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4.4
Trigonometric Functions
of Any Angle
Objectives
 Evaluate trigonometric functions of any angle.
 Find reference angles.
 Evaluate trigonometric functions of real
numbers.
2
Introduction
3
Introduction
Recall that when using the
unit circle to evaluate the
value of a trig function, cos θ
= x and sin θ = y. What we
didn’t point out is that since
the radius (hypotenuse) is 1,
the trig values are really cos
θ = x/1 and sin θ = y/1.
So what if the radius
(hypotenuse) is not 1?
4
Introduction
The definitions of trigonometric functions were restricted to
acute angles. In this section, the definitions are extended to
cover any angle. When  is an acute angle, the definitions
here coincide with those given in the preceding section.
5
Introduction
Because r =
cannot be zero,
it follows that the sine and cosine
functions are defined for any real value
of .
However, when x = 0, the tangent and
secant of  are undefined.
For example, the tangent of 90 is
undefined. Similarly, when y = 0, the
cotangent and cosecant of  are
undefined.
6
Introduction
The previous definitions imply that tan θ and sec θ are not
defined when x = 0. So what values of θ are we talking
about?
 3
,
or 90 ̊, 270 ̊
2 2
They also imply that cot θ and csc θ are not defined when
y = 0. So what values of θ are we talking about?
0, 
or 0 ̊, 180 ̊
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Example – Evaluating Trigonometric Functions
Let (–3, 4) be a point on the terminal side of . Find the
sine, cosine, and tangent of .
Solution:
You can see that x = –3, y = 4,
and
4
-3
8
Example – Solution
cont’d
So, you have the following.
4
5
-3
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Trig of Any Angle
Let (x, y) be a point on the terminal side of an angle θ in
standard position with
2
2
r x y 0
y
sin  
y
r
csc  
cos  
x
r
r
, x0
x
x
cot   , y  0
y
y
tan   , x  0
x
r
, y0
y
sec  
x
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Your Turn:
Let θ be an angle whose terminal side contains the point
(−2, 5). Find the six trig functions for θ.
sin  
5
29
cos  
tan   
2
29
5
2
csc 
sec  
cot   
29
5
29
2
2
5
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Trig of Any Angle
The signs of the trigonometric functions in the four
quadrants can be determined from the definitions of the
functions.
For instance, because cos  = x/r, it follows that cos  is
positive wherever x  0, which is in Quadrants I and IV.
(Remember, r is always positive.)
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Trig of Any Angle
In a similar manner, you can verify the results shown.
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Example:
Given sin θ = 4/5 and tan θ < 0, find cos θ and csc θ.
x  52  42  3
cos  
csc 
3
5
5
4
5
4


-3
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Your Turn:
Given tan θ = -3/2 and sin θ < 0, find cos θ and csc θ.
Solution:
Quadrant IV – cos positive & csc negative
r
13
3
𝑟=
32 + 22 = 13
𝜃
2
𝑥
2
cos 𝜃 = =
𝑟
13
2 13
cos 𝜃 =
13
𝑟
13
csc 𝜃 = − = −
𝑦
3
13
csc 𝜃 = −
3
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Reference Angles
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Reference Angles
The values of the trigonometric functions of angles greater
than 90 (or less than 0) can be determined from their
values at corresponding acute angles called reference
angles.
𝜃’
𝜃
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Reference Angles
The reference angles for  in Quadrants II, III, and IV are
shown below.
 ′ =  –  (radians)
 ′ = 180 –  (degrees)
 ′ =  –  (radians)
 ′ =  – 180 (degrees)
 ′ = 2 –  (radians)
 ′ = 360 –  (degrees)
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Example – Finding Reference Angles
Find the reference angle  ′.
a.  = 300
b.  = 2.3
c.  = –135
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Example (a) – Solution
Because 300 lies in Quadrant IV, the angle it makes
with the x-axis is
 ′ = 360 – 300
= 60.
Degrees
The figure shows the angle  = 300
and its reference angle  ′ = 60.
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Example (b) – Solution
cont’d
Because 2.3 lies between  /2  1.5708 and   3.1416,
it follows that it is in Quadrant II and its reference angle is
 ′ =  – 2.3
 0.8416.
Radians
The figure shows the angle  = 2.3
and its reference angle  ′ =  – 2.3.
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Example (c) – Solution
cont’d
First, determine that –135 is coterminal with 225, which
lies in Quadrant III. So, the reference angle is
 ′ = 225 – 180
= 45.
Degrees
The figure shows the angle
 = –135 and its reference
angle  ′ = 45.
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Reference Angles
When your angle is negative or is greater than one
revolution, to find the reference angle, first find the
positive coterminal angle between 0° and 360° or
0 and 2𝜋.
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Your Turn:
Find the reference angle for each of the following.
1. 213°
2. 1.7 rad
3. −144°
213  180  33
  1.7  1.44
-144 ̊ is coterminal to 216 ̊
216 ̊ - 180 ̊ = 36 ̊
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Trigonometric Functions of
Real Numbers
25
Trigonometric Functions of Real Numbers
To see how a reference angle is used to evaluate a
trigonometric function, consider the point (x, y) on the
terminal side of , as shown in figure below.
opp = | y |, adj = | x |
26
Trigonometric Functions of Real Numbers
How Reference Angles Work:
y
sin  
r
sin  ' 
y
r
Same except
maybe a difference
of sign, depending
on the quadrant
the terminal side
of 𝜃 is in.
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Trigonometric Functions of Real Numbers
To find the value of
a trig function of
any angle:
1. Find the trig
value for the
associated
reference angle.
2. Pick the correct
sign depending
on where the
terminal side
lies.
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Trigonometric Functions of Real Numbers
So, it follows that sin  and sin  ′ are equal, except possibly
in sign. The same is true for tan  and tan  ′ and for the
other four trigonometric functions.
In all cases, the quadrant in which  lies determines the
sign of the function value.
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Trigonometric Functions of Real Numbers
You can greatly extend the scope of exact trigonometric
values.
For instance, knowing the function values of 30 means
that you know the function values of all angles for which
30 is a reference angle.
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Trigonometric Functions of Real Numbers
For convenience, the table below shows the exact values
of the sine, cosine, and tangent functions of special angles
and quadrant angles.
Trigonometric Values of Common Angles
31
Example – Using Reference Angles
Evaluate each trigonometric function.
a. cos
b. tan(–210)
c. csc
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Example (a) – Solution
Because  = 4 /3 lies in Quadrant III, the reference angle
is
as shown in the figure.
Moreover, the cosine is negative in
Quadrant III, so
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Example (b) – Solution
cont’d
Because –210 + 360 = 150, it follows that –210 is
coterminal with the second-quadrant angle 150.
So, the reference angle is  ′ = 180 – 150 = 30, as shown
in the figure.
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Example (b) – Solution
cont’d
Finally, because the tangent is negative in Quadrant II, you
have
tan(–210) = (–) tan 30
=
.
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Example (c) – Solution
cont’d
Because (11 /4) – 2 = 3 /4, it follows that 11 /4 is
coterminal with the second-quadrant angle 3 /4.
So, the reference angle is  ′ =  – (3 /4) =  /4, as shown
in the figure.
36
Example (c) – Solution
cont’d
Because the cosecant is positive in Quadrant II, you have
37
Your Turn:
Evaluate:
1. sin 5𝜋/3

3
2
2. cos (−60°)
1
2
3. tan 11𝜋/6
3

3
38
Your Turn:
Let θ be an angle in Quadrant III such that sin θ = −5/13.
Find a) sec θ and b) tan θ using trig identities.
sin   cos   1
2
2
13
sec 𝜃 = −
12
2
 5
2


cos
 1


 13 
25
169
144
cos 2  
169
12
12
cos  , Q  III  cos  
13
13
cos 2   1 
5
5
tan   13 
12 12

13

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Assignment:
Pg. 294-296: #1 – 107 odd, 111.
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