Download JS-16 Inverse Trigonometric Functions We define inverse

JS-16 Inverse Trigonometric Functions
We 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 

2
to

or from
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  
359

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  
follows that sin-1 y =  - x. A similar discussion
2.
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 
Discussed
cos-1 x
[-1, 1]
[0, ]
Since cos x is MD in
[0, ] and [0, ]
covers acute angles.
tan 1 x
 ,  
  
 , 
 2 2
Same as sin-1 x
cot 1 x
 ,  
(0, )
Same as cos-1 x
sec 1 x
 ,   1,  
   
0, 2    2 , 
same as cos-1 x
(/2 is not attained)
cosec-1 x
 ,   1,  
    
  2 , 0    0, 2 
same as sin-1 x
(0 is not attained)
The following additional remarks on inverse functions will be useful.
(i)
sin-1 x is positive or negative according as x is positive or negative. Same is true for tan-1 x.
(ii)
cos-1 x  0 for all x within its domain
(iii)
cot-1 x > 0 for all x
360
(iv)
sin-1 x, cos-1 x are continuous and differentiable in (-1, 1)
(v)
tan-1 x, cot-1 x are continuous and differentiable every where.
(vi)
sin-1 x, tan-1 x are monotonically increasing while cos-1 x, cot-1 x are decreasing in their respective
domains.
Results
1.
2.
3.
4.
5.
(i) sin-1 x = cos ec 1
1
1
, x  0, x  1 (ii) cos-1 x = sec 1 , x  0,
x
x
(iii)
1

cot 1
if x  0


x
1
tan x  
cot 1 1   if x  0

x
(i)
sin-1 (- x)
=
- sin-1 x, x  1
(ii)
cos-1 (- x)
=
 - cos-1 x,
(iii)
tan-1 (- x)
=
- tan-1 x for all x
(i)
sin(sin-1 x)
=
x,
x 1
(ii)
cos(cos-1 x)
=
x,
x 1
(iii)
tan(tan-1 x)
=
x for all x
(i)
sin-1(sin x)
=  1
(ii)
cos-1(cos x)
= 2n - x
if x   2n  1 , 2n 
= x – 2n
if x   2n,  2n  1  
n
x 1
 x  n  , x  n 

x y
,
1  xy
(i)
tan-1 x + tan-1 y = tan 1
(ii)
tan-1 x + tan-1 y =   tan 1

, n 
2
x, y  0,
x y
,
1  xy
x, y  0,
361

2 
xy  1
xy  1
x 1
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



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 

7.
sin 1 2  1  x 2
8.
sin 1 ( x 1  y 2  y 1  x 2 ) ,
if x, y  0, x 2  y 2  1


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 ),
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

Exercise
1.
Find the principal values of each of the following :
(i)
tan 1  3


(ii)
1
cos 1  
2
(iv)
sec1  2


(v)
 1 
cot 1  

3

(vii)
 1 
sin 1 

 2
(viii)
cos ec 1 2
(iii)
(vi)
(ix)
362
sin 1 1
 1
cos 1   
 2
tan 1  1
(x)
 2 
sec 1  

3

 1
 2
 1
 2
2.
Find the value of tan 1 1  cos 1     sin 1   
3.
Find the value of tan 1 3  sec1  2 
4.
Find the values of each of the following :
5.
6.
(i)
1 

tan 1  2 cos(2sin 1 ) 
2 

(iii)
3π 

tan 1  tan  (iv)
4 

(ii)
2π 

sin 1  sin 
3 

π
3
7π 

cos 1  cos 
6 

 1 
 2 
(v) sin   sin 1    
Prove the following identities in a proper domain.
(i)
sin 1 x  cos1 1  x2
(ii)
cos1 x  sin 1 1  x2
(iii)
 x 
sin 1 x  tan 1 

2
 1 x 
(iv)
 x 
tan 1 x  sin 1 

2
 1 x 
(v)
 1  x2
cos 1 x  tan 1 
 x

(vi)
2sin 1 x  sin 1  2 x 1  x 2 


(vii)
 3a 2 x  x3 
x
  3tan 1  
tan 1 
2
2
a
 a  a  3x  
(viii)
tan 1 x 
(ix)
cos1 x  2sin 1
(xi)
 1  x2  1  x2
tan 1 
 1  x 2  1  x 2




1 x
2
 ab  1
1
 1 x 
cos 1 
 , xε  0,1
2
 1 x 
 bc  1
 ca  1
 cot 1 
 cot 1 
0
(x) cot 1 


 a b 
 bc 
 c  a 
 π 1 1 2
   cos x
 4 2
Prove the following:
(i)
3
8
77
sin 1  sin 1  sin 1
5
17
85
(ii)
sin 1
5
7
253
 sin 1
 cos 1
13
25
325
(iii)
sin 1
1
π
 cot 1 3 
4
5
(iv)
cos 1
4
3
27
 tan 1  tan 1
5
5
11
363
7.
(v)
3
5  33

cos sin 1  sin 1  
5
13  65

(vii)
cos 1
(ix)
tan 1
(xi)
1
1
1
1 π
tan 1  tan 1  tan 1  tan 1 
3
5
7
8 4
(xiii)
3 tan 1
(xv)
1
1
1 π
4 tan 1  tan 1  tan 1

5
70
99 4
(xvii)
2sin 1
cos 1
63
1
3
 2 tan 1  sin 1
65
5
5
4
12
33
 cos 1  cos 1
5
13
65
(viii)
tan 1
1
2 1
3
 tan 1  cos 1
4
9 2
5
1
1
2
 tan 1  tan 1
7
13
9
(x)
1
1
1 π
2 tan 1  tan 1  2 tan 1 
5
7
8 4
1
1 π
2 tan 1  tan 1 
3
7 4
tan 1
m
mn π
 tan 1

n
mn 4
(xvi)
sin 1
12
4
63
 cos 1  tan 1
π
13
5
16
(xviii)
 1  x2 
π
1  2 x 
tan 1 

  tan 
2 
 1 x  2
 2x 
1
1
1
π
 tan 1
 tan 1

(xiv)
4
20
1985 4
3
24
 tan 1
5
7
(xii)
Write each of the following in the simplest form :
(i)
 2x 
sin 1 
(ii)
2 
 1 x 
(iv)
sin  tan 1 x 2  cot 1 x 2 
(vii)
(x)
(xii)
8.
(vi)
sin 1 3 x  4 x3 
(iii)
(v) cos 1  4 x3  3x 
sin 1  x 1  x  x 1  x 2 


(vi)
 cos x 
tan 1 

 1  sin x 
2


x
1  cos x  sin x 
1  3 x  x 
tan
,
x

π
(viii)
(ix)
tan 1 
,
x

a
tan



 1  3x 2 
2
2
 cos x  sin x 


 a x 
2
 1  x 2  1
 1 1  2 x  1

1  1  y 
1
tan  sin 

cos
(xi)
tan 
, x  0



2
2 
x
 1 x  2
 1  y 
2


tan
1
1  cos 3x
1  cos 3x
(xiii)
 1  x2  1  x2 
1
tan 
 (xiv) cot
2
2
 1  x  1  x 
1
Solve the equations :
(i)
sin 1 x  cos 1 x.
(iii)
tan 1 ( x  1)  tan 1 ( x  1)  cot 1 2. (iv)
(ii)
364
tan 1 x  cot 1 x.
cot 1 x  cot 1 2 x 
3
4

1  x2  x

9.
(vi)
cos 1 x  sin 1 x  cos 1 x 3
(viii)
2 cot 1 2  cos 1
(v)
sin 1 x  cos 1 x  sin 1 (3x  2)
(vii)
tan-1(x + 1) + tan-1(x – 1) = tan-1
(ix)
tan 1 x  tan 1 (1  x)  2 tan 1 x  x 2 . (x)
cos 1
(xi)
2
2a
2x
1
1 1  b
sin
tan
 cos
. (xii)
1 a2
1 x2
1 b2
x2 1
2x
4
cot
 tan 1 2

 0.
2x
x 1 3
(xiii)
sin 1 1  x   2sin 1 x 
(xv)
2tan-1 (2x + 1) = cos-1 x
8
31
1

2
(xiv)
2
1 a2
1 1  b

cos
 2 tan 1 x
2
2
1 a
1 b
1
 tan x    cot x 
1
2
cos 1 x  sin 1 x 
(xvi)
3
 cos ec 1 x
5
1
2

52
8

6
Prove that: sec 2 (tan 1 2)  cos ec 2 (cot 1 3)  15

1
costan x 
1

10.
Prove that : (i) sin cot
11.
If cos 1
12.
If sin 1 x  sin 1 y  sin 1 z   prove that
 
x2 1
x2  2
x
y
x 2 2 xy
y2
 cos 1   prove that 2 
cos   2  sin 2  .
a
b
a
ab
b
(i)
x 1  x 2  y 1  y 2  z 1  z 2  2 xyz
(ii)
x 4  y 4  z 4  4x 2 y 2 z 2  2 x 2 y 2  y 2 z 2  z 2 x 2
13.
Prove that
14.
Prove that :
15.
Prove that :
(ii)

x2 1
(ii) cos tan 1 sin cot 1 x 
2
x 2




 3 sin 2 

1  1
tan 1 
  tan  tan     , where     .
2
2
 5  3 cos 2 
4

a
a  2b
 1
 1
tan   cos 1   tan   cos 1  
b
b a
4 2
4 2
(i)
sin-1 x + sin-1 y = cos 1  1  x 2 1  y 2  xy  if x, y  0, x2 + y2  1.


sin-1 x + sin-1 y = sin 1  x 1  y 2  y 1  x 2  if xy > 0, x2 + y2 > 1.


365
(iii)
16.
17.
 
if x  1
 4

 3 if x  1
 4
1 x

1 x
(iv)
tan 1 x  tan 1
(v)
2 tan 1 x  sin 1
(vi)
 1
2  2 x2
sin 1 

 2
2

2
2x
2x
1 1  x

cos
 tan 1
if x  1
2
2
1 x
1 x
1  x2


  sin 1 x 

4

Prove the following numerical equalities
(i)
1
63 
1
sin  sin 1
 
8  2 2
4
(ii)
1
 2 2  
1
sin  sin 1 
   
2
3
 3 

(iii)
3

sin  sin 1  sin 1
5

(iv)
sin 1
(v)
sin-1 (sin 10) = 3 - 10
(vi)
cos-1 (cos 20) = 20 - 6
(vii)
46  6
 33 
1 
sin 1  sin
  cos  cos

7 
7  7


(viii)
4 tan 1
(x)
tan 1
4
 1
5
1
1

 tan 1

5
239 4
(ix)
cos 1
3
8
36 
 sin 1
 sin 1

5
17
85 2
1
 1
 13 
 cos 1     cos 1   
2
 7
 14 
1
1
1
1 
1
1


 tan 1  tan 1  tan 1 
(xi) cos  2 tan 1   sin  4 tan 1 
3
5
7
8 4
7
3


Find the greatest and least value of
(i)
18.
 
 2 if x  0
1
1 1
tan x  tan
 
x
  if x  0
 2
sin-1 x + cos-1 x + tan-1 x
(ii)
1 x 
tan 1 
,0  x  1
1 x 
 x cos 
1  cos 
Reduce to its simplest form tan 1 
  cot 

 1  x sin  
 x  sin  
366
ANSWERS
1.

(i)

(vii)
 4
3
(ii)

3
(iii)

2
(iv)
3
4
(viii)
 6
(ix)
  4 (x)

.
3
3.

4.
(i)
(vi)
 x

4 2
(vii)
sin 1
(x)
1
tan 1 x
2
(xi)
x y
1  xy
(ii)
(v)
3 sin 1 x
(ix)
3 tan 1 x
18.
3
1
cos 1 x 2
2
1
(xv)
1
(iii)
2
(iv)
 3  17
4
(vii)
1
4
(viii)

(xii)
(vi)
0, or 
(x)
a b
1  ab
(xi)
ba
1  ab
(xv)
0
(xvi)
½
(i)
3 
,
4 4
(ii)

,0
4
1
2
(viii)
1
cot 1 x
2

2
x
a
(xiv)
(i)
(ii)
 4
sin 1 x  sin 1 x
2 tan 1 x
8.
(iii)
(iii)
(i)
2

3
3 sin 1 x
7.

(vi)
(ii)
3
4

2
3

4

2.
(xiii)
(v)
19.
(iii)
tan 1 ( x  1)  tan 1 ( x  1)  tan 1


tan 1
25
,
24
3.
(xiii)

1
2
( x  1)  ( x  1)
1
 tan 1
2
1  x 1
2

tan 1
2
1
 tan 1
2
x
2
x2  4  x   2 Both values satisfy given equation .
367
(iv)

4
1
x
(xii)
3x
2
(v)
1, or
(ix)
1
.
2
0
(xiv)
x
2
SOLUTIONS
8.
2
3
1
2
-1
(iv)
(x)
If x  0 then the equation is tan 1

1 1

3
tan x 2 x 
1
4
1 2
2x

2x2  3x  1  0 
cos1
3x
 1
2 x2  1

x
3  17
4

x can not be negative )
Put a  tan  , b  tan  then equation becomes cos1  cos 2   cos1 (cos 2 )  2tan 1 x

11.
1
1
1
3
 tan 1

x
2x
4
2  2  2tan 1 x
 x  tan (   ) 
tan   tan 
1  tan   tan 

a b
.
1  ab
x y
x2
y2
x
y
.  1  2 1  2  cos  .
 cos 1  Taking cosine of both sides
a b
a
b
a
b

x 2 
y2 
 xy

 
 cos    1  2 1  2  which will lead to the desired result.
 ab

 a  b 
2
13.
Put sin 1 x  A,sin 1 y  B ,sin 1 z  C then we are given A  B  C  
We can show by conditional identities sin 2 A  sin 2B  sin 2C  4sin Asin B sin C
 2sin A cos A  sin B cos B  2sin C cos C

 4 xyz
2 x 1  x 2  2 y 1  y 2  2 z 1  z 2  4 xy etc.
The second result can be proved by noting that cos( A  B)   cos C

cos A cos B  sin A sin B   cos C 
 (1  x 2 )(1  y 2 )  x 2 y 2  2 xy
14.
LHS
 tan 1
1  x 2 1  y 2  xy   1  z 2
1  x2 1  y 2
3sin 2
1
 tan 
5  3cos 2 4
3sin 2 tan 
1
4(5  3cos 2 )
 tan 1
368
 1  z 2 etc.
12sin 2  tan  (5  3cos 2 )
20  12cos 2  3sin 2 tan 
Now expression under tan 1

15.
24 tan   5tan   5tan 3   3tan   3tan 3 
20  20 tan 2   12  12 tan 2   6 tan 2 
Take

1
a
a
cos1  then cos 
2
b
b
1  tan
1  tan
17.
(i)

2 

1  tan
1  tan
2
Put sin 1
Now cos
2
2
2
1
8 3
2
4
1

2


  
  
4 2
 




2 1  tan 2 
1  tan   1  tan 
2
2b
2 
2
2
= 
= 




cos 
a
1  tan 2
1  tan 2
2
2

1
 sin 
4

2 tan 3   32 tan 
 tan 
2 tan 2   32

  
LHS  tan     tan
4 2

63
  then . 
8
Our target
(ii)
24 tan 
1  tan 2 

5
tan


3tan

.
2
1  tan 2 
 1  tan 
2
12(1  tan  )
6 tan 2 
20 

1  tan 2 
1  tan 2 
1  cos

2
63
1
 
 0,  and sin   8 , cos 
8
 2
1  cos

2
2

2 ( Note that  ,  belong to  0,   ).


2
2 4
 2

 sin 
4
1
3
4

2
1
2 2
 2 2
2 2
  
Put sin 1  
   then     ,0  and sin   
3 
3
 2 

 
Target  sin  
2
sin

2

1  cos
2
But sin

2
  sin

2
. Also
cos  1  sin 2 
  
    ,0 
 2 
But | cos |  sin 
369

1 1 / 3
2
 cos   1 
8
1
 
9
3
(iii)
Note that : sin 1
3
4
3
3 
 sin 1  sin 1  cos1 
5
5
5
5 2
(iv)
converting to tan 1 we get

sin
2
(v)
45  32
36
 tan 1
60  24
77
 tan 1

1
3
3 8

4
15  tan 1 36
 tan
24
77
1
4  15
3
8
36
Expression  tan
 tan 1  tan 1
4
15
77
1
1
 tan 1

77
77 
 cot 1

36
36
2
First Method : by hit and trial sin10  sin  3  10  and 

2
 3  10 

2
Hence sin 1 sin10  3  10
Second Method :

The general value of sin 1  sin10   n  (1)n (10)
sin (n  (1)n (10)   sin10 for all integers n 
The principal value must lie between 
If n is even n  2m then we get 

10 

2
 2m  10 



 
2
 1.84  m   1.36
(*)
There is no integer satisfying
(*)
10 
5
2
2
and

2

2
 (2m  1) 10 
3
5 1

m  
4
 4

5

 
2
 2m  10 

If n is odd n  2m  1 then 



2
m 1
370

2

or n  3
10

2
 n  (1)n (10) 

1
5 1
m 
4
 4
 (2m  1)   10 




2
1
10 1
 2m  1  
2
 2

2
Hence principal value of sin 1 (sin10)  3 10
Or in short sin 1 (sin10)  n  (1)n (10)
Test few initial values like n   2,  1, 0,1,2 etc .
cos1 (cos20)  2n  20 For n  0 ,
(vi)
 20 which is much out side
0, 
For n 1 we get 4  20 0,   no matter whether we take  sign or  sign .
(viii)
A lay woman approach is much simpler
sin
33
5
5
2
 sin(4  ) = sin
 sin
7
7
7
7
Since
2


 33
lies between 
and
, sin 1  sin
7
7
2
2

Again cos
46
4

 cos  6 
7
7

 cos1 cos
18.
note that 6  20 0,  
6  20
n3
For
f ( x) 
(i)

2
46 4

7
7

 2

7

4

  cos
7

sum 
6
7
 tan 1 x ,  1 x  1
Since f ( x) is strictly monotonic minimum and maximum values are

2
19.


and
4

2


4
Expression  tan 1
 tan 1
ie

4
and
3
4
(ii)
x cos
x  sin 
 tan 1
1  x sin 
cos
x cos 2   (1  x sin  )( x  sin  )
(1  x sin  )cos  x( x  sin  )cos
 tan 1
sin  ( x 2  1  2 x sin  )
cos ( x 2  1  2 x sin  )
show f ( x) is MD .
=
tan 1
 tan 1
 tan 1  tan   
371
x cos
x  sin 

1  sin 
cos
x( x  sin  ) cos
1
(1  x sin  ) cos
( x 2  1)sin   2 x sin 2 
( x 2  1)cos  x sin 2 
372