Download Inverse Trigonometric Functions and Their Derivatives

Inverse Trigonometric Functions and Their
Derivatives
None of the trigonometric functions satisfies the horizontal line
test, so none of them has an inverse.
The inverse trigonometric functions are defined to be the
inverses of particular parts of the trigonometric functions; parts
that do have inverses.
y = sin(x)
π
π
− ≤ x≤
2
2
sin −1(x) = arcsin(x)
y = cos(x)
0 ≤ x ≤π
cos−1(x) = arccos(x)
y = tan(x)
π
π
− ≤ x≤
2
2
tan−1(x) = arctan(x)
y = sec(x)
0 ≤ x ≤π
x ≠π
2
sec−1(x) = arcsec(x)
Evaluating inverse trigonometric functions
The equation y = sin −1(x) is equivalent with the equation
x = sin( y). Thus sin −1(x) should be thought of as the angle
whose sine is x. However, many angles have sine equal to x,
π
and in this case, we want the angle that lies between
and
2
π
.
A
similar
idea
holds
for
all
the
other
inverse
−
2
trigonometric functions. Each is an angle, but you must choose
the particular angle that satisfies the restriction appropriate to
that function.
Example. Find sin−1(−1) and cos−1(1/2)
Solution.
1. Let y = sin−1(−1). Then sin(y) = −1, so y =−π
2
2. Let y = cos−1(1/2). Then cos(y) = 1/2, so y =π
3
We can also find identities involving the inverse trigonometric
functions, by using ordinary trigonometric identities.
Example. Show that cos(sin−1(x)) = 1− x 2
1
The result is obvious
from the diagram:
sin−1( x)
1− x 2
x
Example. Clearly sin(sin−1(x)) = x. Is it also true that
sin−1(sin (x)) = x?
Actually, this is not true because of the restrictions on the inverse
sine. The graph of sin−1(sin (x)) is shown below.
5π
1
then sin(x) =
. But the angle in
4
2
the range −π ≤ x ≤ π having that sine is π not 5π .
4
4
2
2
For example, if x =
All such expressions (inversetrigfn(trigfn) = ) must be treated
with care.
Derivatives of Inverse Trigonometric Functions
Theorem.
d sin −1(x)  =

dx 
1
1− x2
Proof. Let y = sin −1(x) so sin( y) = x. Then
cos( y) dy =1
dx
dy = 1 =
dx cos( y)
1
1−sin 2( y)
=
1
1− x 2
In a similar way we can show that
d  cos−1( x)  =

dx 
Theorem.
−1
1− x 2
d  tan −1(x)  = 1

dx 
1+ x2
Proof. Let y = tan −1(x) so tan( y) = x. Then
sec2 ( y) dy =1
dx
dy = 1 =
1
= 1
dx sec2 ( y ) 1+ tan 2( y ) 1+ x 2
Theorem.
d sec −1(x)  =

dx 
1
| x| x 2 −1
Proof. Let y = sec−1( x) so sec( y) = x. Then
sec( y)tan( y) dy =1
dx
dy =
1
= cos( y) cos( y)
dx sec( y) tan( y)
sin( y)
x
x 2 −1
y
2( y) 1
cos
=
=
sin( y) x2
x =
1
x2 −1 | x| x2 −1
1
We have now shown the following rules for differentiation.
Basic Rule
1. d sin −1( x)  =

dx 
Generalized Rule
1
1− x 2
2. d  tan −1(x)  = 1

dx 
1+ x 2
3. d sec−1(x)  =

dx 
1
| x| x 2 −1
d sin −1(u)  =

dx 
1
1−u
du
2 dx
d  tan −1(u )  = 1 du

dx 
1+u 2 dx
d sec−1(u)  =

dx 
1
du
dx
|u| u 2 −1
The other three inverse trig functions have derivatives that
are the negatives of their respective cofunctions.
Miscellaneous Problems
1
−
1
Problem. Suppose that θ = cos   . Find the exact values
2
of sin(θ ),tan(θ ),cot(θ ),sec(θ ),csc(θ ).
Solution. Construct the following triangle.
2
3
θ
1
so that cos(θ ) =
1
2
The other functions can be read directly
from that triangle.
sin(θ ) = 3
2
tan(θ ) = 3
sec(θ ) = 2
2
cot(θ ) = 1
3
csc(θ ) = 2
3
3
θ
1
 − 1 3  
sec
Problem. Compute
sin . − 4  
 

Solution. Construct the following triangle.
7
θ
−3
4
Then sec(θ) =
4
7
Problem. Complete the identity tan cos −1.(x)  = ?


Solution. Construct the following triangle.
1
1− x 2
θ
Then tan(θ) =
x
1− x 2
x
Problem. Find the derivative of cos−1(2 x +.1)
Solution.
d cos −1(2 x +1)  =−

dx 
1
1− (2 x +1)
d (2x +1) =−
2 dx
2
1−(2 x +1)2
Problem. Find the derivative of tan−1( x .)
Solution.
d  tan −1( x )  =
1
d ( x)= 1
1 =
1

dx 
(1+ x) 2 x 2 x (1+ x)
(1+ ( x ) 2 ) dx
Problem. Find the derivative of sec−1(ex) .
Solution.
d sec−1(e x )  =

dx 
|e x |
d  ex =
1
x= 1
e
 
dx
2
2 x −1
  e x e2x −1
x
e
e
−1
1
( )
Problem. Find the derivative of ln  sin −1( x)  .


Solution.
(
d ln sin −1( x)
dx 
)
1
d  sin −1(x)  = 1
1
=



−
1
dx
 sin −1( x) 1− x2
 sin (x) 
Problem. Find the derivative of
Solution.
d  tan −1( x)  =

dx 

tan −1( x).
1
d  tan −1( x)  =
1
1


2
dx
−
1
−
1

1
+
x
2 tan (x)
2 tan ( x)
Problem. Find the derivative of sin−1( x) +.cos−1( x)
Solution.
Why?
d sin −1(x) + cos−1( x)  =

dx 
+ −1 = 0
1− x 2 1− x 2
1
Because sin−1( x) + cos−1( x) = π
2
3
−
1
Problem. Find the derivative of x2  sin ( x) .


(
d  x2 sin −1( x)

dx 

)
3


3
2
−
1
2
−
1






= 2 x sin (x)  + x 3 sin ( x)  d sin −1(x) 


   
 dx 
3
2 1
−
1
2
−
1
= 2x sin ( x)  + x 3 sin ( x) 

   
 1− x2
2
−
1
sin (x)
3
−
1
= 2 x sin ( x)  +3x2


1− x 2
Solution.
(
)
(
Problem. Find the derivative of sin tan −1.(x)
Solution.
(
)
d sin tan −1(x) 

dx 

= cos tan −1( x)  d  tan −1( x) 


 dx 
= cos tan −1( x)  1

1+ x2
)
Problem. Find dy by implicit differentiation, if
dx
.
x3 + x tan −1( y) = e y
Solution.
3x2 + tan−1( y) + x dy = e y dy
dx
1+ y 2 dx
so
x dy − e y dy =−3x2 − tan−1( y)
dx
1+ y 2 dx
 x
 dy
y
− e  =−3x 2 − tan −1( y )

2
1
+
y

 dx
dy = −3x2 − tan −1( y)
dx  x

y
−e 

1+ y 2
