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The Inverse
Sine, Cosine
and Tangent
Functions
Let's review a few things about inverse functions.
• To have an inverse function, a function must be one-to-one
(remember if a horizontal line intersections the graph of a
function in more than one place it is NOT one-to-one).
•If we have points on a function graph and we trade x and y
places we'll have points on the inverse function graph.
•Functions and their inverses "undo" each other so f
f 1  x
•Since x and y trade places, the domain of the function is the
range of the inverse and the range of the function is the
domain of the inverse
•The graph of a function and its inverse are reflections about
the line y = x (a 45° line).
Is y = sin x a one-to-one function?
No! A
horizontal
line
intersects
its graph
many
times.
If we only
look at part
of the sine
graph
where the x
values go
from -/2 to
/2 and the
y values go
from -1 to 1,
we could
find an
inverse
function.
If we want to find an inverse sine function, we can't have the
sine repeat itself so we are only going to look at part of the sine
graph.
We are going to define the inverse function of sine then to
be y = sin-1x. Remember for inverse functions x and y trade
places so let's take the values we got for sin x and we'll
trade them places for y = sin-1x with our restricted domain.
x
y = sin x
x
y = sin-1 x
Remember we
are only looking


at x values from

1
1 
2
2
-/2 to /2 so
that we can
 1
1

 


have a one-to6 2
2
6
one function
0 0
0 0
that will have an
inverse.

1
1 
-1
2
2 6 So for y = sin x, we can put in
6
numbers between -1 and 1.

 What we get out is the angle
1
1
2
2 between -/2 to /2 that has
that sine value.
Let's graph both of these.
Here is the graph of sin x
between -/2 to /2
x

y = sin x

1 
 1 Notice they are
2
reflections
 1 about a 45° line
 
6 2
0

6

2
0
1
2
1


2
1
Here is the graph of
from -1 to 1
y = sin-1 x
x

2
1



2
6
1
sin-1

2
x
0
0
1
2

6
1

2
Since sin x and sin-1 x are inverse functions they "undo"
each other.
f
1
 f x   sin sin x   x where
1


x

2
2
1
1
f  f x   sin sin x   x where  1  x  1
    
sin sin   
  6  6
1
CAUTION!!! You must be careful that the angle
is between -/2 to /2 to cancel these.
Looking at the unit circle, if we choose from -/2 to /2 it
would be the right half of the circle.
ItWhen
looksyou
likesee
the
1 3
 1 3




0,1
 , 
2, 2 
 2 2 
answer
inverseshould
sine then,
be it





 2 2
 2 2
2





,
11/6
means
butthey'll give
 2 , 2 
2
 2 2 



 3
3
3 
 3 1
remember
you the sine
thevalue
4
5

 , 
 3 1
 2 2
 , 
4


 2 2
6
range
and it's
orasking
what you



6
get
which
outangle
must on
bethe
an
right half
between
of the -/2
unit

0 1,0 angle
 1,0
tocircle
/2, has
so we'd
this sign
use
2
avalue.
coterminal angle
7
in this range which
6
11



3
1
5
 , 
 3 1
is:
  , 
7 6  2 2 
 2 2
4 4


3
 2
2


,

 2

2

 1
3
  , 
 2 2 


3
2
0,1
5
3
4
1
3
 , 
2 2 


 2
2


,

 2

2



 1
sin     
6
 2
1
Notice
y values
(sine
repeat
themselves
This isthe
asking,
where
on values)
the rightnever
half of
the unit
circle is in
this
half the
circle
so the sine function would be 1-to-1 here.
the sine
value
-1/2?
Remember when you have sin-1 x it means
"What angle between -/2 to /2 has a sine
value of x?"
Let's look at the unit
sin
1
2 

2
4
circle to answer this.
 1 3
 ,

 2 2 


0,1

2
3 3
5 4
6

2 2


 2 , 2 


 3 1
 , 
 2 2


1 3
 ,

2 2 



2
3
 2 2


 2 , 2 



 3 1


 2 ,2
4



6


3
sin  
  
3
 2 
1
This is 5/3 but need
 1,0

7
6
 3 1


 2 , 2 


0 1,0
2
5
4 4
3

2
2


 2 , 2 

 1
3
  ,

 2 2 


3
2
0,1
5
3
11
7 6

4

1
3
 ,

2 2 


 3 1


 2 , 2 


2
2

 2 , 2 


This
has a
a sine
This is
is asking,
asking, "What
"What angle
angle from
from -/2
-/2 to
to /2
/2 has
sine
value
2 over
2?"
value of
of square
negativeroot
square
root
3 over 2?"
 3  sin 1  2   
sin  sin

 2 
4



 4
1
Now
This remember
looks like sine
for
inverse
and its inverse
sign we should
only
use
cancel
values
out on
butthe
then
right
half
you'd
of be
thegetting
unit circle.
an
So
answer
this isout
asking,
that was
"Where
not in the
onrange
the right
so
half
you of
must
the be
circle
careful
is the
sine
on these.
value square
Let's work
root
2the
over
stuff
2?"in the
parenthesis first and
see what happens.
 1 3
 ,

 2 2 


0,1

2
3 3
5 4
6

2 2


 2 , 2 


 3 1
 , 
 2 2


1 3
 ,

2 2 



2
3
 2 2


 2 , 2 



 3 1


 2 ,2
4



6
 1,0

7
6
 3 1


 2 , 2 


0 1,0
2
5
4 4
3

2
2


 2 , 2 

 1
3
  ,

 2 2 


3
2
0,1
5
3
11
7 6

4

1
3
 ,

2 2 


 3 1


 2 , 2 


2
2

 2 , 2 


Let's think about an inverse cosine function. If we chose
the right half of the unit circle for cosine values would we
have a one-to-one function?
NO! For example:

  1
cos  cos   
3
 3 2
Let's look at the graph
of y = cos x and see
where it IS one-to-one
 1 3
 ,

 2 2 


0,1

2
3 3
5 4
6

2 2


 2 , 2 


 3 1
 , 
 2 2


1 3
 ,

2 2 



2
3
 2 2


 2 , 2 



 3 1


 2 ,2
4



6
 1,0

7
6
 3 1


 2 , 2 


0 1,0
2
5
4 4
3

2
2


 2 , 2 

 1
3
  ,

 2 2 


3
2
0,1
From 0 to 
5
3
11
7 6

4

1
3
 ,

2 2 


 3 1


 2 , 2 


2
2

 2 , 2 


So cos-1 x is the inverse function of cos x but the domain
is between -1 to 1 and the range is from 0 to .
 f x   cos cos x   x where 0  x  
1
1
f  f x   coscos x   x where - 1  x  1
f
1
1
1 
cos

2 3

3  5
1
cos  
 
 2  6
1
So this is asking where on the
upper half of the unit circle does
the cosine value equal 1/2.
 1 3
 ,

 2 2 


0,1

2
3 3
5 4
6

2 2


 2 , 2 


 3 1
 , 
 2 2


1 3
 ,

2 2 



2
3
 2 2


 2 , 2 



 3 1


 2 ,2
4



6
 1,0

7
6
 3 1


 2 , 2 


0 1,0
2
5
4 4
3

2
2


 2 , 2 

 1
3
  ,

 2 2 


3
2
0,1
5
3
11
7 6

4

1
3
 ,

2 2 


 3 1


 2 , 2 


2
2

 2 , 2 


tan-1 x is the inverse function of tan x but again we must
have restrictions to have tan x a one-to-one function.
f
1
 f  x    tan  tan x   x where 1

x

2
2
1
1
f  f  x    tan  tan x   x where -  x  
 1 3
 ,

 2 2 


0,1

2
3 3
5 4
6

2 2


 2 , 2 


 3 1
 , 
 2 2


1 3
 ,

2 2 



2
3
 2 2


 2 , 2 



 3 1


 2 ,2
4



6
 1,0

7
6
 3 1


 2 , 2 


We'll take tan x from -/2 to /2
0 1,0
2
5
4 4
3

2
2


 2 , 2 

 1
3
  ,

 2 2 


3
2
0,1
5
3
11
7 6

4

1
3
 ,

2 2 


 3 1


 2 , 2 


2
2

 2 , 2 


For help on using your calculator to
compute inverse trig functions, click
here.