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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 trig values are really cos
θ = x/1 and sin θ = y/1.
(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
 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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Let θ be an angle whose terminal side contains the point
(−2, 5). Find the six trig functions for θ.
sin  
5
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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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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
=
32 + 22 = 13

2

2
cos  = =

13
2 13
cos  =
13

13
csc  = − = −

3
13
csc  = −
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.
’

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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.
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 2.
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Find the reference angle for each of the following.
1. 213°
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 |
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Trigonometric Functions of Real Numbers
How Reference Angles Work:
sin  
y
r
sin  ' 
y
r
Same except
maybe a difference
of sign, depending
the terminal side
of  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
Trigonometric Values of Common Angles
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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
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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.
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Example (c) – Solution
cont’d
Because the cosecant is positive in Quadrant II, you have
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Evaluate:
1. sin 5/3

3
2
2. cos (−60°)
1
2
3. tan 11/6
3

3
38
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  = −
12
2
 5
2


cos
 1


 13 
25
169
144
cos 2  
169
12
12
cos  , Q  III  cos  
13
13
cos2   1 
5
5
tan   13 
12 12

13

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Assignment:
Pg. 294-296: #1 – 107 odd, 111.
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