Download Chapter – 9

Inverse Trigonometric Functions
Chapter – 9
Inverse Trigonometric Functions
Introduction: To study inverse trigonometric functions with full rigour is a big challenge and it requires a
purely analytical approach. Almost all results are true in a given interval only. These interval are
discovered when we attempt to prove these results methodically. Unfortunately the results given in most
of the books are not true for all the values within the domain. In any such crisis we may say that LHS is one
of the values of RHS or vice versa.
Let us now define inverse trigonometric functions. Suppose a≤1 and sin x = a then as we know there
are infinite values of x satisfying this equation and which are which are given by x = n  + (-1)n  where  is
the numerically smallest angle (For instance if a = ½ then  = /6) whose since is a. Then we write sin -1 a =
n + (-1)n 
Thus SIN-1 a is multiple valued function taking infinite values for any a satisfying a≤1. We also note that
 is the numerically smallest angle whose sine is a i.e.
a
1.
1

1

5 
, 
and if a   ,     not

2
6
2
6
6 
Principal Values Of Inverse Functions: Recall basic properties of sin x we know that its domain is (, ) and range is [-1, 1], but to attain does not have to run from - to . We easily notice from
the graph of sin x that to attain all the values in [-1, 1], x has to run either from 

to

or from
2
2



3
3
5
to
or from
to
and so on the most natural choice of which is  to
since
2
2
2
2
2
2
(i)
sin x monotonically increasing in this interval
(ii)
Normal acute angles are covered.
This particular  is denoted by sin-1 a. Thus remember
  
whose sine is a where a≤1.”
,
 2 2 
“sin-1 a is the angle in  
 

To invert the equality y = sin x we sin-1 y = x provided y≤1 and x   ,
otherwise we can
 2 2 
  
we proceed as
,
 2 2 
not write it as sin-1 y = x. To invert the equality y = sin x when x   
follows
  
We write sin x = sin x’ where x    ,  which is always possible since any value in [-1, 1] is
 2 2
  
,
through the function sin x. For instance suppose we want to
 2 2 
  3 
invert y = sin x when x  ,
then
 2 2 
attained by an angle in  
Text Book of Trigonometry
44
Inverse Trigonometric Functions

2
x
3
2

3

 x 
2
2


2
  x

2
Also sin x = sin( - x).
  
whence it
,
 2 2 
Therefore the given equality can be written as y = sin( - x) where   x  
2.
follows that sin-1 y =  - x. A similar discussion follows for other inverse trigonometric functions.
Domain And Range Of Inverse Trigonometric Functions:
Function
domain
rangeReason for choosing range
sin-1 x
[-1, 1]
  
  2 , 2 
cos-1 x
[-1, 1]
[0, ]
tan 1 x
 ,  
Same as sin-1 x
cot 1 x
 ,  
  
 , 
 2 2
(0, )
Same as cos-1 x
sec 1 x
 ,   1,  
   
0, 2    2 , 
same as cos-1 x
Discussed
Since cos x is MD in
[0, ] and [0, ]
covers acute angles.
(/2 is not attained)
cosec-1 x
 ,   1,  
    
  2 , 0    0, 2 
same as sin-1 x
(0 is not attained)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
The following additional remarks on inverse functions will be useful.
sin-1 x is positive or negative according as x is positive or negative. Same is true for tan-1 x.
cos-1 x  0 for all x within its domain
cot-1 x > 0 for all x
sin-1 x, cos-1 x are continuous and differentiable in (-1, 1)
tan-1 x, cot-1 x are continuous and differentiable every where.
sin-1 x, tan-1 x are monotonically increasing while cos-1 x, cot-1 x are decreasing in their respective
domains.
Results
1.
1
1
, x  0, x  1 (ii) cos-1 x = sec 1 , x  0,
x
x
1

cot 1
if x  0


x
1
tan x  
cot 1 1   if x  0

x
(i) sin-1 x = cos ec 1
(iii)
Text Book of Trigonometry
45
x 1
Inverse Trigonometric Functions
Proof: (i)
if x  0,1
sin 1 x  cos1 1  x2
if x   1,0 
  cos1 x 1  x2
(ii)
if x  0,1
cos1 x  sin 1 1  x2
if x  1,0
   sin 1 1  x2
Indeed we can relate any inverse trigonometric function with any other inverse trigonometric
function. Suppose we want to express tan-1 x in term of cos-1 x.



2
First of all let x > 0 then   tan 1 x   0,  and from tan  = x we can conclude that
cos  
1
1  x2
 
 0 and    0,  .
 2
1 x
1
1
  
But if x < 0 then     , 0  and cos  
can not be inverted since
is still
2
2
 2 
1 x
1 x
 
positive and cos 1  positive    0, 
 2
 
To invert we write cos  = cos (-) and note that    0, 
 2
1
1
Thus   cos 1
Or tan 1    cos 1
1  x2
1  x2
1

1
if x  0
 cos
1  x2

1
Consequently tan x  
1
 cos 1
if x  0

1  x2
1
 
cot  
If x > 0, put tan-1 x = , then    0,  and tan  = x Or
x
 2
1
1
 
since  0 and    0,  this can be inverted and we can write   cot 1 .
x
x
 2
1
  
Again if x < 0 and we put tan-1 x =  then     , 0  .Now cot   can not be
x
 2 
which can be inverted since
(iii)
1
2
1
  
  
 0 and     , 0   cot 1  negative    ,    we try to write
x
 2 
 2 
 
 
cot  where    ,   . Indeed cot      cot  and      ,  
2 
2 
1
Therefore cot  = x can be written as cot       and we will invert it to get
x
inverted since
Text Book of Trigonometry
46
cot 
=
Inverse Trigonometric Functions
    cot 1
2.
3.
4.
1
x
 tan 1
1
1
 cot 1   .
x
x
-1
- sin x, x  1
(i)
sin-1 (- x)
=
(ii)
cos-1 (- x)
=
 - cos-1 x,
-1
-1
x 1
(iii)
(i)
tan (- x)
sin(sin-1 x)
=
=
- tan x for all x
x,
x 1
(ii)
cos(cos-1 x)
=
x,
x for all x
-1
(iii)
tan(tan x)
=
(i)
sin-1(sin x)
=  1
(ii)
-1
cos (cos x)
n
x 1
 x  n  , x  n 

, n 
2

2 

if x   2n  1 , 2n 
= 2n - x
if x   2n,  2n  1  
= x – 2n




, n  
2
2


  x  2 m 
2
2
Proof (i): If n is even (Say n = 2m) then x   n 
 2 m 


 x  2 m 
2
2

Also sin(x – 2m) = sin x for all x  sin-1sin x = x – 2m = (-1)n (x - n)
If n is odd (n is odd (n = 2m + 1) then






x   n  , n  
 2m  1    x   2m  1  
2
2
2
2




   2m  1    x    2m  1 
2
2



   2m  1    x 
2
2
n
1
 sin  sin x    2m  1   x   1  x  n 
 n  2m  1
(ii)
If x  [(2n – 1), 2n ] then
2n - x  [0, ] and when x  [2n, (2n + 1)], then x – 2n  [0, ]
whence the result (ii) follows.
19 


 3  ,3  
6
2
2


19 
19
1 
 We can calculate sin  sin 6  directly also. Indeed sin 6




1


 1
 sin  3     and sin 1       .
6
2
6

 2


For example sin 1  sin
19 

3  19

 3   
   1 
6 
6
 6




As another example to evaluate sin-1(sin 20) we note that 20  6  , 6  
2
2

(i.e. n = 6)
 sin-1(sin 20) = (-1)6(20 - 6) = 20 - 6
Text Book of Trigonometry
47
Inverse Trigonometric Functions
5.
x y
,
x, y  0, xy  1
1  xy
x y
(ii)
tan-1 x + tan-1 y =   tan 1
,
x, y  0, xy  1
1  xy
 
Proof: (i) Put tan-1 x = , tan-1 y = , then ,   0,  and since
 2
tan   tan 
x y
tan      

0
**
 xy  1
1  tan  tan  1  xy
x y
 
      0,  and this can be inverted to get  +  = tan 1
1  xy
 2
tan-1 x + tan-1 y = tan 1
(i)
whence (i) follows.
If xy > 1 then (**) can not be inverted since tan( + ) < 0 and inversion will yield
 +  = tan 1
x y
1  xy
  
,0
 2 
 angle in (0, ) = an angle in  
 
,   since tan( + ) < 0
2 
x y
  

        , 0  and tan( +  - ) =
1  xy
 2 
x y
which can be inverted and we get       tan 1
1  xy
x y
Or
tan-1 x + tan-1 y =   tan 1
1  xy
We note that     
6.
sin-1 x + cos-1 x = /2,
x 1
tan-1 x + cot-1 x = /2,
for all x
sec-1 x + cosec-1 x = /2, x  1
7.

sin 1 2  1  x 2


1

1
  2sin x if x   1,  
2



 1 1 

  2sin 1 x if x   
,

2 2



 1 
   2sin 1 x if x  
,1

 2 

On putting x = sin , LHS = sin-1(sin 2)
(*)

1 
  
 . Then     2 ,  4 


2



 
  
 2   ,  
  2   ,      2    , 0 
2

 2 
2 
1
1
1
 sin  sin 2  sin  sin    2     2sin x
when x   1, 
Text Book of Trigonometry
48
Inverse Trigonometric Functions

1 
  
 then     4 , 4  


 2 2
1
1
sin (sin 2)
 2  2sin x .
Again if x   

1
  
2    , 
 2 2
,

  
 
 
,1 then   ,   2   ,      2   0, 
 2
2 
 4 2
 2 
1
1
1
   2sin x.
sin (sin 2)
 sin  sin (  2)     2
 1
Finally when x  

sin 1 ( x 1  y 2  y 1  x 2 ) ,
if x, y  0, x 2  y 2  1


8.
sin 1 x  sin 1 y     sin 1 ( x 1  y 2  y 1  x 2 ) if x  0, y  0, x 2  y 2  1
 1
2
2
if x , y [0,1]
cos ( xy  1  x  1  y ),
 
Proof : (i) If we put sin 1 x   ,sin 1 y  then  ,  0, 
( x , y  0)
 2
and x 2  y 2  1

sin 2   sin 2  1
 cos 2  cos 2  0

cos( +  ) cos ( - )  0
 

cos(   )  0
( cos (   )  0 ) 
    0 , 
 2

the equality sin (   )
= sin   cos   cos  sin 
= x 1  y 2  y 1  x2
(*)
can be inverted and we can get     sin 1 ( x 1  y 2  y 1  x 2 )
(ii)
If x 2  y 2 1 then following the same steps we will arrive at
cos(   )  0

(iii)
 
     , 
2 
 
  (   )   ,  
2 

The equality (*) can be inverted and we get
  (   )
Or

 sin 1 [ x 1  y 2  y 1  x 2 ]
sin 1 x  sin 1 y    [ sin 1 x 1  y 2  y 1  x 2 ]



Put cos 1 x   , cos 1 y  then  ,    0,  since x , y 0,1 
2

Now cos (+ ) = cos cos - sin sin = xy - 1  x 2
(

x , y (0,1) , x  1  x 2 , y  1  y 2 )


The equality can be inverted since  +    ,  
2 
Thus we get ,     cos 1 ( xy  1  x 2
Or
1  y2  0
cos 1 x  cos 1 y
Text Book of Trigonometry
1  y2 )
 cos 1 ( xy  1  x 2
49
1  y2 )
Inverse Trigonometric Functions
9.
 1 2 x
1  x  1
 sin 1  x 2

2x

1
2 tan x     sin 1
if x  1
1  x2

2x

1
 sin 1  x 2   if x  1

Put x = tan  on RHS. If -1  x  1, then 



4
4
 


 2 
2
2

1  2 tan  
1
1
  sin 
  sin  sin 2   2  2 tan x
2

 1  tan  


 
 
Again if x > 1 then   
 2   ,      2   0, 
4
2
2 
 2
2x
      2 tan 1 x   2 tan 1 x
 sin-1 (sin 2) = sin-1 (sin( - 2)) =   sin 1
2
1 x

Finally     2  0  sin -1 (sin 2) = sin -1 (sin(-- 2)) = - - 2 tan -1 x
2
2x
  sin 1
       2 tan 1 x     2 tan 1 x  LHS.
2
1 x
 2x
2
1 x
Now sin 1 
Solved Examples
1.
Prove the following numerical equalities

1
 1 
(ii)
cot  cos 1      
2 2
 3 

1
1  2 6

(iii)
(iv)
cos  2 cos 1  sin 1  
5
5
5

1
14

(v)
(vi)
sin  2 tan 1   cos tan 1 2 2 
3
15

 1
 
Solution: (i) Take cos 1     t , then t   ,   and cos t
 3
2 
1 3
cos t
1
cot t 


sin t  1  cos2 t
2 2
(i)

(ii)
Put

1
 3  2
sin  cot 1     
5
 4 
2
33 


sin 1  cos

10 
5

4
5
16 
sin 1  sin 1  sin 1

5
13
65 2
1
  and
3
 3
 
cot 1     t then t   ,   ,
 4
2 
t
1  cos t
3

, But cot t   and
4
2
2
3

t
2
  
cos t  
t   ,    whence sin 

5
2
5
 2 

and LHS = sin
Text Book of Trigonometry
50
Inverse Trigonometric Functions


1
1
1
1

 sin 1  cos 1   cos   cos 1 
5
5
5
5
2
(iii)
LHS = cos  cos 1
(iv)
24
2 6
1
1


  sin  cos 1    1     
5
5
5
5

33
3 
3

  3 
We have cos
 cos  3     cos
  sin   
10
10 
10

 2 10 

 
  sin  sin   
5
 5
2


whence sin 1  cos
(v)
33 

  
1 
  sin  sin      
10 
5
  5 
1
 , tan 1 2 2   then
3
2 tan 
1
LHS = sin  2   cos  

2
1  tan 
1  tan 2 
Put tan 1


 
 sin ce    0, 2  we have taken  sign with radical 




14
1

on substituting tan   , tan   2 2
15
3
(vi)
First method of proving this equality could be applying formulae set (9) several
We should prefer proceeding independently.
Put sin 1
4
 x,
5
sin 1
5
 y,
13
sin 1
16
z
65
We have to show x + y + z = /2
It is sufficient to show that x + y = /2 – z
We have sin(x + y) = sin x cos y + cos x sin y =
4 12 3 5 63
   
5 13 5 13 65
( x, y are acute angles)
16 



 z   cos z  cos  sin 1 
65 
2


and sin 
2
 16 
 1   
 65 


 sin  x  y   sin   z 
2

 65  16 
2
 65
2
2
81  49
 65
2
x + y + z = /2
x + y – z = /2
(A)
(B)
If n = 2 we get
x+y+z=
(C)
Text Book of Trigonometry
63
65
n 

 x  y  n   1   z 
2

If n = 0 we get
If n = 1 we get
If n = 3 we get

5
2
5
x+y–z=
2
(D)
51
times.
Inverse Trigonometric Functions
3
, (C) and (D) are not possible. For n  4 and n < 0
2
3
relations obtained will again be impossible since 0 < x + y + z <
2
Since max(x + y + z) =
the
(B) is not possible either (why ?) therefore it can be concluded that
x+y+z=
3
essentially
2
NOTE: A superior solution will be to convert every term in tan-1 . Indeed
4 5

1 4
1 5
1 63
1
3
12  tan 1 16
LHS = tan
 tan
 tan
 tan
4 5
3
12
16
63
1 
3 12
63
16
63
63 
 tan 1
 tan 1
 tan 1
 cot 1

16
63
16
16 2
2.
Solve the following equations
x 1
x 1 
 tan 1

x2
x2 4
(i)
tan 1
(iii)
4 sin-1 x + cos-1 x = 
Solution: (i)
sin-1 x + sin-1 2x =
(iv)
1

sin  cos 1 x   1
5

On taking tan of both sides we get
x 1

tan  tan 1
 tan 1
x2

We easily get x  
are positive.
1
2
x 1 x 1

x 1 
x

2
x  2 1
 1 
x2
 x  1 x  1
1
 x  2  x  2 
for x  
1
2
both expressions
 The equation is satisfied for x  
1
2
x 1
x 1
and
x2
x2
.
1

is possible. Thus solutions are 
.
4
2

The equation can be written as sin 1 2 x   sin 1 x
3


 sin  sin 1 2 x   sin   sin 1 x 
3

3
1


2x 
1  x2  x
 2x = sin   cos  sin 1 x   cos sin  sin 1 x  
2
2
3
3
tan 1     tan 1    
(ii)

3
(ii)
5x
3
3

1  x2
 x
2
2
28
3
Now for x = 
. LHS = sin-1 (-ve) + sin-1 (-ve) = negative
28

Text Book of Trigonometry
52
Inverse Trigonometric Functions

x 
(iii)
3
is not a solution. Thus the only solution is
28
x 
3
(check !!)
28
The equation is sin-1 x + cos-1 x + 3 sin-1 x = 


 3sin 1 x  
2
 sin 1 x 

1
 x
6
2
Thus x = ½ is a solution.
(iv)
We must have
1

cos 1 x 
5
2
 cos 1 x 
which is impossible since max cos-1 x = 
NOTE:- The student may conclude that x 
5
2
5
 0 which is not correct since it
2
satisfy the original equation.
3.
If cos-1 x + cos-1 y + cos-1 z = , prove that x2 + y2 + z2 + 2xyz = 1
Solution: We have cos-1 x + cos-1 y =  - cos z.
On taking cos of both sides we get
cos(cos-1 x) cos(cos-1 y) – sin(cos-1 x) sin(cos-1 y) = cos( - cos-1 z)
 xy  1  x 2 1  y 2  z
 xy  z  1  x 2 1  y 2
On squaring this yields x2 + y2 + z2 = 1 + 2xyz
4.
Show that if a 
Solution:
1
1
the equation (sin-1 x)3 + (cos-1 x)3 = a3 has no solution if a 
32
32
Since sin-1 x + cos-1 x = /2. The equation reduces to
 
t 3    t   a3 where t = sin-1 x
2 
3
which is same as 12 t2 - 62t + (1 – 8a) 3 = 0 which will not have real roots.
If 36 4 – 48 4(1 – 8a) < 0 which reduces to a 
5.


1
.
32
Prove that sin cot 1 cos 2  tan 1 cos 2  tan 2 


0    
2



Thus sin cot 1 cos 2  tan 1 cos 2  sin   2 tan 1 cos 2 
2

2
1  tan  1  cos 2


 sin   2   cos 2 

 tan 2 
2
2
1

cos
2

1

tan



x
3 1  x2 
1
6.
Evaluate cos 1 x  cos 1  
 where 1  x 
2

2
2


1
 
Solution:
Let cos-1 x = . Then x = cos  
    as  1  x  
2
 3
Solution:
Let tan 1 cos 2   . Then cos 2 = tan2 

Text Book of Trigonometry

53
does not
Inverse Trigonometric Functions
x
3 1  x 2 cos 
3 1  cos 2 



2
2
2
2


 

2
 cos cos   sin sin 
 3      sin   1  cos  
3
3


 cos    
3

x
3 1  x2 

 


 cos 1  
  cos 1  cos        
2

2
3 
3




 2 
 
 3      0    3  3 
x
3 1  x2 
 

Hence cos 1 x  cos 1  
       
2

2
3 3



pq  1
qr  1
rp  1
7.
Prove that cot 1
 cot 1
 cot 1
  , if p > q > r.
pq
qr
rp
pq  1
Solution: Since p > q, cot 1
(*)
 cot 1 q  cot 1 p
p 1
qr  1
Since q  r , cot 1
(**)
 cot 1 r  cot 1 q
qr
rp  1
Again since p  r , therefore , cot 1
(***)
   cot 1 p  cot 1 r
rp
pq  1
qr  1
rp  1
Thus form
(*), (**) and (***) , cot 1
 cot 1
 cot 1
pq
qr
rp
1
1
1
1
1
1
 (cot q  cot p)  (cot r  cot q)  (  cot p  cot r )   .
Thus
8.

Prove that sin 1 x  sin 1 y    sin 1 x 1  y 2  y 1  y 2
 if
x, y  [-1, 0], x2 + y2 > 1
  
  
,0,     ,0
 2 
 2 
  
  , 
 2 2
Solution: sin-1 x = , sin-1 y = , then    
Also        , 0  ,

x, y   1, 0
The condition x2 + y2 > 1 yields sin2  + sin2  > 1

cos 2  cos 2  0
 cos      0
 cos     cos      0
(*)

cos      0
Now sin ( + ) = sin cos  + cos sin


  
   either belongs to   ,   or   , 0 
2

 2 


  
But from (*) cos(+)<0. thus       ,   and then   (  )    , 0  .
2

 2 

Text Book of Trigonometry
54
Inverse Trigonometric Functions
= x 1  y 2  y 1  x2
Now (**) becomes sin(  (  ))

 +  = -  - sin -1 [ x 1  y 2  y 1  x 2 ] etc .
x y
  if x, y < 0, xy > 1.
1  xy
  
Solution: tan 1 x   , tan 1 y  , then ,    , 0 
 2 
Also xy> 1  tan  tan  1  sin  sin   cos  cos 
( cos  cos   0)



cos(+ )<0   +    ,   ( initially    ( , 0))
2

tan   tan 
x y
x y
Now tan (  ) 
 tan (    ) 

0
etc .
1  tan   tan  1  xy
1  xy
  
(iv)
Put x = tan  , then x > -1 
 , 
 4 2



   
 
and LHS  tan 1 (tan  )  tan 1          =   ,    liein   ,  
4
4
4
4

 2 2 
  
Again if x < - 1 , then     ,  
 2 4
9.
Prove that tan-1 x + tan-1 y = tan 1



(*)
LHS  tan 1 (tan )  tan 1  tan     
4




  3 




Now the first term is still  but as     ,  , tan 1  tan             
4
2 4 
4

4






  
tan 1 tan       if   ,    whence RHS of
 2 
3


=      
.

4
4

 ak 
Prove that tan 

 bc 
1
1
1
 ck  2
10.
 tan     where k = a+b+c, a, b, c > 0
 ab 
ak
bk

k
bc
ac
Solution: The sum of first two terms =  + tan 1
( x, y  0, xy  1 as xy   1)
k
c
1
c
ck
whence the result follows.
   tan 1
ab
x
3  3x 2  
1 
11.
Prove that cos 1 x  cos 1  
  is x   ,1
2

2
2 

 3
1
Text Book of Trigonometry
2
 bk 
 tan  
 ac 
(*)
1
2
1
55
Inverse Trigonometric Functions



Solution: Put x = cos, then    0,  = ( x(1/ 2, 1))
3

1

3
LHS  cos 1 (cos )  cos 1  cos  
sin  
2
2








=   cos 1  cos cos   sin sin  
   cos 1  cos     
3
3


3




 
 
    
  0,  

3
3
 3
 3
3x
4x
Solve sin 1
 sin 1
 sin 1 x
5
5

12.
Solution:
On taking sine of both sides , we get
3x
16 x 2
9 x2
1
 1
x
5
25
25
Or
x(3 25  16 x 2  4 25  9 x 2 )  25 x
One root is obviously x = 0 the other roots can be obtained by solving the equation
3 25  16 x 2  4 25  9 x 2  25
by means of substitutions x2 = t. We will finally obtain x =  1. Roots are to be checked.
13.
Find the latest and greatest value of (sin-1 x)3 + (cos-1 x)3
Solution: Put cos 1 x  t , then 0 < t <  and the function ( Say f(t)) is given by
 
 
 
f (t )    t   t 3
 f (t )  3t 2  3   t 
 3  t  
2 
2 
 4
 
 

f is MD in  0,  and MI in  ,  
4 
 4


There is an absolute minimum at
and for absolute maximum we can compare the
4
3
2
values of f(t) at boundary points t = 0 , t =  , we easily obtain
3
 
f (min)  f    ,
 4  32
73
.
8
1 x
1  x2
14.
If 0 < x < 1 and A  2 tan 1
. Show that A + B = 
, B  sin 1
1 x
1  x2

( 0  x 1)
Solution: On putting x  tan  , then 0   
4


A  2 tan 1 tan     , B  sin 1 (cos 2)

4





  

A

=  2 ,
B   2  Note that both   and  2   0, 
2
2
4
2

 2 
f (max)  f () 
whence A + B =  .
Text Book of Trigonometry
56
Inverse Trigonometric Functions
1

1 3 x  x
if 0  x 
or x  3
 tan 1  3x 2
3
2x
1
1
15.
Prove that tan x  tan
 
3

1  x2
1
1 3 x  x
if
 x 1
   tan
2
1  3x
3

1

Solution: If 0  x 
, then x = tan  0    .
6
3
1
1
LHS  tan (tan )  tan (2)   2  3
RHS  tan 1 (tan 3)  3  LHS  RHS .


2
3
If x  3, then x = tan


 
 2   ,   3 
3
2
3
2
1
1

LHS
 tan (tan )  tan (2)

=  + tan-1 tan(2 - ) =3 -  (   2    0 )
3


RHS
 tan 1 (tan 3)  tan 1  (tan  3   )   3  
 0     3  
2

 LHS  RHS .
Exercise
1.
Prove that
(i)
(ii)
(iii)
(iv)
(v)
(vi)
2.
sin-1 x + sin-1 y = cos 1  1  x 2 1  y 2  xy  if x, y  0, x2 + y2  1.


2
2
sin-1 x + sin-1 y = sin  x 1  y  y 1  x  if xy > 0, x2 + y2 > 1.


 
 2 if x  0
1
tan 1 x  tan 1 
x 

if x  0
 2
 
if x  1
 4
1
1 1  x
tan x  tan
 
1 x
 3 if x  1
 4
1
2
2x
2x
1 1  x

cos
 tan 1
if x  1
2
2
1 x
1 x
1  x2
 1
2  2 x2 

1
sin 

  sin 1 x 
 2

2
4


2 tan 1 x  sin 1
Prove the following numerical equalities
Text Book of Trigonometry
57
Inverse Trigonometric Functions
1
63 
1
sin  sin 1
 
8 
2
4
3
4

sin  sin 1  sin 1   1
5
5

(i)
(iii)
sin-1 (sin 10) = 3 - 10
(v)
(iv)
(vi)
1
 2 2  
1
sin  sin 1 
   
2
3
 3 

3
8
36 
sin 1  sin 1
 sin 1

5
17
85 2
cos-1 (cos 20) = 20 - 6
46  6
 33 
1 
sin 1  sin
  cos  cos

7 
7  7


1
1

1
 1
 13 
(ix)
4 tan 1  tan 1

cos 1  cos 1     cos 1   
5
239 4
2
 7
 14 
1
1
1
1 
tan 1  tan 1  tan 1  tan 1 
3
5
7
8 4
1
1




cos  2 tan 1   sin  4 tan 1 
7
3


(vii)
(viii)
(x)
(xi)
3.
(ii)
Solve the following equation
52
8
(i)
sin 1 1  x   2sin 1 x 

2
(ii)
 tan 1 x    cot 1 x  
(iii)
sin-1 x – cos-1 x = sin-1 (3x – 2)
(iv)
2tan-1 (2x + 1) = cos-1 x
(v)
cos 1 x  sin 1 x 
(vi) tan-1(x + 1) + tan-1(x – 1) = tan-1

6
2
2
8
31
x2  1
2x
2
 tan 1 2

2
x 1
x 1 3
[Hint: Put x = tan, consider cases x (, 1) , x  (1,1), x (1,  ) ]
Solve cos 1
(vii)
4.
Find the greatest and least value of

 tan 1 x provided |x|  1.]
2

1 x 
1
1
  tan 1 x.
tan 1 
 , 0  x  1 [Hint: Function  tan 1  tan x
4
1 x 
sin-1 x + cos-1 x + tan-1 x [Hint: Function 
(i)
(ii)
5.
Use the fact that tan 1 x is monotonically increa sin g. on R.]
If sin-1 x + sin-1 y + sin-1 z =  prove that x4 + y4 + z4 + 4x2y2z2 = 2(x2y2 + y2z2 + z2x2)
[Hint: Solution: We have sin 1 x  sin 1 y    sin 1 z take cos of both sides , we get
1  x2

6.
7.
8.
1  y 2  xy  1  z 2
1  x2
1  y2
 xy  1  z 2 . Now square twice to obtain the required result.]
 x cos 
1  cos 
Reduce to its simplest form tan 1 
[Ans. ]
  cot 

 1  x sin  
 x  sin  
If tan-1 y = 4 tan-1 x, find y as an algebraical function of x.
Hence prove that tan 22 30 is a root of the equation x4 – 6x2 + 1 = 0
m tan  a   
n tan 
1
nm

If
, then Prove that   a  tan 1 
tan a  

2
2
2
cos 
cos  a   
nm

Text Book of Trigonometry
58
Inverse Trigonometric Functions
9.
10.
pq  1
qr  1
rq  1
 cot 1
 cot 1
 2, if p  q  r
pq
qr
rp
[Hint: LHS
  cot 1 q  cot 1 p   cot 1 r  cot 1 q  cot 1 p  cot 1 r  2. ]
pq
qr
rp
If p > 0 > q > r and pq > -1 > pr, show that tan 1
 tan 1
 tan 1

1  pq
1  qr
1  rp
Prove that cot 1
11.
If sin-1 x + sin-1 y + sin-1 z =  prove that x 1  x 2  y 1  y 2  z 1  z 2  2 xyz
12.
[Hint: Put sin 1 x  ,sin 1 y  ,sin 1 y   and use
sin 2 A  sin 2 B  sin 2C  4sin A sin B sin C ]
Using the principal values, express the following as a single angle:
 142 
1
1
3tan 1    2 tan 1    sin 1 

2
5
 65 5 
142
142
1
11
1
5
[Hint: Change sin 1
to tan 1
. Also 3 tan 1  tan 1 , 2 tan 1  tan 1 . ]
2
2
5
12
31
65 5
13.
1 1 2 3 4  1 1 1 3 4  1 1
cot
 tan
  , and
2
3
6
3
3
5
3
2  1 1 1 2 2  1 1
 tan
 .
3
36
3
3
Prove that (i)
(ii)
1 1
tan
2
Answers
3.
(i)
(v)
4.
(i)
0
½
3 
,
4 4
Text Book of Trigonometry
(ii)
-1
(ii)

,0
4
(iii)
59
1, ½
(iv)
0