Download 29. Inverse Trigonometric Functions

Mathematics
Session
Inverse Trigonometric Functions
Session Objectives
Session Objectives
1. Basic Concepts of inverse
trigonometric functions
• Definition
• Domain and Range
2. Properties of Inverse Trigonmetric Function
3. Conversion of one form of Inverse Trig. Fn. to other
Forms
4. Identities containing inverse trigonometric functions
Basic Concepts - Definition
Inverse of a Function :
Function must be bijective
Trigonometric functions are periodic.
sin x is periodic with period equal to 2π
Not
Bijective
Hence , inverse of sin x
should not be valid ????
Basic Concepts - Definition
However, trigonometric functions
are bijective for particular value
sets in the domain.
sin x is bijective in [-π/2 , π/2 ]
and in [π/2 ,3π/2 ] ……. for x  R
sin-1 x is valid in these value sets
Inverse trigonometric function - principal value set
Smallest Numerical Angle
[-π/2 , π/2 ]
Basic Concepts - Definition
Inverse trigonometric functions 
inverse circular function.
arc sin x 
sin-1x ( principal value )
Inverse Trigonometric function sin –1 x
Domain of sin
–1x
( value which x can take ) is [-1,1]
Y
3/2
Range of sin
/2
-1
1
-/2
X
–1
x
(values which sin–1x can
take ) is [- /2, /2 ]
Inverse Trigonometric function cos–1 x
Domain of cos
–1x
( value which x can take ) is [-1,1]
Y

/2
-1
1
-
X
Range of cos
–1
x
(values which cos–1x can
take ) is [0, ]
Inverse Trigonometric function tan–1 x
Domain of tan
–1x
( value which x can take ) is (- , )

/2
X
-/2
-
Range of tan
–1
x
(values which sec–1x can
take ) is (-/2, /2)
Inverse Trigonometric function sec–1 x
Domain of sec –1x ( value which x
can take ) is (- -1] U [1, )
Y

/2
-1
Range of sec
1
X
-/2
-
–1
x
(values which sec–1x
can take ) is [0, ]
excl. x = /2
Inverse Trigonometric function cot–1 x
Domain of cot
–1x
( value which x can take ) is (- , )
Y

/2
X
-/2
-
Range of cot
–1
x
(values which cot–1x can
take ) is (0, )
Inverse Trigonometric function Domain and Range
Function
Domain
Range
sin-1x
[-1,1]
[-/2, /2]
cos-1x
[-1,1]
[0, ]
Function
Domain
Range
tan-1x
(- , )
(-/2, /2)
sec-1x
(-, -1] U [1, )
[0, ] excl. /2
cosec-1x
(-, -1] U [1, )
[-/2, /2] excl. 0
cot-1x
(- , )
(0, )
Inverse Trigonometric function
– Properties
Always remember to keep the
constraint of domain and range ,
while solving inverse trigonometric
functions.
sin( sin-1 x) = x
and cos (cos-1 x) = x if
tan( tan-1 x) = x
x if
x is in ( -, )
sin-1 ( sin x) = x
if
x is in [-/2, /2]
x is in [-1,1]
cos-1 ( cos x) = x
if
x is in [0, ]
sec-1 ( sec x) = x
if
x is in [0, ] excl. x = /2
Inverse Trigonometric function
– Properties
sin-1 (-x) = -sin-1 (x) if
x is in [-1,1]
cos-1 (-x) = - cos-1x if x is in [-1,1]
tan-1(-x) = - tan-1x if x is in ( -, )
cot-1(-x) =  - cot-1x if x is in (-,)
cosec-1(-x) = - cosec-1x if x is in (-,-1] U [1,)
sec-1(-x) =  - sec-1x if x is in (-,-1] U [1,)
Inverse Trigonometric function
– Properties
Always remember to keep the constraint of
domain and range , while solving inverse
trigonometric functions.
sin-1 (-x) = -sin-1 (x) if
x is in [-1,1]
Let y = sin-1(-x) ; constraint :
sin y = - x
y is in [-/2, /2]
 x = - sin y = sin ( -y )
sin-1(-x) = sin-1 ( sin (-y))
sin-1(-x) = -y
 sin-1(-x) = -sin-1 x
Class Exercise - 1
Find the principal value of
2 
 sin 3 


1 
sin
Solution :
2 

Let sin1  sin
 

3 

2 

 sin    sin

3


 sin  
3
2
  
where    , 
 2 2
  

3
Class Exercise - 2
Find the principal value of
sin –1 ( sin 5 )
Solution :
5
Wrong
Let y = sin-1(sin 5).Hence y is in [-/2,/2]
Now , sin 5 = sin [(5/). ] = sin ( 1.59)
= - sin (2 - 1.59)
= sin ( 1.59  -2)
sin 5 = sin ( 5 - 2 )
sin-1(sin 5) = sin-1 ( sin ( 5 - 2 ))
=5- 2
in [-/2, /2]
Other important properties
sin-1 x+ cos-1 x = /2 ;
if x is in [-1,1]
1
tan
1
x  tan
1 
y  tan
xy 
 1  xy 


If x > 0 , y > 0 and xy < 1
1
tan
1
tan
1
x  tan
1
x  tan
1 
xy 
 1  xy  If x<0,y<0 and xy < 1


y     tan
1 
y    tan
If x > 0 , y > 0 and xy > 1
xy 
 1  xy 


Class Exercise - 5
Find the value of
1  1 
tan
1  1 
 2   tan
 
3
 
Solution :
1
1
Let tan1     and tan1    
2
3
1
1
  
 tan     and tan     and ,     , 
2
3
 2 2


1 1



2 3

Now tan       
 1   1 . 1  
2 3

   

 tan       1
Class Exercise - 5
Find the value of
1
1
tan1    tan1  
2
3
Solution :  tan       1
     

4
why
    
5
4
Class Exercise - 9
In triangle ABC if A = tan-12
and B = tan-1 3 , prove that
C = 450
Solution :
For triangle ABC , A+B+C = 
tan A  2; tanB  3
tan  A  B   tan    C 
tan A  tanB
  tanC
1  tan A.tanB
  1   tanC  tanC  1

32
  tanC
1  3.2
Inverse Trigonometric function
– Conversion
To convert one inverse function to
other inverse function :
1. Assume given inverse function as some angle
( say  )
2.
Draw a right angled triangle satisfying the angle.
Find the third un known side
3. Find the trigonometric function from the triangle in
step 2. Take its inverse and we will get  = desired
inverse function
Conversion - Illustrative
Problem
The value of cot-1 3 + cosec-1  5 is
(a) /3
(b) /2
( c) /4
(d) none
Step 1
Assume given inverse function as some
angle ( say  )
Let cot-1 3 + cosec-1  5 = x + y,
Where x = cot-13 ; cot x = 3 and
y = cosec-1  5 ; cosec y =  5
Conversion - Illustrative
Problem
The value of cot-1 3 + cosec-1  5 is
(a) /3
(b) /2
( c) /4
(d) none
Step 2
Draw a right angled triangle satisfying the
angle. Find the third unknown side
 10
x
3
5
y
2
1
1
cot x = 3 , tan x = 1/3
cosec y =  5 , tan y = 1/2
Conversion - Illustrative
Problem
The value of cot-1 3 + cosec-1  5 is
(a) /3
Step 3
(b) /2
( c) /4
(d) none
tan x = 1/3 ,tan y = 1/2
Find the trigonometric function from the
triangle in step 2. Take its inverse and we will
get  = desired inverse function










1  1 

3
2

tan( x  y) 

1
1
1 . 
3 2
tan ( x+ y) = 1
 x + y = /4
tan( x  y)   tan x  tan y 
1 tan x tan y 






Conversion - Illustrative
Problem
Prove that
sin cot-1 tan cos-1 x = x
Solution :
Step 1
Assume given inverse function as some
angle ( say  )
Let
y = sin cot-1 tan cos-1 x
And
cos-1 x = , cos  = x
Hence , y
= sin cot-1 tan 
Conversion - Illustrative
Problem
Prove that sin cot-1 tan cos-1 x = x
Step 2
y
= sin cot-1 tan 
Draw a right angled triangle satisfying the
angle. Find the third unknown side
1

1 x 2
cos  = x, tan  =
x
Hence , y
= sin cot-1
1 x2
x
1 x2
x
Conversion - Illustrative
Problem
Prove that sin cot-1 tan cos-1 x = x
y
Step 2
= sin
cot-1
1 x2
x
Draw a right angled triangle satisfying the
angle. Find the third unknown side
1
x

1 x 2
Let
cot-1
and
y
1  x 2 = , cot  =
x
= sin 
From the adjoining triangle , sin  = x
Hence y = x = R.H.S.
1 x2
x
Class Exercise - 3
Find the value of

1 1 
tan  2 tan

3



1 1 
Let  2 tan


3

 1
 tan 
2 3
1

2.  
2 tan
3
2 , tan  
As tan  

1
1  tan2
1
2
9
3
tan  
4
Solution :
Class Exercise - 4
Find the value of
1
1 5 
tan  cos

2
3 

Solution :
1
5
1 5
cos
   cos2 
2
3
3
5
1
1

cos
2

3 5
3
 tan2  

 tan2  
1  cos 2
5
3 5
1
3
2
3 5
3 5
 tan   
2
 tan  
2
4




Class Exercise - 6
Prove that


tan1  x   cot 1  x  1  tan1 x2  x  1
Class Exercise - 6
Prove that


tan1  x   cot 1  x  1  tan1 x2  x  1
Solution :
Let tan1 x   and cot 1  x  1  
1
 tan   x and cot    x  1  tan  
x 1
And L.H.S. of the given identity is
1
x 1
 tan      
 1 
1  x. 

 x  1
x
+
tan   tan 
as tan      
1  tan .tan 
Class Exercise - 6
Prove that


tan1  x   cot 1  x  1  tan1 x2  x  1
Solution : given identity is +
1
x 1
and tan      
 1 
1  x. 

 x  1
x
x2  x  1
 tan      
 x  1  1

 tan       x2  x  1
      tan1  x2  x  1
Class Exercise - 7
Solve the equation
1  5 
sin
1  12 
 x   sin
 

 x  2


Solution :
5
 12 
Let sin1     and sin1 


x
 x 
5
 12 
 sin     and sin   

x
x
 



 given equation is    
2
cos       0  cos .cos   sin .sin 
Class Exercise - 7
Solve the equation
5
 12  
sin1    sin1 
 2
x
x
 


Solution :
cos       0  cos .cos   sin .sin 
25  x2
5
As sin      cos  
x
x
144  x2
 12 
and sin   
 cos  

x
 x 
 25  x2   144  x2  5
   12 





.
  .

x
x

 
 x  x 

 

Class Exercise - 7
Solve the equation
5
 12  
sin1    sin1 
 2
x
x
 


Solution :
 25  x2

x


  144  x2
.
x
 
 
 5
    .  12 
 x  x 

 25  x2 . 144  x2  60
 x4  169x2  0



2
 25  x2 . 144  x2  60 
 x  0,  13 , x  0
Now If x   13 ,  ,   0, hence x  13
Class Exercise - 8
If sin-1 x + sin-1 (1- x) = cos-1x,
the value of x could be
(a) 1, 0
(b) 1,1/2
(c) 0,1/2
(d) 1, -1/2
Class Exercise - 8
If sin-1 x + sin-1 (1- x) = cos-1x,
the value of x could be
(a) 1, 0
(b) 1,1/2
(c) 0,1/2
(d) 1, -1/2
Solution :
Let sin1 x   and sin1 1  x   
 sin   x and sin   1  x 
and given equation is + = cos-1x  cos (+) = x
As sin   x  cos   1  x2
2
and sin   1  x   cos   1  1  x 

2x  x2
Class Exercise - 8
If sin-1 x + sin-1 (1- x) = cos-1x,
the value of x could be
(a) 1, 0
(b) 1,1/2
(c) 0,1/2
(d) 1, -1/2
Solution : cos (+) = x
sin   x , cos   1  x2 , sin   1  x  , cos  
2x  x2
  1  x2  .  2x  x2   x. 1  x   x

 

  1  x2  .  2x  x2   2x  x2

 

2
2
2
2
 1  x . 2x  x  2x  x

 2x  x2

 1  x   2x  x   0
2
2

 

Class Exercise - 8
If sin-1 x + sin-1 (1- x) = cos-1x,
the value of x could be
(a) 1, 0
(b) 1,1/2
Solution :

(c) 0,1/2
 2x  x2
(d) 1, -1/2
 1  x   2x  x   0
2
2
 x. 2  x  1  2x    0
1
 x  0, ,2
2
x  2 as x  sin 
1
 x  0,
2
Class Exercise - 10
If cos-1 x + cos-1 y + cos-1z = ,
Then prove that x2+y2+z2 = 1 - 2xyz
Solution :
Let cos1x  A , cos1 y  B and cos1 z  C
 cos A  x ,cosB  y and cosC  z
and given : A+B+C = 
Now, L.H.S. = cos2A + cos2B +cos2C
= cos2A + 1- sin2B +cos2C
= 1+(cos2A - sin2B) +cos2C
Class Exercise - 10
If cos-1 x + cos-1 y + cos-1z = ,
Then prove that x2+y2+z2 = 1 - 2xyz
Solution : Given : A+B+C = 
L.H.S. = 1+(cos2A - sin2B) +cos2C
= 1+ cos(A+B).cos(A–B) +cos2C
= 1+ cos( – C).cos(A–B) +cos2C
= 1+ cosC [– cos(A–B) +cosC ]
= 1+ cosC [– cos(A–B) – cos(A+B)]
Class Exercise - 10
If cos-1 x + cos-1 y + cos-1z = ,
Then prove that x2+y2+z2 = 1 - 2xyz
Solution : Given : A+B+C = 
L.H.S. = 1+ cosC [– cos(A–B) – cos(A+B)]
= 1– cosC [ cos(A–B) + cos(A+B)]
= 1– cosC [2cos A. cos B]
= 1– 2xyz = R.H.S.
Thank you