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Chapter 5
Trigonometric
Functions
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All rights reserved
© 2010
2011 Pearson Education, Inc. All rights reserved
1
SECTION 5.3 Trigonometric Functions of Any Angle: The Unit Circle
OBJECTIVES
1
2
3
4
5
Evaluate trigonometric functions of any angle.
Determine the signs of the trigonometric
functions in each quadrant.
Find a reference angle.
Use reference angles to find trigonometric
function values.
Define the trigonometric functions using the
unit circle.
TRIGONOMETRIC FUNCTIONS OF ANGLES
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3
DEFINITION OF THE TRIGONOMETRIC
FUNCTIONS OF ANY ANGLE θ
Let P(x,y) be any point on the terminal ray of an
angle  in standard position (other than the
2
2
origin) and let r  x  y . We define
y
sin  
r
x
cos 
r
y
tan   ,
x
x  0 
r
csc   ,
y
r
sec   ,
x
x
cot   ,
y
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y  0 
x  0 
y  0 
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EXAMPLE 1
Finding Trigonometric Function Values
Suppose  is an angle whose terminal side
contains the point P(–1,3). Find the exact
values of the six trigonometric functions of .
Solution
r x y
2

 1
2
2
 32
 10
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EXAMPLE 1
Finding Trigonometric Function Values
Solution continued
Now, with x = –1, y = 3, and r = 10, we have
y
3
3 10
sin   

r
10
10
r
10
csc   
y
3
x 1
10
cos   

r
10
10
r
10
sec   
  10
x
1
y 3
tan   
 3
x 1
x 1
1
cot   

y 3
3
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TRIGONOMETRIC FUNCTION VALUES OF
QUADRANTAL ANGLES
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COTERMINAL ANGLES
Because the value of each trigonometric function
of an angle in standard position is completely
determined by the position of the terminal side,
the following statements are true.
1. Coterminal angles are assigned identical values
by the six trigonometric functions.
2. The signs of the values of the trigonometric
functions are determined by the quadrant
containing the terminal side.
3. For any integer n, θ, and θ + n360° are coterminal
angles and θ and θ + 2πn are coterminal angles.
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TRIGONOMETRIC FUNCTION VALUES OF
COTERMINAL ANGLES
 in degrees
 in radians
sin   sin   n360
sin   sin   2n
cos  cos  n360
cos  cos  2n
These equations hold for any integer n.
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SIGNS OF THE TRIGONOMETRIC
FUNCTIONS
Suppose the angle  is not
quadrantal and its terminal
side contains the point
(x,y). The only value other
than x and y used in
defining the trigonometric
functions, r  x 2  y 2, is
always positive. Therefore,
the signs of x and y
determine the signs of the
trigonometric functions.
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EXAMPLE 4
Determining the Quadrant in Which an
Angle Lies
If tan θ > 0 and cos θ < 0, in which quadrant
does θ lie?
Solution
Because tan θ > 0, θ lies either in quadrant I or
in quadrant III. However, cos θ > 0 for θ in
quadrant I; so θ must lie in quadrant III.
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EXAMPLE 5
Evaluating Trigonometric Functions
3
Given that tan   , and cos  0, find the
2
exact value of sin  and sec  .
Solution
Since tan θ > 0 and cos θ < 0, θ lies in Quadrant
III; both x and y must be negative.
y 3 3
tan   

x 2 2
r x y 
2
2
2   3
2
2
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 4  9  13
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EXAMPLE 5
Evaluating Trigonometric Functions
Solution continued
With x  2, y  3, and r  13, we can find
sin  and sec  .
y 3
3 13
sin   

r
13
13
r
13
13
sec   

x 2
2
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DEFINITION OF A REFERENCE ANGLE
Let  be an angle in standard position that is
not a quadrantal angle.
The reference angle for  is the positive
acute angle  (“theta prime”) formed by the
terminal side of  and the x-axis.
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DEFINITION OF A REFERENCE ANGLE
Quadrant I
Quadrant II
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DEFINITION OF A REFERENCE ANGLE
Quadrant III
Quadrant IV
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EXAMPLE 6
Identifying Reference Angles
Find the reference angle  for each angle .
a.  = 250º
b.  =
c.  = 5.75
Solution
a. Because 250º lies in quadrant III,
 =   180º. So  = 250º  180º = 70º.
b. Because
lies in quadrant II,  = π  .
So  = π
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EXAMPLE 6
Identifying Reference Angles
Solution continued
c. Since no degree symbol appears in θ = 5.75,
 has radian measure. Now
≈ 4.71 and
2π ≈ 6.28. So  lies in quadrant IV and
 = 2π  . So
 = 2π – 5.75 ≈ 6.28 – 5.75 = 0.53.
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USING REFERENCE ANGLES TO FIND
TRIGONOMETRIC FUNCTION VALUES
Step 1 Assuming that  > 360º or θ < 0°, find a
coterminal angle for  with degree
measure between 0º and 360º.
Otherwise, go to Step 2.
Step 2 Find the reference angle  for the angle
resulting from Step 1. Write the
trigonometric function of  .
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USING REFERENCE ANGLES TO FIND
TRIGONOMETRIC FUNCTION VALUES
Step 3 Choose the correct sign for the
trigonometric function value of θ based
on the quadrant in which it lies. Write
the given trigonometric function of θ in
terms of the same trigonometric
function of θ with the appropriate sign.
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EXAMPLE 8
Using the Reference Angle to Find Values
of Trigonometric Functions
Find the exact value of each expression.
59
a. tan 330º
b. sec
6
Solution
a. Step 1 0º < 330º < 360º; find its reference angle.
Step 2 330º is in Q IV; its reference angle  is
   360º 330º  30º.
3
tan    tan 30º 
3
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EXAMPLE 8
Using the Reference Angle to Find Values
of Trigonometric Functions
Solution continued
Step 3 In Q IV, tan θ is negative, so
3
tan 330º   tan 30º  
.
3
b. Step 1
59 11  48 11


 8
6
6
6
59
11
is between 0 and 2π coterminal with
.
6
6
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EXAMPLE 8
Using the Reference Angle to Find Values
of Trigonometric Functions
Solution continued
11

is in Q IV; its reference angle  is
Step 2
6
11 
   2 
 .
6
6
 2 3
sec    sec 
6
3
Step 3 In Q IV, sec θ > 0; so
59
11
 2 3
sec
 sec
 sec 
.
6
6
6
3
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TRIGONOMETRIC FUNCTIONS AND THE
UNIT CIRCLE
A circle with radius 1 centered at the origin of a
rectangular coordinate system is a unit circle.
In a unit circle, s = rθ = 1·θ = θ, so the radian
measure and the arc length of an arc intercepted
by a central angle in a unit circle are
numerically identical.
The correspondence between real numbers and
endpoints of arcs on the unit circle is used to
define the trigonometric functions of real
numbers, or the circular functions.
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UNIT CIRCLE DEFINITIONS
OF THE TRIGONOMETRIC FUNCTIONS
OF REAL NUMBERS
Let t be any real number and let P = (x,y) be the
point on the unit circle associated with t. Then
sin t  y
cos t  x
y
tan t  ( x  0)
x
1
csc t  ( y  0)
y
1
sec t  ( x  0)
x
x
cot t  ( y  0)
y
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