
= − 2 3 3 , csc − 5π
= −2, cot − 5π
= 3
6
6
6
(7)
Try It:
Exercise 6
(Solution on p. 28.)
Use reference angles to nd all six trigonometric functions of
−
7π
4 .
3 Using Even and Odd Trigonometric Functions
To be able to use our six trigonometric functions freely with both positive and negative angle inputs, we
should examine how each function treats a negative input. As it turns out, there is an important dierence
among the functions in this regard.
Consider the function f
(x) = x2 ,
shown in Figure 5. The graph of the function is symmetrical about
the y -axis. All along the curve, any two points with opposite x -values have the same function value. This
2
matches the result of calculation: (4)
2
2
2
= (−4) ,(−5) = (5) ,
and so on. So f
(x) = x2 is
a function such that two inputs that are opposites have the same output. That means f
Figure 5:
The function
f (x) = x2
is an even function.
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an
even function,
(−x) = f (x) .
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Now consider the function f
8
(x) = x3 , shown in Figure 6.
The graph is not symmetrical about the y -axis.
All along the graph, any two points with opposite x -values also have opposite y -values.
an
So f
(x) = x3 is
odd function, one such that two inputs that are opposites have outputs that are also opposites.
means f
(−x) = −f (x) .
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That
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The function
is an odd function.
Figure 6:
f (x) = x
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unit circle with a positive
We can test whether a trigonometric function is even or odd by drawing a
and a negative angle, as in Figure 7. The sine of the positive angle is y. The sine of the negative angle is
The
sine function,
−y.
then, is an odd function. We can test each of the six trigonometric functions in this
fashion. The results are shown in Table 2.
Figure 7
sin t = y
sin (−t) = −y
sin t 6= sin (−t)
sec t =
sec (−t) =
1
x
1
x
sec t = sec (−t)
cos t
=x
cos (−t) = x
cos t = cos (−t)
csc t =
csc (−t) =
1
y
1
−y
csc t 6= csc (−t)
Table 2
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tan (t)
=
y
x
tan (−t) = − xy
tan t 6= tan (−t)
cot t =
cot (−t) =
x
y
x
−y
cot t 6= cot (−t)
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A General Note:
An
11
An
even function is one in which f (−x) = f (x) .
odd function is one in which f (−x) = −f (x) .
Cosine and secant are even:
cos (−t) = cos t
(8)
sec (−t) = sec t
Sine, tangent, cosecant, and cotangent are odd:
sin (−t) = −sin t
tan (−t) = −tan t
(9)
csc (−t) = −csc t
cot (−t) = −cot t
Example 4
Using Even and Odd Properties of Trigonometric Functions
If the secant of angle t is 2, what is the secant of
− t?
Solution
Secant is an even function. The secant of an angle is the same as the secant of its opposite. So if
the secant of angle t is 2, the secant of
− t is
also 2.
Try It:
Exercise 8
If the cotangent of angle t is
√
(Solution on p. 28.)
3,
what is the cotangent of
− t?
4 Recognizing and Using Fundamental Identities
We have explored a number of properties of trigonometric functions. Now, we can take the relationships a
step further, and derive some fundamental identities. Identities are statements that are true for all values of
the input on which they are dened. Usually, identities can be derived from denitions and relationships we
already know. For example, the Pythagorean Identity we learned earlier was derived from the Pythagorean
Theorem and the denitions of sine and cosine.
A General Note:
We can derive some useful
identities from the six trigonometric functions.
The other four trigonometric functions can be related back to the sine and cosine functions using
these basic relationships:
=
sin t
cos t
(10)
sec t =
1
cos t
(11)
tan t
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1
sin t
csc t =
cot t
=
1
=
tan t
(12)
cos t
(13)
sin t
Example 5
Using Identities to Evaluate Trigonometric Functions
√
√
sin (45 ◦ ) = 22 , cos (45 ◦ ) = 22 , evaluate tan (45 ◦ ) .
√
1
5π
Given sin
= 2 , cos 5π
= − 23 , evaluate sec 5π
6
6
6 .
a. Given
b.
Solution
Because we know the sine and cosine values for these angles, we can use identities to evaluate the
other functions.
a.
tan (45 ◦ ) =
=
sin(45 ◦ )
◦)
√ cos(45
2
2
√
2
2
(14)
=1
b.
sec
5π
6
=
1
cos( 5π
6 )
1√
− 23
√
= −21 3
−2
=√
3 √
= −233
=
(15)
Try It:
Exercise 10
Evaluate csc
(Solution on p. 28.)
7π
6
.
Example 6
Using Identities to Simplify Trigonometric Expressions
Simplify
sec t
tan t .
Solution
We can simplify this by rewriting both functions in terms of sine and cosine.
(16)
By showing that
sec t
tan t can be simplied to csc t,we have, in fact, established a new identity.
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sec t
= csc t
tan t
(17)
Try It:
Exercise 12
(Solution on p. 28.)
Simplify tan t (cos t) .
4.1 Alternate Forms of the Pythagorean Identity
We can use these fundamental identities to derive alternative forms of the
2
sin t = 1. One
form is obtained by dividing both sides by cos
cos2 t
cos2 t
+
sin2 t
cos2 t
2
=
2
Pythagorean Identity, cos2 t +
t:
1
cos2 t
2
(18)
1 + tan t = sec t
The other form is obtained by dividing both sides by
cos2 t
sin2 t
+
sin2 t :
sin2 t
sin2 t
=
1
sin2 t
2
(19)
cot2 t + 1 = csc t
A General Note:
1 + tan2 t = sec2 t
(20)
cot2 t + 1 = csc2 t
(21)
Example 7
Using Identities to Relate Trigonometric Functions
12
13 and t is in quadrant IV, as shown in Figure 8, nd the values of the other ve
trigonometric functions.
If cos (t)
=
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Figure 8
Solution
We can nd the sine using the Pythagorean Identity, cos
2
t + sin2 t = 1, and the remaining functions
by relating them to sine and cosine.
12 2
13
+ sin2 t = 1
sin2 t = 1 −
sin2 t = 1 −
sin2 t =
sin t =
sin t =
sin t =
12 2
13
144
169
25
169q
(22)
25
± 169
√
25
± √169
5
± 13
The sign of the sine depends on the y -values in the quadrant where the angle is located. Since the
angle is in quadrant IV, where the y -values are negative, its sine is negative,
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−
5
13 .
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The remaining functions can be calculated using identities relating them to sine and cosine.
tan t =
sec t =
sin t
cos t
1
cos t
=
=
5
− 13
5
= − 12
12
13
1
12
13
1
sin t
csc t =
= −15
13
1
1
cot t = tan
=
5
t
− 12
=
=
13
12
(23)
− 13
5
= − 12
5
Try It:
Exercise 14
If sec (t)
=
(Solution on p. 28.)
− 17
8 and 0
< t < π,
nd the values of the other ve functions.
As we discussed in the chapter opening, a function that repeats its values in regular intervals is known
as a
periodic function.
The trigonometric functions are periodic. For the four trigonometric functions,
sine, cosine, cosecant and secant, a revolution of one circle, or 2π, will result in the same outputs for these
functions. And for tangent and cotangent, only a half a revolution will result in the same outputs.
Other functions can also be periodic.
For example, the lengths of months repeat every four years.
If x represents the length time, measured in years, and f
then f
(x + 4) = f (x) .
(x)
represents the number of days in February,
This pattern repeats over and over through time. In other words, every four years,
February is guaranteed to have the same number of days as it did 4 years earlier. The positive number 4 is
the smallest positive number that satises this condition and is called the period. A
period is the shortest
interval over which a function completes one full cyclein this example, the period is 4 and represents the
time it takes for us to be certain February has the same number of days.
A General Note:
such that f
The
period P of a repeating function f is the number representing the interval
(x + P ) = f (x)
for any value of x.
The period of the cosine, sine, secant, and cosecant functions is 2π.
The period of the tangent and cotangent functions is π.
Example 8
Finding the Values of Trigonometric Functions
.
Find the values of the six trigonometric functions of angle t based on Figure 9
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Figure 9
Solution
√
3
2
cos t = x = − 12
√
√
− 3
sint
tan t = cost
= − 12 = 3
2
1
sec t = cost
= −11 = −2
2
√
1
csc t = sint
= 1√3 = − 2 3 3
− 2
√
1
cot t = tant
= √13 = 33
sin t = y = −
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Try It:
Exercise 16
(Solution on p. 28.)
.
Find the values of the six trigonometric functions of angle t based on Figure 10
Figure 10
Example 9
Finding the√ Value of Trigonometric Functions
If sin (t)
=−
3
2 and cos (t)
= 21 ,nd sec (t) , csc (t) , tan (t) ,
Solution
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cot (t) .
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sec t =
csc t =
1
cos t
1
sin t
=
=
tan t =
sin t
cos t
=
cot t =
1
tan t
=
1
1
2
=2
1√
3
− √
2
− 23
1
2
1
√
− 3
−
√
2 3
3
(25)
√
=− 3
√
=−
3
3
Try It:
Exercise 18
If sin (t)
=
2
2 and cos (t)
(Solution on p. 28.)
√
√
=
2
2 ,nd sec (t) , csc (t) , tan (t) , and cot (t) .
5 Evaluating Trigonometric Functions with a Calculator
We have learned how to evaluate the six trigonometric functions for the common rst-quadrant angles and
to use them as reference angles for angles in other quadrants. To evaluate trigonometric functions of other
angles, we use a scientic or graphing calculator or computer software. If the calculator has a degree mode
and a radian mode, conrm the correct mode is chosen before making a calculation.
Evaluating a tangent function with a scientic calculator as opposed to a graphing calculator or computer
algebra system is like evaluating a sine or cosine: Enter the value and press the TAN key. For the reciprocal
functions, there may not be any dedicated keys that say CSC, SEC, or COT. In that case, the function must
be evaluated as the reciprocal of a sine, cosine, or tangent.
If we need to work with degrees and our calculator or software does not have a degree mode, we can
enter the degrees multiplied by the conversion factor
of 30
◦
,
π
180 to convert the degrees to radians. To nd the secant
we could press
(for a scientic calculator):
1
π COS
30 × 180
(26)
or
(for a graphing calculator):
How To:
cosecant.
1
cos
30π
180
Given an angle measure in radians, use a scientic calculator to nd the
1.If the calculator has degree mode and radian mode, set it to radian mode.
2.Enter: 1 /
3.Enter the value of the angle inside parentheses.
4.Press the SIN key.
5.Press the = key.
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Given an angle measure in radians, use a graphing utility/calculator to nd
the cosecant.
How To:
1.If the graphing utility has degree mode and radian mode, set it to radian mode.
2.Enter: 1 /
3.Press the SIN key.
4.Enter the value of the angle inside parentheses.
5.Press the ENTER key.
Example 10
Evaluating the Secant Using Technology
Evaluate the cosecant of
5π
7 .
Solution
For a scientic calculator, enter information as follows:
1 / ( 5
× π
csc
5π
7
/ 7 ) SIN =
(28)
≈ 1.279
(29)
Try It:
Exercise 20
(Solution on p. 28.)
Evaluate the cotangent of
Media:
−
π
8.
Access these online resources for additional instruction and practice with other trigono-
metric functions.
•
•
•
•
Determining Trig Function Values
1
More Examples of Determining Trig Functions
Pythagorean Identities
3
Trig Functions on a Calculator
4
1 http://openstaxcollege.org/l/trigfuncval
2 http://openstaxcollege.org/l/moretrigfun
3 http://openstaxcollege.org/l/pythagiden
4 http://openstaxcollege.org/l/trigcalc
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6 Key Equations
Tangent function
tan t =
Secant function
sec t =
Cosecant function
csc t =
Cotangent function
cot t =
sint
cost
1
cost
1
sint
1
tan t
=
cos t
sin t
Table 3
7 Key Concepts
•
The tangent of an angle is the ratio of the y -value to the x -value of the corresponding point on the
unit circle.
•
The secant, cotangent, and cosecant are all reciprocals of other functions. The secant is the reciprocal
of the cosine function, the cotangent is the reciprocal of the tangent function, and the cosecant is the
reciprocal of the sine function.
•
•
•
The six trigonometric functions can be found from a point on the unit circle. See Example 1.
Trigonometric functions can also be found from an angle. See Example 2.
Trigonometric functions of angles outside the rst quadrant can be determined using reference angles.
See Example 3.
•
•
•
•
•
•
A function is said to be even if f
(−x) = f (x)
and odd if f
(−x) = −f (x) .
Cosine and secant are even; sine, tangent, cosecant, and cotangent are odd.
Even and odd properties can be used to evaluate trigonometric functions. See Example 4.
The Pythagorean Identity makes it possible to nd a cosine from a sine or a sine from a cosine.
Identities can be used to evaluate trigonometric functions. See Example 5 and Example 6.
Fundamental identities such as the Pythagorean Identity can be manipulated algebraically to produce
new identities. See Example 7.
•
•
The trigonometric functions repeat at regular intervals.
The period P of a repeating function f is the smallest interval such that f
(x + P ) = f (x)
for any value
of x.
•
•
The values of trigonometric functions of special angles can be found by mathematical analysis.
To evaluate trigonometric functions of other angles, we can use a calculator or computer software. See
Example 10.
8 Section Exercises
8.1 Verbal
Exercise 21
On an interval of
(Solution on p. 28.)
[0, 2π) ,
can the sine and cosine values of a radian measure ever be equal? If so,
where?
Exercise 22
What would you estimate the cosine of π degrees to be? Explain your reasoning.
Exercise 23
(Solution on p. 28.)
For any angle in quadrant II, if you knew the sine of the angle, how could you determine the cosine
of the angle?
Exercise 24
Describe the secant function.
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Exercise 25
21
(Solution on p. 28.)
Tangent and cotangent have a period of π. What does this tell us about the output of these
functions?
8.2 Algebraic
For the following exercises, nd the exact value of each expression.
Exercise 26
tan π6
Exercise 27
(Solution on p. 28.)
sec π6
Exercise 28
csc π6
Exercise 29
(Solution on p. 28.)
cot π6
Exercise 30
tan π4
Exercise 31
(Solution on p. 28.)
sec π4
Exercise 32
csc π4
Exercise 33
(Solution on p. 28.)
cot π4
Exercise 34
tan π3
Exercise 35
(Solution on p. 28.)
sec π3
Exercise 36
csc π3
Exercise 37
(Solution on p. 29.)
cot π3
For the following exercises, use reference angles to evaluate the expression.
Exercise 38
tan 5π
6
Exercise 39
(Solution on p. 29.)
sec 7π
6
Exercise 40
csc 11π
6
Exercise 41
(Solution on p. 29.)
cot 13π
6
Exercise 42
tan 7π
4
Exercise 43
sec 3π
4
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(Solution on p. 29.)
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Exercise 44
csc 5π
4
Exercise 45
(Solution on p. 29.)
cot 11π
4
Exercise 46
tan 8π
3
Exercise 47
(Solution on p. 29.)
sec 4π
3
Exercise 48
csc 2π
3
Exercise 49
(Solution on p. 29.)
cot 5π
3
Exercise 50
tan 225 ◦
Exercise 51
(Solution on p. 29.)
sec 300 ◦
Exercise 52
csc 150 ◦
Exercise 53
(Solution on p. 29.)
cot 240 ◦
Exercise 54
tan 330 ◦
Exercise 55
(Solution on p. 29.)
sec 120 ◦
Exercise 56
csc 210 ◦
Exercise 57
(Solution on p. 29.)
cot 315 ◦
Exercise 58
If sin t
= 43 ,
and
t is
in quadrant II, nd cos t, sec t, csc t, tan t, cot t.
Exercise 59
If cos t
= − 13 ,
(Solution on p. 29.)
and
t is
in quadrant III, nd sin t, sec t, csc t, tan t, cot t.
Exercise 60
If tan t
=
12
5 , and 0
Exercise √61
If sin t
=
◦
π
2 , nd sin t, cos t, sec t, csc t, and
= 12 ,
nd sec t, csc t, tan t, and
≈ 0.643 cos 40 ◦ ≈ 0.766
sec 40
Exercise √63
If sin t
=
cot t.
(Solution on p. 29.)
3
2 and cos t
Exercise 62
If sin 40
≤t<
◦
cot t.
, csc 40 ◦ , tan 40 ◦ , and cot 40 ◦ .
(Solution on p. 29.)
2
2 , what is the sin (−t)?
Exercise 64
If cos t
= 12 ,
what is the cos (−t)?
Exercise 65
If sec t
= 3.1,
(Solution on p. 29.)
what is the sec (−t)?
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Exercise 66
If csc t
= 0.34,
what is the csc (−t)?
Exercise 67
If tan t
(Solution on p. 29.)
= −1.4,
what is the tan (−t)?
Exercise 68
If cot t
= 9.23,
what is the cot (−t)?
8.3 Graphical
For the following exercises, use the angle in the unit circle to nd the value of the each of the six trigonometric
functions.
Exercise 69
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(Solution on p. 29.)
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Exercise 70
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Exercise 71
25
(Solution on p. 29.)
8.4 Technology
For the following exercises, use a graphing calculator to evaluate.
Exercise 72
csc 5π
9
Exercise 73
(Solution on p. 29.)
cot 4π
7
Exercise 74
π
sec 10
Exercise 75
(Solution on p. 29.)
tan 5π
8
Exercise 76
sec 3π
4
Exercise 77
csc π4
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(Solution on p. 29.)
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Exercise 78
tan 98
◦
Exercise 79
(Solution on p. 29.)
cot 33 ◦
Exercise 80
cot 140 ◦
Exercise 81
(Solution on p. 29.)
sec 310 ◦
8.5 Extensions
For the following exercises, use identities to evaluate the expression.
Exercise 82
If tan (t)
≈ 2.7,
and sin (t)
≈ 0.94,
nd cos (t) .
and cos (t)
≈ 0.61,
nd sin (t) .
and cos (t)
≈ 0.95,
nd tan (t) .
≈ 0.5,
nd csc (t) .
Exercise 83
If tan (t)
≈ 1.3,
(Solution on p. 29.)
Exercise 84
If csc (t)
≈ 3.2,
Exercise 85
If cot (t)
≈ 0.58,
(Solution on p. 30.)
and cos (t)
Exercise 86
Determine whether the function f
Exercise 87
Determine whether the function
Exercise 88
Determine whether the function
(x) = 2sin x cos x is
even, odd, or neither.
(Solution on p. 30.)
f (x) = 3sin2 x cos x + sec x
f (x) = sin x − 2cos2 x
Exercise 89
Determine whether the function f
is even, odd, or neither.
is even, odd, or neither.
(Solution on p. 30.)
(x) = csc2 x + sec x is
even, odd, or neither.
For the following exercises, use identities to simplify the expression.
Exercise 90
csc t tan t
Exercise 91
(Solution on p. 30.)
sec t
csc t
8.6 Real-World Applications
Exercise 92
1
600 d ,
Use the equation to nd
The amount of sunlight in a certain city can be modeled by the function h
where h represents the hours of sunlight, and d is the day of the year.
nd
how many hours of sunlight there are on February 10, the 42
= 15cos
day of the year. State the period
of the function.
Exercise 93
represents the hours of sunlight, and d
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(Solution on p. 30.)
1
500 d , where h
is the day of the year. Use the equation to nd how many
The amount of sunlight in a certain city can be modeled by the function
h = 16cos
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th
hours of sunlight there are on September 24, the 267
day of the year. State the period of the
function.
Exercise 94
The equation P
onds.
= 20sin (2πt) + 100 models
the blood pressure, P, where
(a) Find the blood pressure after 15 seconds.
t
represents time in sec-
(b) What are the maximum and minimum
blood pressures?
Exercise 95
(Solution on p. 30.)
The height of a piston, h, in inches, can be modeled by the equation y
the crank angle. Find the height of the piston when the crank
= 2cos x+6, where x represents
◦
angle is 55 .
Exercise 96
The height of a piston, h,in inches, can be modeled by the equation y
= 2cos x+5,where x represents
55 ◦ .
the crank angle. Find the height of the piston when the crank angle is
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Solutions to Exercises in this Module
Solution √to Exercise
(p. 3)
√
2
2 , cos t
sin t = −
2
2 , tan t
=
= −1, sec t =
√
Solution √to Exercise (p. 5)
sin π3 =
cos π3 =
tan π3 =
sec π3
csc π3
cot π3
3
2
1
2
√
3
=2
=
=
√
2 3
3
√
3
3
Solution to Exercise
(p. 7) √
√
−7π
4
−7π
4
sin
sec
=
=
2
2 , cos
√
2, csc
−7π
4
−7π
4
=
=
2
2 , tan
√
Solution
to Exercise (p. 11)
√
−
√
2, csc t = − 2, cot t = −1
2, cot
−7π
4
−7π
4
= 1,
=1
3
Solution to Exercise (p. 12)
−2
Solution to Exercise (p. 13)
sin t
Solution to Exercise (p. 15)
8
cos t = − 17
, sin t =
csc t =
17
15 , cot t
=
15
17 , tan t
8
− 15
= − 15
8
Solution to Exercise (p. 17)
sin t = −1, cos t = 0, tan t = Undened
sec t =
Undened,csc t
= −1, cot t = 0
Solution
√ (p. 18)
√ to Exercise
sec t =
2, csc t =
2, tan t = 1, cot t = 1
Solution to Exercise (p. 19)
≈ −2.414
Solution to Exercise (p. 20)
π
4 and the terminal side of the angle is in quadrants I and III. Thus,
π 5π
,
,
the sine and cosine values are equal.
4 4
Yes, when the reference angle is
at x
=
Solution to Exercise (p. 20)
Substitute the sine of the angle in for y in the Pythagorean Theorem x
negative solution.
Solution to Exercise (p. 21)
The outputs of tangent and cotangent will repeat every π units.
Solution
to Exercise (p. 21)
√
2 3
3
Solution
to Exercise (p. 21)
√
3
Solution
to Exercise (p. 21)
√
2
Solution to Exercise (p. 21)
1
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2
+ y 2 = 1. Solve
for x and take the
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Solution to Exercise (p. 21)
2
Solution
to Exercise (p. 21)
√
3
3
Solution
to Exercise (p. 21)
√
−233
Solution
to Exercise (p. 21)
√
3
Solution
to Exercise (p. 21)
√
− 2
Solution to Exercise (p. 22)
−1
Solution to Exercise (p. 22)
−2
Solution
to Exercise (p. 22)
√
−
3
3
Solution to Exercise (p. 22)
2
Solution
to Exercise (p. 22)
√
3
3
Solution to Exercise (p. 22)
−2
Solution to Exercise (p. 22)
−1
Solution to√Exercise (p. 22)
Ifsin t
√
√
= − 2 3 2 , sec t = −3, csc t = − 3 4 2 , tan t = 2 2, cot t =
Solution to Exercise
(p. 22)
√
√
sec t = 2, csc t =
2 3
3 ,
tan t =
Solution
to Exercise (p. 22)
√
−
3, cot t =
√
2
4
√
3
3
2
2
Solution to Exercise (p. 22)
3.1
Solution to Exercise (p. 23)
1.4
Solution√ to Exercise
(p. 23)
√
sin t =
2
2 , cos t
=
2
2 , tan t
= 1, cot t = 1, sec t =
Solution √to Exercise (p. 25) √
sin t = −
3
2 , cos t
= − 12 , tan t =
Solution to Exercise (p. 25)
√
3, cot t =
0.228
Solution to Exercise (p. 25)
2.414
Solution to Exercise (p. 25)
1.414
Solution to Exercise (p. 26)
1.540
Solution to Exercise (p. 26)
1.556
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√
2, csc t =
3
3 , sec t
√
2
√
= −2, csc t = − 2 3 3
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Solution to Exercise (p. 26)
sin (t) ≈ 0.79
Solution to Exercise (p. 26)
csct ≈ 1.16
Solution to Exercise (p. 26)
even
Solution to Exercise (p. 26)
even
Solution to Exercise (p. 26)
sin t
cos t
= tan t
Solution to Exercise (p. 26)
13.77 hours, period: 1000π
Solution to Exercise (p. 27)
7.73 inches
Glossary
Denition 1: cosecant
the reciprocal of the sine function: on the unit circle, csc t
Denition 2: cotangent
= y1 , y 6= 0
the reciprocal of the tangent function: on the unit circle, cot t
Denition 3: identities
= xy , y 6= 0
statements that are true for all values of the input on which they are dened
Denition 4: period
the smallest interval P of a repeating function f such that f
(x + P ) = f (x)
Denition 5: secant
the reciprocal of the cosine function: on the unit circle, sec t
Denition 6: tangent
the quotient of the sine and cosine: on the unit circle, tan t
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= x1 , x 6= 0
= xy , x 6= 0