Sect. 4-7
Inverse Trig Functions
But first, let’s start with two non-trig functions that
are inverses.
f ( x)  x 2
f 1 ( x)  x
But something is not quite right
with this pair. Do you know
what is wrong?
The graph of x2 will not pass the horizontal line test
and does not really have an inverse because it is not
1-1.
2
y

x
.
Consider the graph of
y

f(2) = 4 and f(-2) = 4
so what is an inverse
function supposed
to do with 4?


f 1 (4)  2 or f 1 (4)  2 ?




By definition of an inverse function, two different
outputs cannot be generated for the same input, so as
2
is, y  x does not have an inverse.
x
If we consider only half of the graph for x > 0,
it now passes the
2
y

x
horizontal line test
y=x
and we do have an
4
inverse:
f ( x)  x for x  0
2
y x
2
1
f ( x)  x




Note how each graph reflects across the line y = x onto
its inverse.
x
A similar restriction on the domain is necessary to
create an inverse function for each trig function.
y
y = sin(x)

y = 1/2









x

We are going to build the inverse function of y = sinx
from the red highlighted section of the sine curve.
This section picks up
all the outputs of the
sine from –1 to 1.
y

It includes:
• the origin
• Quadrant I angles
that generate the
positive ratios
• negative angles in
Quadrant IV that
generate the negative
ratios.









x

The special angles on the curve are plotted.
x





2

3

4

6
0

6

4

3

2
f ( x)
1
3

2
2

2
1

2
0
1
2
2
2
3
2
1
y
y = sin(x)







x

x





2

3

4

6
0

6

4

3

2
sin( x)
1
3

2
2

2
1

2
0
1
2
2
2
3
2
1
The new
table that
generates
the graph
of the
inverse is
found by
simply
switching
the x- and
y-values.
We call
sine’s
inverse
arcsine.
x
sin 1 ( x)
1

3
2
2

2
1

2
0
1
2
2
2
3
2


1



2

3

4

6
0

6

4

3

2
The domain of
the chosen
section of
sine is   ,   ,
 2 2
so the range
of arcsine is
  
  2 , 2 
.
The range of
the chosen
section of
sine is [-1 ,1],
so the domain
of arcsine is
[-1, 1].
Note how each point on the original graph gets “reflected”
across the line y = x.

 ,
2
  
1 to 1, 
  2
y = arcsin(x)
y
y = sin(x)
 3   3  
 ,
 

 3 2  to  2 , 3 

 


 2   2  
 ,
 

 4 2  to  2 , 4 

 



etc.
You will see the
inverse listed
as both:
arcsin( x) and sin 1 ( x)

x
Examples:



1

sin
arcsin(1)  or sin (1) 
  1
2
2
2
 3 
 3
3

1

sin

arcsin 
  or sin 

 
2
3
2
3 2




Unless you are
instructed to
use degrees,
assume that
inverse trig
functions will
output real
numbers (in
radians).
Note: For the trig function, the input is the angle and the
output is the ratio, but for the inverse trig function, the
input is the ratio and the output is the angle.
You Try…
Use the unit circle to
answer the following.
1
arcsin   
2
 
30  or 
 6
o
1
because sin  30  
2
o


3
o
sin  
  60  or  
3

 2 
1
3
because sin  60   
2
o
The other inverse trig functions are generated by using
similar restrictions on the domain of the trig function.
What do you
think would be
a good domain
restriction for
the cosine?
Congratulations if
you realized that
the restriction we
used on the sine
is not going to
work on the
cosine.
y = cos(x)


y








x

The chosen section for the cosine is in the red frame. This
section includes all outputs from –1 to 1 and all inputs in
the first and second quadrants.
Since the domain and range for the section are 0,   and 1,1,
the domain and range for the inverse cosine are 1,1 and 0 ,  .
y = cos(x)
y
y
y = arccos(x)













x






x
y
y = arccos(x)
Examples:








To solve arccos(-1) determine
which angle in QI or QII whose
cosine is -1:
arccos(-1) =

x
You Try…

2
 2
o
o
45
(
or
)
because
sin
45

arccos 
 
4
2
2


 2
arccos 
  135o (or 3 ) because cos 135o    2
4
2
 2 
The other trig functions require similar restrictions on
their domains in order to generate an inverse.
Like the sine function, the domain of the section of the
  
tangent that generates the arctan is  ,  .
 2 2
y
y
y=arctan(x)


y=tan(x)








x









  
D    ,  and R   ,  
 2 2
  
D   ,   and R    , 
 2 2

x

OK, lets try a few more.
tan 1 ( 3) 
tan 1 ( 3) 

3
60 (or ) because tan 60 
 3
3
1

3
60o (or  ) because tan  60o  
 3
3
1
o
You Try…
tan 1 (0) 
tan 1 (1) 
o
The table below will summarize the parameters we have
so far. Remember, the angle is the input for a trig function
and the ratio is the output. For the inverse trig functions
the ratio is the input and the angle is the output.
arcsin(x)
Domain
Range
arccos(x) arctan(x)
1  x  1 1  x  1   x  


2
x

2


0 x   x
2
2
When x<0, y=arcsin(x) will be in which quadrant?
y<0 in IV
When x<0, y=arccos(x) will be in which quadrant? y>0 in II
When x<0, y=arctan(x) will be in which quadrant? y<0 in IV
Finding an angle.
(Figuring out which ratio to use and getting to
use the 2nd button and one of the trig buttons.)
Ex: Find . Round to four decimal places.
17.2
tan  
9
2nd tan
17.2

9
)
17.2

9
  62.3789
Make sure you are in degree mode
(not radians).
Ex: Find . Round to three decimal places.

23
7
7
cos  
23
2nd cos
7

23
  72.281
)
Ex: Find . Round to three decimal places.

200
sin  
400
200
nd
2
sin
200

  30
400 )
When trying to find a side,
use sin, cos, or tan.
When trying to find an angle,
use sin-1, cos-1, or tan-1.
Definition of an
Inverse Function
A function, f, has an inverse
function, g, if and only if
f(g(x)) = x and g(f(x)) = x, for
every x in domain of g and
in the domain of f.
Composition of Trig functions
with Inverse Trig Functions.
 1   3  
sin  sin 
   ?

2 
 44444
1444442
43
so
 1   3  
 3




sin sin 
 sin  



2
2




First, what do we know about
 ?
We know that  is an angle whose sine is
 3
.
2
Did you suspect from the beginning that this was the
answer because that is the way inverse functions are
SUPPOSED to behave? If so, good instincts but….
Consider a slightly different setup:
 

arcsin sin 120 
 3
  60.
arcsin 

2


This is also the
composition of two
inverse functions but…
Did you suspect the answer was going to be 120
degrees? This problem behaved differently because
the first angle, 120 degrees, was outside the range of
the arcsin. So use some caution when evaluating the
composition of inverse trig functions.
The table below will summarize the properties for
composition of functions.
If
1  x  1 and    y   , then
If
1  x  1 and 0  x   , then
2
2
sin(arcsin x) = x and arcsin(sin y) = y.
cos(arccos x) = x and arccos(cos y) = y.


If x is a real number and   y  , then
2
2
tan(arctan x) = x and arctan(tan y) = y.
Examples
If possible, find the
exact value.
cos(cos 1  1)  1
  5  

arcsin  sin 
  
3
  3 
tan  arctan  5   5
Example:
If
 5
y  arc sec 
 find tan y.
 2 
Solution: For this problem we use the right triangle
5
1
By the Pythagorean Theorem,
this makes the opposite side.
2
5  2  5 4  1 1
2
2
tan(y) =
1
2
Example:
3

Let arctan
  x , find sec x.
 5 
3

 3  
sec  arctan     sec( x)
tan x 
5
 5 

Solution: Use the right triangle on the coordinate graph
Now using the triangle we can
find sec x after we find the hyp.
 32  52
34
 9  25  34
34
sec( x ) 
5
Example:


Find the exact value of tan arccos 2 .
3
adj 2
2
Let u = arccos , then cos u 
 .
3y
hyp 3
3
32  22  5
u


2
opp
2
tan arccos  tan u 
 5
3
adj
2
x
Example:
• Calculate sin (arctan 4/3).
You Try…
• Calculate sec[arctan(-3/5)]
Find the exact
value of each
expression
without a
calculator. If
your answer
is an angle,
express it in
radians.
 1 
1. sin 1  
 2 
2. arcsin  1
10. sec 1 2 
 1 
11. arccos

 2
3. tan 1  1
 1 
4. arctan 

 3
5. arcsin  0 
   
12. arcsin  sin    
  2 
 1 
6. cos  

2



 1  
14. tan  arccos  
 2 

1

7. arctan  3

8. sin 1  1
 3
9. cos 

2


1
 
13. arcsin sin 270

  
15. arccos cos
  3

 

 1  1  
16. sin  cos    
 2 

Answers
 1   
1. sin   
6
 2 
2. arccos  1  
  3  5

9. cos 
 6
2


1

3. tan  1 
4
 1  
4. arctan 

 3 6
1
5. arcsin 0   0
 1  
6. cos 

 2 4

7. arctan  3 
3

1
8. sin  1 
2
1
Negative ratios for arccos
generate angles in Quadrant II.
y
1
2
 3

x
1


The reference angle is


6
6  5
 
so the answer is     
6
6 6
6
10. sec 1 2   cos 1 1 / 2   

11. arccos

y
3
 1  3

2 4
2
3
60
   

12. arcsin  sin      
2
  2 
 

  
15. arccos cos
  3
x


1 
   arccos  

2 3
 1  1  
 2
16. sin  cos      sin 
 2 
 3


-1

13. arcsin sin 270  arcsin  1  90 
2

 1  
 2 
14. tan  arccos    tan 
 3
y
 2 
 3 


14.
3


 2
15.
1
x

2
 3