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Inverse of Transcendental
Functions
1- Inverse of Trigonometric Functions
2- Inverse of Exponential Functions
3- Inverse of Hyperbolic Functions
1- Inverse of Trigonometric Functions
Since the trigonometric functions are not one-to-one, so they
don’t have inverse functions. However, if we restrict their
domains, then we may obtain one-to-one functions that have
the same values as the trigonometric functions and that have
inverse over these restricted domains.
For example, the function y  sin x is not one –to-one on
its natural domain R. However, when the domain is
restricted to the interval –π/2 to π/2, it becomes one-to-one.
y
Graph of y  sin x
1
y
x
2 3 / 2   / 2
 /2

1
1
x
 /2
 / 2
1
1
y
y  sin x
 /2
x
1
1
 / 2
3 / 2
2
Important Rules
*
1
y  sin x  sin y  x

1

* sin sin x  x , if  1  x  1
* sin
1
sin x   x ,
if 

2
x 

2
Example
Find the domain of


f  x   sin 1 x 2  1
Solution
D :  1  x2  1  1
 0  x2  2
D:  2  x  2
y
Graph of y  cos x
1
y
2 3 / 2   / 2
 /2

3 / 2
1
1
 /2

x
1
1
y  cos x
y

 /2
x
1
1
2
x
Important Rules
*
1
y  cos x  cos y  x

1

* cos cos x  x , if  1  x  1
* cos
1
 cos x   x ,
if 0  x  
y
Graph of y  tan x
x
3 / 2
y
 /2
 / 2
x
1
y  tan x
y
 /2
x
 / 2

 / 2
 /2

3 / 2
Important Rules
*
1
y  tan x  tan y  x

1

* tan tan x  x , if    x  
* tan
1
 tan x   x ,
tan   
1

2
if 

2
x 

2
tan    
1

2
Example
1 
1 
lim tan 

x2
 x2
Evaluate
Solution
x2

1


x2
1  
lim tan 


x2
 x2 2
1 
Notes
sin x   sin x 
1
cos x   cos x 
1
tan x   tan x 
1
sin
1
1/ 2   / 6
1
1
1
 sin x 
1
1

 csc x
sin x
1
1

 sec x
cos x
1
1

 cot x
tan x
 cos x 
 tan x 
cos
tan 1 1   / 4
1


3/2 
 /6
Important Rules
*
*
*
1
1
1/ x 
1
1
1/ x 
1
1
1/ x 
csc x  sin
sec x  cos
cot x  tan
1
Proof
csc x  sin
1
1/ x 
y  csc x
1
1
1
 
 sin y
x csc y
 x  csc y
 sin 1/ x   sin sin y 
1
1
y  csc x  sin 1/ x 
1
1
Example
Evaluate the given inverse function
i ) sec
1
 3
ii ) cot
1
 2.474
Solution
i ) sec
1
 3  cos
1  1 
   1.910633236
 3
1 
ii ) cot  2.474   tan 
  0.3840267299
 2.474 
1
1 
2- Inverse Exponential Functions
x
Every exponential function of the form a is a one-to-one
function. It therefore has an inverse function, which is
called the logarithmic function with base a and is denoted
by log a x .
y
ax
log a x
1
1
Domain:
(0, )
x
Range:
R  (, )
The Natural Logarithmic Function
The logarithm with base e is called the natural logarithm and
has a special notation
loge x  ln x
y  ln x
y e x
y
1
x
1
Domaim : (0, )
Rnge : R
Basic Properties of Natural Logarithmic Function
ln e  x
x
e
lnx y   ln x  ln y
ln x
x
lnx / y   ln x  ln y
   r ln x
ln x

ln 0  
r
ln  
Example
Solve the following equations for x
a) e
53 x
 10


5  3x  ln 10 
e
 e5

ln x 2 1
x
2

1  e
5
x  e 1
2
1
x   5  ln10   0.8991
3

b ) ln x  1  5
Solution
ln e 53x  ln 10 

2
5
x  e  1  12.141382.
5
Example
Sketch the function
f  x   ln  x  2   1
Solution
y
y
x
y
x=2
x
x
3- Inverse Hyperbolic Functions
The hyperbolic functions sinh x is one-to-one functions
1
and so they have inverse functions denoted by sinh x


1) sinh 1 x  ln x  x 2  1 ,
1

x 

2) cosh x  ln x  x  1 , x  1
1 x
3) tanh x  ln
,
1 x
1
2
1  x  1
1


sinh x  ln x  x  1 ,
Proof (1)
2
e e
 x  sinh y  x 
2
y
y  sinh x
1
x R
y
 e  e  2x  e  2x  e  0
y
e 
y 2
y
 2 xe   1  0
y
2x  4x  4
e  
2
2
y
y
y
e  x  x 1
y


2
y  ln  x  x  1 


2
Proof (3)
tanh
y  tanh x
1
1
1 x
x  ln
,
1 x
 x  tanh y
1 x 1
x 
e
e
e
y
e
y

x e
y
e
 e 1  x   e 1  x 
y
y
 
 e
y
1  x

1  x 
y

e
y
 xe
y
y
y
e
y
e
y
e
y
 xe
  1  x   1  x 
 e
y
2

y  ln 


1  x  
1  x  
y
Important Rules
1
sec h x  cosh
1
csc h x  sinh
1
1
1
coth x  tanh
1 / x 
1 / x 
1
1 / x 