Trigonometric
Functions of Any
Angle
4.4
Definitions of Trigonometric
Functions of Any Angle
• Let  is be any angle in standard position, and let P = (x,
y) be a point on the terminal side of . If r = x2 + y2 is the
distance from (0, 0) to (x, y), the six trigonometric
functions of  are defined by the following ratios.
y
x
y
sin   ,
cos   ,
tan   , x  0
r
r
x
r
r
x
csc   ,y  0 sec   , x  0 cot   ,y  0
y
x
y
Example
Let P = (-3, -4) be a point on the terminal side of . Find each of the six
trigonometric functions of .
Solution The situation is shown below. We need values for x, y, and r to
evaluate all six trigonometric functions. We are given the values of x and y.
Because P = (-3, -4) is a point on the terminal side of , x = -3 and y = -4.
Furthermore,
y
5

-5
x y r
2
2
2
5
r
-5
P = (-3, -4)
x = -3
y = -4
x
Example Cont.
Solution
Now that we know x, y, and r, we can find the
six trigonometric functions of .
y 4
4
x 3
3
y 4 4
sin   
  , cos  
  , tan   

r
5
5
r
5
5
x 3 3
r
5
5
r
5
5
x 3 3
csc   
  , sec   
  , cot   

y 4
4
x 3
3
y 4 4
The bottom row shows the
reciprocals of the row above.
Example
Let tan θ = -2/3 and cos θ > 0. Find each of the six trigonometric
functions of .
x2  y2  r 2
tan  
2 y

3
x
3
cot  
2
cos  
3
3 13

13
13
sec  
13
3
sin  
 2  2 13

13
13
csc  
13
2
(2) 2  (3) 2  r 2
13  r
We have to be in
Quadrant IV
The Signs of the Trigonometric
Functions
y
Quadrant II
Sine and
cosecant
positive
Quadrant I
All functions
positive
x
Quadrant III
tangent and
cotangent
positive
Quadrant IV
cosine and
secant
positive
All Students Take Calculus
Definition of a Reference Angle
• Let  be a nonacute angle in standard
position that lies in a quadrant. Its reference
angle is the positive acute angle ´ prime
formed by the terminal side or  and the xaxis.
Example
Find the reference angle , for the following
b
angle:
 =315º
Solution:
315
a
 =360º - 315º = 45º
a
45
b
P(a, b)
Example
Find the reference angles for:
345
360  345  15
 135
 135  360  225  180  45
5
6
5 6 5 




6
6
6
6
11
4
11
3
3 
 2 
 

4
4
4
4
Using Reference Angles to Evaluate
Trigonometric Functions
• The values of a trigonometric functions of a
given angle, , are the same as the values
for the trigonometric functions of the
reference angle, ´, except possibly for the
sign. A function value of the acute angle, ´,
is always positive. However, the same
functions value for  may be positive or
negative.
A Procedure for Using Reference Angles
to Evaluate Trigonometric Functions
• The value of a trigonometric function of any
angle  is found as follows:
• Find the associated reference angle, ´, and
the function value for ´.
• Use the quadrant in which  lies to prefix
the appropriate sign to the function value in
step 1.
Example
Use reference angles to find the exact value of
the following trigonometric functions.
a. sin 135°
Solution
a. We use our two-step procedure to find sin 135°.
Step 1 Find the reference angle,  ´, and sin  ´.
135º terminates in quadrant II with a
reference angle  ´ = 180º – 135º = 45º.
y
135°
45°
x
Solution
Example cont.
The function value for the reference angle is sin 45º = 2 / 2.
Step 2 Use the quadrant in which è lies to prefix the
appropriate sign to the function value in step 1. The angle
135º lies in quadrant II. Because the sine is positive in
quadrant II, we put a + sign before the function value of the
reference angle. Thus, sin135= +sin45=2 / 2
Example
• Evaluate:
cos
4
3

cot
3
cos
cot
4

1
  cos  
3
3
2


1
1
 3
  cot  



3
3
3
3
tan
3