CHAPTER 6: INVERSE CIRCULAR FUNCTIONS AND TRIGONOMETRIC
EQUATIONS
6.1
INVERSE CIRCULAR FUNCTIONS

Horizontal Line Test
o Any horizontal line will intersect the graph of a one-to-one function in at
most one point

Inverse Function
o The inverse function of the one-to-one function f is defined as
f 1 

 y, x   x, y   f 
Summary of Inverse Functions
1.
In a one-to-one function, each x-value corresponds to only one y-value and
each y-value corresponds to only one x-value.
2.
If a function f is one-to-one, then f has an inverse function f 1.
3.
The domain of f is the range of f 1 , and the range of f is the domain of
f 1.
4.
The graphs of f and f 1 are reflections of each other across the line
y  x.
5.
To find f 1 ( x) from f ( x), follow these steps:
Step 1 Replace f ( x) with y and interchange x and y.
Step 2 Solve for y.
Step 3 Replace y with f 1 ( x).

Inverse Sine Function
o Since the graph of f ( x)  sin x is not one-to-one, we restrict the domain
  
to   , 
 2 2

This interval contains enough of the graph of the sine function to
include all possible values of y.

This interval is an accepted convention that is adopted by scientific
and graphing calculators.
o The inverse circular functions are used in calculus to solve certain types of
related rates problems and to integrate certain rational functions
x


2


4
0

4

2
x
-1

2
2
0
2
2
1
y=sinx
-1

(x,y)
 

  , 1 
 2

0
 
2
  , 

2 
 4
(0, 0)
2
2
 2 
 ,

4 2 
1
 
 ,1
2 
2
2
y  sin 1 x
(x,y)
or
y  arcsin x


2


4


 1,  
2

0

4

2

2 
,  
 
4
 2
(0, 0)
 2 
, 

2
4

 
 1, 
 2

The graph is continuous over the entire domain,

Its x-intercepts are of the form

Its period is

The graph is symmetric with respect to the
origin, so the function is an odd function. For all
x in the domain, sin( x)   sin( x).
 ,  

n ,  n  .
2 .
The inverse sine function is increasing and
continuous on its domain
1,1

Its x-intercept is 0, and its y-intercept is 0.

The graph is symmetric with respect to the
origin, so the function is an odd function.

Inverse Cosine Function
o Since the graph of f ( x)  cos x is not one-to-one, we restrict the domain
to  0,  

This interval contains enough of the graph of the cosine function to
include all possible values of y.

This interval is an accepted convention that is adopted by scientific
and graphing calculators.
x
0

4
y=cosx
1
(x,y)
(0,1)
2
2

2
3
4
0
 2 
 ,

4 2 
 
 ,0
2 

2
2
-1

x
 3
2
 , 

2 
 4
 , 1
y  cos1 x

The graph is continuous over the entire domain,

Its x-intercepts are of the form
 ,  
 2n  1

2
,  n  .
2 .

Its period is

The graph is symmetric with respect to the yaxis, so the function is an even function. For all
x in the domain, cos(  x)  cos( x).
(x,y)
or
y  arccos x
1
2
2
0

2
2
-1
0

4

2
3
4

(1, 0)
 2 
, 

2
4

 
 0, 
 2

2 3 
, 
 
2
4 

 1,  

The inverse cosine function is decreasing
and continuous on its domain  1,1

Its x-intercept is 1, and its y-intercept is


2
Its graph is not symmetric with respect to
the y-axis or the origin.
.

Inverse Tangent Function
 Domain is  ,   .




  
Range is   ,  .
 2 2
The inverse tangent function is increasing and continuous
on its domain  ,   .
Its x-intercept is 0, and its y-intercept is 0.


The lines y  and y   are horizontal asymptotes.
2
2
6.2
TRIGONOMETRIC EQUATIONS I

Solving by Linear Methods
o Conditional equations (equations that are satisfied by some values but not
others) with trigonometric or circular functions can usually be solved
using algebraic methods and trigonometric identities.

Example: Solve each equation for exact solutions over the interval
0, 2  .
Solve: sin x  2  3
Solution:
sin x  2  3
sin x  1
x

2
(subtract 2 from each side)
 
 
2
o Solving by Factoring

Example: Solve each equation for exact solutions over the interval
0, 2  .
Solve: 2 cos 2 x  cos x  1
Solution:
2 cos 2 x  cos x  1
2 cos 2 x  cos x  1  0 (subtract 1 from both sides)
 2 cos x  1 cos x  1  0
(factor)
2 cos x  1  0 or cos x  1  0 (zero property)
1
or cos x  1
2
7 11
x
,
or x  0
6 6
 7 11 
,
0,

 6 6 
cos x  
o Solving by Quadratic Methods

Example: Solve each equation for exact solutions over the interval
0,360  .
Solve: 3cot 2   3cot   1  0
Solution:
3cot 2   3cot   1  0
cot  

(3)  (3) 2  4(3)(1)
2(3)
3  9  12
6
3  21
6
So, cot   1.2638 or cot   0.2638

1
1
 0.7913 or tan  
 3.7913
1.2638
0.2638
Evaluting the inverse tangent we have   38.4 or   180  38.4  218.4
and tan  
Evaluting the inverse tangent we have   75.2 =284.8 or   180  75.2  104.8

Solving a Trigonometric Function by Squaring
o Example: Solve each equation for exact solutions over the interval
0,360  .

Solve: sin   cos  1
sin   cos   1
 sin   cos  
2
 1
2
sin 2   2sin  cos   cos 2   1
 sin
2
  cos 2    2sin  cos   1
1  2sin  cos   1
2sin  cos   0
sin  cos   0
sin   0 or cos   0
  0 ,180 or   90 , 270

Solving a Trigonometric Function by Using Identities
o Example: Solve each equation for exact solutions over the interval
0, 2  .

Solve: tan x  1  3  3 cot x
Solution:
tan x  1  3  3 cot x
tan x  1  3 
3
tan x
tan 2 x  tan x  3 tan x  3
tan 2 x  tan x  3 tan x  3  0
tan x  tan x  1  3  tan x  1  0
 tan x  3   tan x  1  0
tan x  3 or tan x  1
 4
3 7
x ,
or x 
,
3 3
4 4

Solving a Trigonometric Equation
1.
2.
3.
4.
5.
6.3
Decide whether the equation is linear or quadratic in form, so you can
determine the solution method.
If only one trigonometric function is present, solve the equation for that
function.
If more than one trigonometric function is present, rearrange the equation
so that one side equals zero. Then try to factor and set each factor equal to
zero to solve.
If the equation is quadratic in form, but not factorable, use the quadratic
formula. Check that solutions are in the desired interval.
Try using identities to change the form of the equation. It may be helpful
to square both sides of the equation first. If this is done, check for
extraneous solutions.
TRIGONOMETRIC EQUATIONS II

Solving an Angle Using a Half-Angle Identity
x
x
o Example: Solve sin  cos over the interval 0, 2  .
2
2
Solution:
sin
x
x
x
 cos  0  x  2  0   
2
2
2
2
x 
x

 sin    cos 
2 
2

x
x
sin 2  cos 2
2
2
x
x
sin 2  1  sin 2
2
2
x
2sin 2  1
2
x
1
sin 2 
2 2
x
1
sin 
2
2
sin
x
2

2
2
2
x 
2 
 x

2 4
4
2