π
a.
b. 15°
c. 160°
d. 7
e. 340°
4
6
17. Find an angle θ with 0 < θ < 360° or 0 < θ < 2π that has the same cosine value as: π
π
a.
b. 80°
c. 140°
d. 4
e. 305°
3
3
18. Find an angle θ with 0 < θ < 360° or 0 < θ < 2π that has the same cosine value as: π
π
a.
b. 15°
c. 160°
d. 7
e. 340°
4
6
19. Find the coordinates of the point on a circle with radius 15 corresponding to an angle
of 220°.
20. Find the coordinates of the point on a circle with radius 20 corresponding to an angle
of 280°.
21. Marla is running clockwise around a circular track. She runs at a constant speed of 3
meters per second. She takes 46 seconds to complete one lap of the track. From her
starting point, it takes her 12 seconds to reach the northernmost point of the track. Impose
a coordinate system with the center of the track at the origin, and the northernmost point
on the positive y-axis. [UW]
a) Give Marla’s coordinates at her starting point.
b) Give Marla’s coordinates when she has been running for 10 seconds.
c) Give Marla’s coordinates when she has been running for 901.3 seconds.
Section 5.4 The Other Trigonometric Functions 333
Section 5.4 The Other Trigonometric Functions
In the previous section, we defined the sine and cosine functions as ratios of the sides of a
right triangle in a circle. Since the triangle has 3 sides there are 6 possible combinations
of ratios. While the sine and cosine are the two prominent ratios that can be formed,
there are four others, and together they define the 6 trigonometric functions.
Tangent, Secant, Cosecant, and Cotangent Functions
For the point ( x, y) on a circle of radius r at an angle of θ , we
can define four additional important functions as the ratios of the
( x, y)
sides of the corresponding triangle:
The tangent function:
y
tan(θ ) =
r
x
y
θ
The secant function:
r
sec(θ ) =
x
x
The cosecant function:
r
csc(θ ) =
y
The cotangent function:
x
cot(θ ) =
y
Geometrically, notice that the definition of tangent corresponds with the slope of the line
segment between the origin (0, 0) and the point ( x, y). This relationship can be very
helpful in thinking about tangent values.
You may also notice that the ratios defining the secant, cosecant, and cotangent are the
reciprocals of the ratios defining the cosine, sine, and tangent functions, respectively.
Additionally, notice that using our results from the last section,
y
r sin(θ )
sin(θ )
tan(θ ) = =
=
x r cos(θ ) cos(θ )
Applying this concept to the other trig functions we can state the other reciprocal
identities.
Identities
The other four trigonometric functions can be related back to the sine and cosine
functions using these basic relationships:
sin(θ )
θ
tan(θ ) =
1
sec(θ ) =
1
csc(θ ) =
1
cos( )
cot(θ ) =
=
cos(θ )
cos(θ )
sin(θ )
tan(θ ) sin(θ )
334 Chapter 5
These relationships are called identities. Identities are statements that are true for all
values of the input on which they are defined. Identities are usually something that can
be derived from definitions and relationships we already know, similar to how the
identities above were derived from the circle relationships of the six trig functions. The
Pythagorean Identity we learned earlier was derived from the Pythagorean Theorem and
the definitions of sine and cosine. We will discuss the role of identities more after an
example.
Example 1
π
Evaluate tan(45 )
° and
5
sec
.
6
Since we know the sine and cosine values for these angles, it makes sense to relate the
tangent and secant values back to the sine and cosine values.
2
sin(45 )
°
2
tan(45 )
° =
=
= 1
cos(45 )
°
2 2
Notice this result is consistent with our interpretation of the tangent value as the slope
of the line passing through the origin at the given angle: a line at 45 degrees would
indeed have a slope of 1.
5π
1
1
− 2
−
sec
2 3
=
=
=
, which could also be written as
.
6
5π − 3
3
cos
3
2
6
Try it Now π
1. Evaluate
7
csc
.
6
Just as we often need to simplify algebraic expressions, it is often also necessary or
helpful to simplify trigonometric expressions. To do so, we utilize the definitions and
identities we have established.
Section 5.4 The Other Trigonometric Functions 335
Example 2 sec(θ)
Simplify
.
tan(θ )
We can simplify this by rewriting both functions in terms of sine and cosine
sec(
1
θ )
cos(θ )
=
To divide the fractions we could invert and multiply
tan(θ ) sin(θ )cos(θ)
1 cos(θ )
=
cancelling the cosines,
cos(θ ) sin(θ )
1
=
( = csc
simplifying and using the identity
sin θ )
(θ )
sec(θ )
By showing that
can be simplified to c (
sc θ ), we have, in fact, established a new
tan(θ )
sec(θ )
identity: that
( = csc .
tan θ )
(θ )
Occasionally a question may ask you to “prove the identity” or “establish the identity.”
This is the same idea as when an algebra book asks a question like “show that
( x − )
1 2
2
= x − 2 x +1.” In this type of question we must show the algebraic
manipulations that demonstrate that the left and right side of the equation are in fact
equal. You can think of a “prove the identity” problem as a simplification problem where
you know the answer: you know what the end goal of the simplification should be, and
just need to show the steps to get there.
To prove an identity, in most cases you will start with the expression on one side of the
identity and manipulate it using algebra and trigonometric identities until you have
simplified it to the expression on the other side of the equation. Do not treat the identity
like an equation to solve – it isn’t! The proof is establishing if the two expressions are
equal, so we must take care to work with one side at a time rather than applying
an operation simultaneously to both sides of the equation.
Example 3
+
α
Prove the identity 1 cot( ) = sin(α) + cos(α) .
csc(α)
Since the left side seems a bit more complicated, we will start there and simplify the
expression until we obtain the right side. We can use the right side as a guide for what
might be good steps to make. In this case, the left side involves a fraction while the
right side doesn’t, which suggests we should look to see if the fraction can be reduced.
336 Chapter 5
Additionally, since the right side involves sine and cosine and the left does not, it
suggests that rewriting the cotangent and cosecant using sine and cosine might be a
good idea.
1+cot(α) Rewriting the cotangent and cosecant
csc(α)
cos(α)
1+ sin(α)
=
1
To divide the fractions, we invert and multiply
sin(α)
cos(α) sin(α)
= 1+
Distributing,
sin(α)
1
sin(α) cos(α) sin(α)
= 1⋅
+
⋅
Simplifying the fractions,
1
sin(α)
1
= sin(α) + cos(α)
Establishing the identity.
α
Notice that in the second step, we could have combined the 1 and cos( ) before
sin(α)
inverting and multiplying. It is very common when proving or simplifying identities for
there to be more than one way to obtain the same result.
We can also utilize identities we have previously learned, like the Pythagorean Identity,
while simplifying or proving identities.
Example 4
cos2 (θ )
Establish the identity
= −
.
1+ sin(
1 sin
θ )
(θ )
Since the left side of the identity is more complicated, it makes sense to start there. To
simplify this, we will have to reduce the fraction, which would require the numerator to
have a factor in common with the denominator. Additionally, we notice that the right
side only involves sine. Both of these suggest that we need to convert the cosine into
something involving sine.
Recall the Pythagorean Identity told us cos2 (θ ) + sin2 (θ ) = 1. By moving one of the
trig functions to the other side, we can establish:
sin2 (θ ) = 1− cos2 (θ )
and
cos2 (θ ) = 1− sin2 (θ )
Utilizing this, we now can establish the identity. We start on one side and manipulate:
Section 5.4 The Other Trigonometric Functions 337
cos2 (θ )
Utilizing the Pythagorean Identity
1+ sin(θ )
1− sin2 (θ )
=
Factoring the numerator
1+ sin(θ )
(1− sin(θ ) (1+ sin(θ )
=
Cancelling the like factors
1+ sin(θ )
= 1− sin(θ )
Establishing the identity
We can also build new identities from previously established identities. For example, if
we divide both sides of the Pythagorean Identity by cosine squared (which is allowed
since we’ve already shown the identity is true),
cos2 (θ ) + sin2 (θ )
1
=
Splitting the fraction on the left,
cos2 (θ )
cos2 (θ )
cos2 (θ ) sin2 (θ )
1
+
=
Simplifying and using the definitions of tan and sec
cos2 (θ ) cos2 (θ ) cos2 (θ )
1+ tan2 (θ ) = sec2 (θ ) .
Try it Now
2. Use a similar approach to establish that cot2 (θ ) +1 = csc2 (θ ) .
Identities
Alternate forms of the Pythagorean Identity
1+ tan2 (θ ) = sec2 (θ )
cot2 (θ ) +1 = csc2 (θ )
Example 5
If
2
tan(θ ) = and θ is in the 3rd quadrant, find cos(θ ) .
7
There are two approaches to this problem, both of which work equally well.
Approach 1
Since
y
tan(θ ) = and the angle is in the third quadrant, we can imagine a triangle in a
x
circle of some radius so that the point on the circle is (-7, -2). Using the Pythagorean
Theorem, we can find the radius of the circle:
2
2
2
( 7
− ) + (− )
2 = r , so r = 53 .
338 Chapter 5
Now we can find the cosine value:
x
− 7
cos(θ ) = =
r
53
Approach 2
Using the 1+ tan2 (θ ) = sec2 (θ ) form of the Pythagorean Identity with the known
tangent value,
1+ tan2 (θ ) = sec2 (θ )
2 2
1
+ = sec2 (θ )
7
53 =sec2(θ)
49
53
53
sec(θ ) = ±
= ±
49
7
Since the angle is in the third quadrant, the cosine value will be negative so the secant
value will also be negative. Keeping the negative result, and using definition of secant,
53
sec(θ ) = −
7
1
53
= −
Inverting both sides
cos(θ )
7
7
7 53
cos(θ ) = −
= −
53
53
Try it Now
π
3. If
7
sec(φ) = − and < φ < π , find tan(φ) and sin(φ) .
3
2
Important Topics of This Section
6 Trigonometric Functions:
Sine
Cosine
Tangent
Cosecant
Secant
Cotangent
Trig identities
Section 5.4 The Other Trigonometric Functions 339
Try it Now Answers
1. -2
2.
cos2 (θ ) + sin2 (θ ) =1
sin2 θ
2
2
θ
θ
cos ( ) sin ( )
1
+
=
sin2 (θ ) sin2 (θ ) sin2 (θ )
cot2 (θ ) +1 = csc2 (θ )
3.
40
sin(φ) =
40
tan(φ) =
7
− 3
340 Chapter 5
Section 5.4 Exercises
π
1. If
θ = , find exact values for sec(θ ),csc(θ ), tan(θ ), cot (θ ) .
4
π
2. If
7
θ =
, find exact values for sec(θ ),csc(θ ), tan (θ ), cot (θ ) .
4
π
3. If
5
θ =
, find exact values for sec(θ ),csc(θ ), tan (θ ), cot (θ ) .
6
π
4. If
θ = , find exact values for sec(θ ),csc(θ ), tan(θ ), cot (θ ) .
6
π
5. If
2
θ =
, find exact values for sec(θ ),csc(θ ), tan (θ ), cot (θ ) .
3
π
6. If
4
θ =
, find exact values for sec(θ ),csc(θ ), tan (θ ), cot (θ ) .
3
7. Evaluate: a. sec(135°) b. csc(210°) c. tan (60°) d. cot (225°)
8. Evaluate: a. sec(30°) b. csc(315°) c. tan (135°) d. cot (150°)
9. If
(θ ) 3
sin
= , and θ is in quadrant II, find cos(θ ), sec(θ ),csc(θ ), tan (θ ), cot (θ ).
4
10. If
(θ ) 2
sin
= , and θ is in quadrant II, find cos(θ ), sec(θ ),csc(θ ), tan (θ ), cot (θ ).
7
11. If
(θ )
1
cos
= − , and θ is in quadrant III, find
3
sin (θ ), sec(θ ),csc(θ ), tan (θ ), cot (θ ) .
12. If
(θ ) 1
cos
= , and θ is in quadrant I, find sin (θ ), sec(θ ),csc(θ ), tan (θ ), cot (θ ) .
5
π
13. If
(θ ) 12
tan
=
, and 0 ≤ θ < , find sin (θ ), cos(θ ),sec(θ ), csc(θ ), cot (θ ) .
5
2
π
14. If tan (θ ) = 4, and 0 ≤ θ < , find sin (θ ), cos(θ ),sec(θ ), csc(θ ), cot (θ ) .
2
Section 5.4 The Other Trigonometric Functions 341
15. Use a calculator to find sine, cosine, and tangent of the following values:
a. 0.15
b. 4
c. 70°
d. 283°
16. Use a calculator to find sine, cosine, and tangent of the following values:
a. 0.5
b. 5.2
c. 10°
d. 195°
Simplify each of the following to an expression involving a single trig function with no
fractions.
17. csc( t) tan ( t)
18. cos( t)csc( t)
sec( t)
19.
csc( t)
cot ( t)
20.
csc( t)
sec( t) − cos( t)
21.
sin ( t)
tan ( t)
22.
sec( t) − cos( t)
1+ cot ( t)
23.
1+ tan ( t)
1+ sin ( t)
24.
1+ csc( t)
2
sin ( t)
2
+ cos ( t)
25.
2
cos ( t)
2
1− sin ( t)
26.
2
sin ( t)
342 Chapter 5
Prove the identities.
2
sin (θ )
27.
= −
θ
1+ cos(θ ) 1 cos( )
28.
2
1
tan ( t) =
−1
2
cos ( t)
29. sec( a) − cos( a) = sin ( a) tan ( a)
2
1+ tan ( b)
30.
2
= csc ( b)
2
tan ( b)
2 ( x)
2
csc
− sin ( x)
31.
=
x
x
csc( x) + sin ( x)
cos( )cot ( )
sin (θ ) − cos(θ )
32.
=
θ
θ
sec(θ ) − csc(θ ) sin ( )cos( )
2
csc (α ) −1
33.
=1+ sin α
2
csc (α ) − csc(α )
( )
34. 1+ cot ( x) = cos( x)(sec( x) + csc( x))
1+ cos( u)
sin ( u)
35.
=
sin ( u)
1− cos( u)
1− sin t
36.
2 ( t)
( )
1
2sec
=
+
2
cos ( t) 1−sin ( t)
4 (γ )
4
sin
− cos (γ )
37.
=
γ +
γ
sin (γ ) − cos(γ )
sin ( ) cos( )
(1+cos( A))(1−cos( A))
38.
=
A
sin ( A)
sin ( )
Se