Download 2.11 Inverse Trigonometric Functions

Section 2.11 Notes Page 1
2.11 Inverse Trigonometric Functions
1
. We put in an angle and get a value as a result. In inverse trig
2
1
functions we put in the value and get an angle: sin 1  30  . So here we put in the value of one half and got
2
30 degrees as a result. We are not allowed to put any number into our inverse trig functions. There are
restrictions on the domain that are given in the following table:
From trigonometry we know that sin 30  
1
y  sin x
Domain
1  x  1
y  cos x
1  x  1
y  tan1 x
  x  
1
y  cot x
  x  
y  sec1 x
x 1
1
1
y  csc x
x 1
Range



y

2
2
0 y 

y

2
2
0 y 
0 y  , y 


2
y

2

2
, y0
We are not going to look at the graphs of the inverse trig functions in this class, however that is where the
domain and range is derived from.
 3
.
EXAMPLE: Find sin 1 

 2 

3

that has a value of
.” You will need to
2
2
2

remember your unit circle or table for this one. This corresponds to an angle of 60 degrees, or , which is the
3
answer.
What this is really asking is: “find an angle between 
and
 1 
EXAMPLE: Find cos 1 
.
 2
What this is really asking is: “find an angle between 0 and  that has a value of
1
.” If you rationalize, then
2
1
2


. This corresponds to an angle of 45 degrees, or , which is the answer.
2
4
2
Section 2.11 Notes Page 2
 3
.
EXAMPLE: Find tan 1 

3



3

that has a value of
.” You will need to
2
3
2

remember your unit circle or table for this one. This corresponds to an angle of 30 degrees, or , which is the
6
answer.
What this is really asking is: “find an angle between 
and
EXAMPLE: Find: cot 1 ( 3 ) .
What this is asking us to do is to find the angle that gives us a cotangent value of 0. Our answer needs to be
between 0 and  since that is our range for an inverse cotangent. Since cotangent is cosine divided by sine, we
5
can find a point on the unit circle such that the cosine over sine gives us a value of  3 . The answer is
.
6
EXAMPLE: Find: csc1 ( 2 ) .
What this is asking us to do is to find the angle that gives us a cosecant value of 0. Our answer needs to be
between   / 2 and  / 2 since that is our range for an inverse cosecant. Since cosecant is 1 divided by sine,
7
we can find a point on the unit circle such that 1 divided by sine gives us a value of  2 . The answer is
.
4
EXAMPLE: Find: sec1 (2) .
What this is asking us to do is to find the angle that gives us a secant value of 2. Our answer needs to be
between 0 and  since that is our range for an inverse secant. Since secant is 1 divided by cosine, we can find

a point on the unit circle such that 1 divided by cosine gives us a value of 2 . The answer is .
3
 7
.
EXAMPLE: Find: cos 1 

 2 
7
as a decimal is 1.32. If we look on the table in this section, we see that the domain must be
2
 7
 is undefined.
between -1 and 1. Since 1.32 falls outside of this, then this means cos 1 

2


The number
Section 2.11 Notes Page 3
EXAMPLE: Write in algebraic form: sectan 4 x .
1
These problems involve drawing a triangle and labeling the sides with algebraic expressions. For all these
problems we will assume that x is positive and the triangle should be drawn in the first quadrant. We can
4x 

rewrite our problem as: sec tan 1  We know that the adjacent side is 1 and the opposite side is 4x. We can
1 

use the Pythagorean theorem to find the hypotenuse: c 2  (4 x) 2  (1) 2 . So we have c  16 x 2  1
16 x 2  1
4x
The secant on the outside of our problem tells us how to write

our answer. From our drawing, secant is 16 x 2  1 over 1 so we
write our answer as:
1


sec tan1 4 x  16 x 2  1


EXAMPLE: Write in algebraic form: cot sin 1 ( x  1) .
These problems involve drawing a triangle and labeling the sides with algebraic expressions. For all these
problems we will assume that x is positive and the triangle should be drawn in the first quadrant. We can
x 1

rewrite our problem as: sec sin 1
 We know that the opposite side is x – 1 and the hypotenuse is 1. We
1 

can use the Pythagorean theorem to find the hypotenuse: 12  ( x  1) 2  a 2 . So we have 1  x 2  2 x  1  a 2 .
Solving for a 2 we get a 2  2 x  x 2 , so a  2 x  x 2 . Now we can draw the triangle.
1
x–1
The cotangent on the outside of our problem tells us how to write

our answer. From our drawing, cotangent is
2x  x 2
so we write our answer as:


cot sin 1 ( x  1) 
2 x  x 2 over x – 1
2x  x2
where x  1 .
x 1
Derivative of inverse trig functions
Let’s find the derivative of sin 1 u . In order to do this we will draw a triangle like in the previous problems.
u
We can rewrite this problem as sin 1 . Then we know the hypotenuse is 1 and the opposite side is u.
1
Then we can use the Pythagorean Theorem to find the remaining side. You will get this triangle:
1

1 u2
u
Section 2.11 Notes Page 4
On our problem, let’s let y  sin u . The definition of the inverse sine tells us that sin y  u . Now let’s take
the derivative of both sides with respect to x using implicit differentiation:
1
d
sin y   du
dx
dx
You will need to use the chain rule on the left side.
cos y  y   u 
Now solve for y 
y 
y 
y 
u
cos y
From our triangle we know that cos y 
u
Now simplify and we have our answer.
1 u2
1
u
1 u
1 u2
. Plug this in for cos y .
1
So if y  sin 1 u then y  
2
u
1 u2
.
EXAMPLE: If y  sec1 u , find y  .
In order to do this we will draw a triangle like in the previous problems. We can rewrite this problem as
u
sec 1 . Then we know the hypotenuse is u and the adjacent side is 1.
1
Then we can use the Pythagorean Theorem to find the remaining side. You will get this triangle:
u 2 1
u

1
The definition of the inverse secant tells us that sec y  u . Now let’s take the derivative of both sides with
respect to x using implicit differentiation:
d
sec y  du
dx
dx
sec y tan y  y   u 
y 
y 
u
sec y tan y
u
u u 1
2
You will need to use the chain rule on the left side.
Now solve for y 
We know that sec y  u and tan y  u 2  1 from our triangle. Now substitute.
So if y  sec1 u then y  
u
u u 2 1
.
Section 2.11 Notes Page 5
Derivatives of Inverse Trig Functions
d
u
sin 1 u 
dx
1  u2
d
u
cos 1 u  
dx
1  u2
d
u
tan 1 u 
dx
1  u2






d
u
cot 1 u  
dx
1  u2


d
u
csc1 u  
dx
u u2 1

d
u
sec1 u 
dx
u u2 1



EXAMPLE: Find the derivative: y  sec1 (2 x) .
We will let u  2 x . Then u   2 . We just need to substitute these into the formula
y 
2
2 x (2 x)  1
2
. Now we can simplify: y 
1
x 4 x2  1
u
u u 2 1
. You will get:
. This is as far as we can go.
EXAMPLE: Find the derivative: y  tan1 5  2 x .
1

1
1
We will let u  5  2 x . Then u  (5  x) 2 (2) . This simplifies to u  
. Now we need to
2
5  2x
1
u
5  2 x . Simplifying gives us:
substitute these into the formula
. You will get: y  
2
1  u2
1  5  2x


1
1
1
1
1
y   5  2 x =  5  2 x = 
. So our final answer is: y  
.

6  2x
1  5  2x
(6  2 x ) 5  2 x
5  2x 6  2x
1
EXAMPLE: Find the derivative: y  x  tan 1 (2 x)  ln(1  4 x 2 ) .
4
We need to use the product rule on this one. When we take the derivative of tan1 (2 x) we will let u  2 x .
u
Then u   2 and we will use the formula the formula
. For the second term since we have a natural log
1 u2
u
we will let u  1  4 x 2 . Then u   8x . To do a derivative of the ln we will use the formula
. Using the
u
2
1
8x
 tan 1 (2 x)  (1)  
product rule we get: y  x 
. Now we can simplify:
2
1  ( 2 x)
4 1  4x2
2
2x
y  x 
 tan 1 (2 x) 
. The first and last terms cancel and then we have: y  tan1 (2 x) .
2
1  4x
1  4x2

1
Section 2.11 Notes Page 6

EXAMPLE: Find the derivative: y  ln cot 2 x .
3
Since we have a natural log we will let u  cot 1 4 x3 . Then u  
6 x2
 
1  2 x3
2
. This simplifies: u  
6x2
. To
1  4 x6
6x2
u
4 x 6 . Now we can
take the derivative of the ln we will use the formula
. This will give us: y  1 
u
cot 1 4 x3
6 x2
simplify: y   1 3
. Nothing more we can do on this one.
cot 2 x  1  4 x6



 x
EXAMPLE: Find the derivative: y  25 sin 1    x 25  x 2 . Write your answer as a single fraction.
5
u
x
1
. Then u   . You will use
. For the second term you will need to
5
5
1 u2
use the product rule combined with the chain rule. Putting this all together you will have:
1
1

1
5
y   25 
 x  (25  x 2 ) 2 (2 x)  25  x 2 (1)
Now we need to simplify.
2
2
 x
1  
5
For the first term we will let u 
y 
5
1
y 
y 
y 
y 
y 
x2
25
5
25  x
25
2
5
25  x
5
x2

2
25
25  x 2
25  x 2



x2
25  x
25  x 2
25  x 2
.
2
x2
25  x
2
x2
25  x 2
25  x 2  (25  x 2 )
2x 2
 25  x 2
.
For 1 
x2
25  x 2
we can get common denominators:
.
25
25
 25  x 2
This simplified further.
 25  x 2
The first term can be simplified further.

25  x 2
1
Multiply the last term by
25  x 2
25  x 2
to get common denominators.
Now simplify the numerator and we are done.