Transcript Inverse Functions

The Inverse
Trigonometric
Functions
Let's again 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.
It looks
like
the
When
you
see
1 3
 1 3
 , 
0,1
 , 
2 2 
answersine
should
 2 2 
inverse
then,be
it





 2 2
 2 2
2


11/6 but



,
means
they'll give
 2 , 2 
2
 2 2 



 3
3
3 
remember
the
 3 1
you
the
sine
value
4
5

 , 
 3 1
 2 2
 , 
4
range
what you


 2 2
6
and
it's or
asking



get out
must
6
which
angle
onbe
thean
angle
between
-/2
right
half
of
the
unit

0 1,0
 1,0
to /2,
sothis
we'd
use
circle
has
sine
2
a coterminal angle
value.
7
in this range which
6
11  3 1 
5
 , 
 3 1
is:

, 
4
7 6  2 2 


2
2 
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 


Definition: The Inverse Sine Function
For  1  x  1
y  sin
1
x 
provided
sin y   x and   / 2  y   / 2.
Another notation for inverse sine is
arcsin( x) which is read "arc sine of x"
Definition: The Inverse Cosine Function
For  1  x  1
y  cos
1
x 
provided
cos y   x and 0  y   .
Another notation for inverse cosine is
arccos( x) which is read "arc cosine of x"
Definition: The Inverse Tangent Function
For any real number x,
y  tan
1
x 
provided
tan y   x and   / 2  y   / 2.
Another notation for inverse tangent is
arctan( x) which is read "arc tangent of x"
We've talked about inverse functions for sine, cosine and
tangent and saw that we must restrict the range so we
have a one-to-one function. Here is a summary:
1
sin x
Restricted to
angles on right half
of unit circle listed
from -/2 to /2
1
cos x
Restricted to
angles on upper
half of unit circle
listed from 0 to 
1
tan x
Restricted to
angles on right half
of unit circle listed
from -/2 to /2
What about the inverses of the reciprocal trig functions?
1
csc x
1
sin x
Restricted to
angles on right half
of unit circle listed
from -/2 to /2
1
sec x
1
cos x
Restricted to
angles on upper
half of unit circle
listed from 0 to 
1
cot x
1
tan x
Restricted to
angles on upper
right half
of unit
half
of circle
unit circle
listed
from -/2
listed
fromto0/2
to 
Cosecant and secant have same restrictions as their reciprocal
functions but cotangent does not. Cosecant excludes 0 since it
would be undefined there and secant excludes /2.
We've seen in this section how to work some of these if
the values are exact values on the unit circle and if not
how to use the calculator. Now we need to find exact
values even if not on the unit circle.
This means the

 1  1   o
2 2
tan  cos     

 2 2
 3  a

1
cosine of some
angle  (with
reference angle
) is -1/3 so
label adjacent
and
hypotenuse.
3
Inverse cosine is quadrants I
2 2
and II and since the cosine
 
value is negative, the
-1
terminal side must be in
drop down line to x axis
quadrant II.
The secret is to draw a triangle in the correct quadrant
and label the sides you know, figure out the other side
by Pythagorean Theorem, and then find the trig
function you need.
If you need the exact value of a reciprocal function you
can find the "flip" over the value you have and find the
inverse of the reciprocal function (being careful that
you meet the restrictions).
 1  2
sec  2  cos    
3
 2
1
1
 1 3
 , 
 2 2 


0,1

2
2
3 3
5 4
6
 2 2


 2 , 2 


This is asking what angle
(on the upper half of the
unit circle) has a cosine
value of -1/2 or in other
words, cos  = -1/2 so
what is ?
 3 1
 , 
 2 2


 1,0
1 3
 , 
2 2 



3
 2 2
 , 
 2 2 



 3 1
 , 
 2 2
4



6

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 


If you use your calculator to find one of the reciprocal
functions you use the same strategy of "flipping" over
the given value and then you can use the inverse trig
function button for the reciprocal.
1
csc 5  sin    0.20
5
1
1
Now use your calculator and compute this.
We've talked about inverse functions for sine, cosine and
tangent and saw that we must restrict the range so we
have a one-to-one function. Here is a summary:
sin 
1
Restricted to
angles on right half
of unit circle listed
from -/2 to /2
cos 
1
tan 
Restricted to
angles on upper
half of unit circle
listed from 0 to 
Restricted to
angles on right half
of unit circle listed
from -/2 to /2
1
What about the inverses of the reciprocal trig functions?
csc 
sec 
cot 
sin 
1
cos 
tan 
Restricted to
angles on upper
half of unit circle
listed from 0 to 
Restricted to
angles on upper
right half
of unit
half
of circle
unit circle
listed
from -/2
listed
fromto0/2
to 
1
1
Restricted to
angles on right half
of unit circle listed
from -/2 to /2
1
1
1
Cosecant and secant have same restrictions as their
reciprocal functions but cotangent does not.
For help on using your calculator to
compute inverse trig functions, click
here.
Also in the video
clips I show you
how to use the
calculator.
We've seen in the last section how to work some of
these if the values are exact values on the unit circle
and if not how to use the calculator. Now we need to
find exact values even if not on the unit circle.
This means the
 1  1   o
2 2
tan  cos     

 2 2
 3  a

1
x
cosine of some
angle x (with
reference angle
x’ ) is -1/3 so
label adjacent
and
hypotenuse.
3
Inverse cosine is quadrants
2 2
x’ x
I and II and since the cosine
value is negative, the
-1
terminal side must be in
drop down line to x axis
quadrant II.
The secret is to draw a triangle in the correct quadrant
and label the sides you know, figure out the other side
by Pythagorean Theorem, and then find the trig
function you need.
If you need the exact value of a reciprocal function you
can find the "flip" over the value you have and find the
inverse of the reciprocal function (being careful that
you meet the restrictions).
 1  2
sec  2  cos    
3
 2
1
1
 1 3
 , 
 2 2 


0,1

2
2
3 3
5 4
6
 2 2


 2 , 2 


This is asking what angle
(on the upper half of the
unit circle) has a cosine
value of -1/2 or in other
words, cos  = -1/2 so
what is x?
 3 1
 , 
 2 2


 1,0
1 3
 , 
2 2 



3
 2 2
 , 
 2 2 



 3 1
 , 
 2 2
4



6

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 


If you use your calculator to find one of the reciprocal
functions you use the same strategy of "flipping" over
the given value and then you can use the inverse trig
function button for the reciprocal.
1
csc 5  sin    0.20
5
1
1
Now use your calculator and compute this.
Definition: The Inverse Cosecant,
Secant and Cotangent Functions
 1
csc x   sin   for x  1
 x
1
-1
 1
sec x   cos   for x  1
 x
1
cot
1
-1
x  

2
 tan
1
x 
More on Cotangent
 -1  1 
 tan  x 

 -1  1 
1
cot x    tan    
 x


 2

for x  0
for x  0
for x  0
Function Gallery:
Inverse Trigonometric Functions