MATHS
Inverse Trigonometric Functions Introduction : The student may be familiar about trigonometric functions viz sin x, cos x, tan x, cosec x, sec x, cot x with respective domains R, R, R – {(2n + 1) /2}, R – {n}, R – {(2n + 1) /2}, R – {n} and respective ranges [–1, 1], [–1, 1], R, R – (–1, 1), R – (–1, 1), R. Correspondingly, six inverse trigonometric functions (also called inverse circular functions) are defined.
sin–1x : The symbol sin–1x or arcsinx denotes the angle so that sin = x. As a direct meaning, sin–1x is not a function, as it does not satisfy the requirements for a rule to become a function. But by a suitable choice [–1, 1] as its domain and standardized set [–/2, /2] as its range, then rule sin–1 x is a single valued function. Thus sin–1x is considered as a function with domain [–1, 1] and range [–/2, /2]. The graph of y = sin–1x is as shown below, which is obtained by taking the mirror image, of the portion of the graph of y = sin x, from x = –/2 to x = /2, on the line y = x.
cos –1 x :
By following the discussions, similar to above, we have cos–1 x or arccos x as a function with domain [–1, 1] and range [0, ].
The graph of y = cos–1x is similarly obtained as the mirror image of the portion of the graph of y = cos x from x = 0 to x =
tan –1 x :
We get tan–1 x or arctanx as a function with domain R and range (–/2, /2). Graph of y = tan–1x
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MATHS y
/2 o
x
/2
cosec –1 x :
cosec–1x or arccosec x is a function with domain R – (–1, 1) and range [–/2, /2] – {0}. Graph of y = cosec–1x
sec –1 x :
sec–1x or arcsec x is a function with domain R – (–1, 1) and range [0,] – {/2}. Graph of y = sec–1x
cot–1 x :
cot–1x or arccot x is a function with domain R and range (0, ) Graph of y = cot–1x
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MATHS
1 1 1 1 . Example # 1 : Find the value of tan cos tan 3 2
Solution :
1 1 1 1 = tan tan cos tan 3 2
= tan = 3 6 6
1 3
.
Example # 2 : Find domain of sin–1 (2x2 – 1) Solution :
Let y = sin–1 (2x2 – 1) For y to be defined – 1 (2x2 – 1) 1 0 2x2 2 0 x2 1 x [–1, 1].
Self practice problems :
(2)
1 1 Find the value of sin 3 sin 2 Find the value of cosec [sec–1 (– 2 ) + cot–1 ( –1)]
(3)
Find the domain of y = sec–1 (x2 + 3x + 1)
(4)
x2 Find the domain of y = cos–1 2 1 x
(5)
Find the domain of y = tan–1 ( x 2 1)
(1)
Answers :
(1) (3)
1 (2) –1 (– , – 3] [ – 2, – 1] [0, )
(4)
R
(5)
(– , –1] [1, )
Property 1 : “–x” The graphs of sin–1x, tan–1 x, cosec –1x are symmetric about origin. Hence we get sin–1 (–x) = – sin–1x tan–1 (–x) = – tan–1x cosec –1 (–x) = – cosec –1x. Also the graphs of cos –1x, sec –1x, cot–1x are symmetric about the point (0, /2). From this, we get cos –1 (–x) = – cos –1x sec –1 (–x) = – sec –1x cot–1 (–x) = – cot–1x.
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MATHS
Property 2 : T(T –1 ) (i)
sin (sin1 x) = x,
1 x 1
Proof : Let = sin–1x. Then x [–1, 1] & [–/2, /2]. sin = x, by meaning of the symbol sin (sin–1 x) = x Similar proofs can be carried out to obtain (ii) (iii) (iv) (v) (vi)
cos (cos 1 x) = x, tan (tan1 x) = x, cot (cot1 x) = x, sec (sec 1 x) = x, cosec (cosec 1 x) = x,
1 x 1 xR xR x 1, x 1 |x| 1
The graph of y = sin (sin–1x) cos (cos –1x)
The graph of y = tan (tan–1x) cot (cot–1x)
The graph of y = cosec (cosec –1x) sec (sec –1x)
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MATHS Property 3 : T –1 (T) (i)
x [2n / 2, 2n / 2] 2n x, sin–1 (sin x) = ( 2n 1) x, x [( 2n 1) / 2, ( 2n 1) / 2], n Z
Proof : If x [2n – /2, 2n + /2], then –2n + x [–/2, /2] and sin (–2n + x) = sin x. Hence sin–1 (sin x) = –2n + x for x [2n – /2, 2n + /2]. Proof of 2nd part is left for the students. Graph of y = sin–1 (sin x)
(ii)
x [2n, ( 2n 1)] 2n x, cos–1 (cos x) = 2n x, x [( 2n 1) , 2n], n
Graph of y = cos–1 (cos x) y
=
y
y
x
x
–
–
2p
=
2p
=
y
+
x
y=
x
(iii)
x 2
tan–1 (tan x) = – n + x, n – /2 < x < n + /2, n Z Graph of y = tan–1 (tan x)
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MATHS
(iv)
cosec –1 (cosec x) is similar to sin–1 (sin x) Graph of y = cosec –1 (cosec x)
(v)
sec –1 (sec x) is similar to cos –1 (cos x) Graph of y = sec –1 (sec x)
(vii)
cot–1 (cot x) = –n + x, x (n , (n + 1) ), n Z Graph of y = cot–1 (cot x)
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MATHS
Remark :
sin (sin–1x), cos (cos –1x), .... cot (cot–1x) are aperiodic (non periodic) functions where as sin–1 (sin x), ..., cot–1(cot x) are periodic functions.
Property 4 : “1/x” (i)
cosec –1(x) = sin–1(1/x), |x| 1
Proof : Let cosec –1 x = 1/x = sin sin–1(1/x) = sin–1 (sin ) = (as [–/2, /2] – {0}) = cosec –1x (ii) sec –1 x = cos –1 (1/x), |x| 1 (iii)
tan 1(1/ x ), x0 cot–1x = 1 tan (1/ x ), x 0
Property 5 : “/2” , 1 x 1 2 Proof : Let A = sin–1x and B = cos –1x sin A = x and cos B = x sin A = cos B sin A = sin (/2 – B) A = /2 – B, because A and /2 – B [–/2, /2] A + B = /2. (i)
sin1 x + cos1 x =
Similarly, we can prove
, xR 2
(ii)
tan1 x + cot1 x =
(iii)
cosec 1 x + sec 1 x =
, x 1 2
1 3 . Example # 3 : Find the value of cosec cot cot 4 Solution :
cot (cot–1 x) = x, x R
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MATHS
3 1 3 = cot cot 4 4 1 3 3 = cosec cot cot 4 = cosec 4
Example # 4
2.
3 . Find the value of tan–1 tan 4
Solution :
tan–1 (tan x) = x
if x , 2 2
As
3 , 4 2 2
3 3 , 2 2 4
3 3 tan–1 tan 4 4
graph of y = tan–1 (tan x) is as :
from the graph we can see that if
3 < x< , 2 2
then tan–1 (tan x) = x –
3 tan–1 tan 4
=
3 – =– 4 4
Example # 5 : Find the value of sin–1 (sin7) and sin–1 (sin (–5)). Solution.
Let y = sin–1 (sin 7) sin–1 (sin 7) 7 as 7 , 2 2
2 < 7 <
5 2
graph of y = sin–1 (sin x) is as :
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MATHS
From the graph we can see that if 2 x
5 , then 2
y = sin–1(sin x ) can be written as : y = x – 2 sin–1 (sin 7) = 7 – 2 Similarly if we have to find sin–1 (sin(–5)) then 3 – 2 < – 5 < – 2 from the graph of sin–1 (sin x), we can say that sin–1 (sin(–5)) = 2 + (–5) = 2 – 5 Example # 6 : Find the value of cos–1 {sin( – 5)} Solution :
Let y = cos–1 {sin(–5)} = cos–1 (– sin 5) = – cos–1 (sin 5)
(cos–1 (– x) = – cos–1x, |x| 1)
= – cos–1 cos 5 2
Note that :
..........(i)
– 2 < 5 < – 2
graph of cos–1 (cos x ) is as :
From the graph we can see that if – 2 x – then cos–1 (cosx) = x + 2
5 5 from the graph cos–1 cos 5 = 5 + 2 = 2 2 2
from (i), we get
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MATHS
5 5 y=– 2
y=5–
3 . 2
2 Example # 7 : Find the value of tan cot 1 3
Solution :
2 Let y = tan cot 1 3
........(i)
cot–1 (–x) = – cot–1x, x R (i) can be written as
2 y = tan cot 1 3 2 y = – tan cot 1 3
1 x
cot–1 x = tan–1
3 y = – tan tan 1 2
if
x>0
y=–
3 2
3 Example # 8 : Find the value of sin tan 1 . 4
Solution :
3 sin tan 1 = sin 4
3 1 3 sin = 5 5
1 1 5 Example # 9 : Find the value of tan cos 3 2
Solution :
1 1 5 Let y = tan cos 3 2 5 = 3
..........(i)
5 0, and cos = 3 2
Let
cos–1
(i) becomes y = tan 2
1 cos tan2 = = 1 cos 2
..........(ii)
5 3 5 (3 5 ) 2 3 = = 5 3 5 4 1 3 1
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MATHS tan
=± 2
3 5 2
0, 4 2
.........(iii)
tan
>0 2
3 5 = 2 2
from (iii), we get y = tan
Example # 10 : Find the value of cos (2cos–1x + sin–1x) when x =
Solution :
1 5
1 1 1 1 1 cos 2 cos 1 sin 1 = cos cos 1 sin 1 cos 1 5 5 5 5 5
1 1 = – sin = cos cos 2 5
=–
Aliter : Let
cos 1
1 = 5
2
=–
cos =
1 and 0, 5 2
sin =
24 5
.........(i)
2 6 . 5
24 sin–1 (sin ) = sin–1 5 0, 2 equation (ii) can be written as 24 = sin–1 5
1 1 5
1 cos 1 5
..........(ii) sin–1 (sin ) =
1 = cos–1 5
24 1 cos–1 = sin–1 5 5 Now equation (i) can be written as
1 24 y = – sin sin 5
24 [–1, 1] 5
from equation (iii), we get y = –
........(iii)
1 24 = sin sin 5
24 5
24 5
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MATHS Example # 11 : Solve sin–1 (x2 – 2x + 1) + cos–1(x2 – x) =
Solution :
sin–1(f(x)) + cos–1(g(x)) =
2
x2 – 2x + 1 = x2 – x
2
f(x) = g(x) and –1 f(x), g(x) 1
x = 1, accepted as a solution
Self practice problems : (6)
1 Find the value of cos sin sin 6
(8)
Find the value of cos–1 (cos 13)
(9)
5 Find sin–1 (sin ), cos–1(cos), tan–1 (tan ), cot–1(cot) for , 3 2
(10)
Find the value of cos–1 (– cos 4)
(11)
7 Find the value of tan–1 tan 8
(12)
1 Find the value of tan–1 cot 4
(13)
2 Find the value of sec cos 1 3
(14)
1 Find the value of cosec sin1 3
(15)
Find the value of sin (2cos–1x + sin–1x) when x =
(16)
Solve the following equations (i) 5 tan–1x + 3 cot–1x = 2 (ii)
(17)
1 41 Evaluate tan cos ec 4
(18)
16 Evaluate sec cot 1 63
(19)
1 1 3 Evaluate sin cot 2 4
(20)
1 Evaluate tan 2 tan 1 5 4
(21)
Solve sin–1(x2 – 2x + 3) + cos–1(x2 – x) =
2
Answers :
(6)
3 2
1 3 Find the value of sin cos cos 4
(7)
1 5
4 sin–1x = – cos–1x
(7)
not defined
(9)
3 – , – 2, – 3, – 2
(8)
13 – 4
(10)
4–
(11)
8
(12)
1 4 2
(13)
3 2
(14)
– 3
(15)
1 5
(16).
(i)
(ii)
x=
x=1
1 2
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MATHS (17)
4 5
(21)
No solution
65 16
(18)
(19)
2 5 5
(20)
7 17
Property 6 : Identities on addition and subtraction:
(i)
1 2 2 & (x 2 y 2 ) 1 sin x 1 y y 1 x , x 0, y 0 sin x + sin y sin1 x 1 y 2 y 1 x 2 , x 0, y 0 & x 2 y 2 1 1
1
Proof : Let A = sin–1 x and B = sin–1y where x, y [0, 1]. sin (A + B) = x
1 y 2 + y
1 x 2
sin–1 sin (A + B) = sin–1 x 1 y 2 y 1 x 2
sin–1 x 1 y 2 y 1 x 2 for 0 A B / 2 A B x2 y2 1 sin1 x sin1 y, = = (sin1 x sin1 y ), x 2 y 2 1 ( A B) for / 2 A B
(ii)
sin–1x – sin–1y = sin–1 x 1 y 2 y 1 x 2 ; x, y [0, 1]
(iii)
2 2 cos–1 x + cos–1y = cos–1 xy 1 x 1 y ; x, y [0, 1]
(iv)
1 2 2 cos xy 1 x 1 y ; 0 x y 1 cos–1x – cos–1y = cos 1 xy 1 x 2 1 y 2 ; 0 y x 1
(v)
(vi)
/2 /2 1 x y tan 1 xy tan–1x + tan–1y = 1 x y tan 1 xy
if if
x, y 0 & xy 1 x, y 0 & xy 1
if
x, y 0 & xy 1
if
x, y 0 & xy 1
xy tan–1x – tan–1y = tan–1 1 xy , x 0, y 0
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MATHS Notes :(i)
x 2 + y2 1 & x, y 0 x 2 + y2 1 & x, y 0
and
2
0 sin1 x + sin1 y
sin1 x + sin1 y 2 ; xy > 1 and x, y 0 < tan1 x + tan1 y < 2 2
(ii)
xy < 1 and x, y 0 0 tan1 x + tan1 y <
(iii)
For x < 0 or y < 0 these identities can be used with the help of property “– x” i.e. change x or y to x or y which are positive.
Example # 12 : Show that sin–1
3 84 15 + sin–1 = – sin–1 5 85 17 2
Solution :
2
3 3 15 8226 15 > 0, > 0 and + = >1 5 17 5 7225 17
3 15 + sin–1 = – sin–1 5 17
sin–1
3 225 15 9 1 1 5 289 17 25
3 8 15 4 84 . = – sin–1 = – sin–1 . 5 17 17 5 85
Example # 13 : Evaluate cos–1
Solution :
Let z = cos–1
12 4 63 + sin–1 – tan–1 13 5 16
4 4 = – cos–1 5 2 5
sin–1
z = cos–1
z=
12 4 63 + sin–1 – tan–1 13 5 16
12 63 1 4 – tan–1 + cos . 2 5 13 16
12 63 1 4 cos 1 – tan–1 – cos 5 13 16 2
4 12 4 12 > 0, > 0 and < 5 13 5 13
cos–1
equation (i) can be written as
4 12 – cos–1 = cos–1 5 13
4 12 16 1 5 13 25
.........(i)
1
144 63 –1 169 = cos 65
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MATHS z=
63 63 – tan–1 – cos–1 65 16 2
63 63 – tan–1 z = sin–1 65 16
63 63 = tan–1 sin–1 65 16
from equation (ii), we get
63 63 – tan–1 z = tan–1 16 16
Example # 14 : Evaluate tan–1 9 + tan–1
Solution :
9 > 0,
tan–1 9 + tan–1
.........(ii)
z=0
5 4
5 5 > 0 and 9 > 1 4 4
5 9 5 π 3 4 = + tan–1 = + tan–1 (– 1) = – = . 5 4 4 4 1 9. 4
Example # 15 : Define y = cos–1 (4x3 – 3x) in terms of cos–1 x and also draw its graph. Solution : Part - 1: Let y = cos–1 (4x3 – 3x) Domain : [–1, 1] and range : [0, ] Let cos–1 x = [0, ] and x = cos y = cos–1 (4 cos3 – 3 cos ) y = cos–1 (cos 3) ...........(i)
Fig.: Graph of cos–1 (cos x)
[0, ] 3 [0, 3] to define y = cos–1 (cos 3), we consider the graph of cos–1 (cos x) in the interval [0, 3]. Now, from the above graph we can see that
(i)
if 0 3 cos–1 (cos 3) = 3 from equation (i), we get y = 3 if 3
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MATHS 3
y = 3
if
0
y = 3 cos–1x
if
1 x1 2
(ii)
if 3 2 cos–1 (cos 3) = 2– 3 from equation (i), we get y = 2 – 3 if 3 2
y = 2 – 3
if
2 < 3 3
y = 2 – 3cos–1 x
if
–
1 1 x< 2 2
(iii)
3 3 cos–1 (cos 3) = – 2 + 3 from equation (i), we get y = – 2 + 3 if 3 3
y = – 2 + 3
if
2 3
y = – 2 + 3 cos–1 x
if
– 1 x –
from (i), (ii) & (iii), we get
1 2
1 1 ; x 1 3 cos x 2 1 1 y = cos–1 (4x3 – 3x) = 2 3 cos 1 x ; x 2 2 2 3 cos 1 x ; 1 x 1 2 Part-2 : For y = cos–1 (4x3 – 3x) domain : [–1, 1] range : [0, ] (i)
if
1 x 1 , y = 3 cos–1x. 2
dy = dx
dy <0 dx
3 1 x
2
= – 3(1 – x2)–1/2
if
...........(i)
1 x , 1 2
1 decreasing if x , 1 2
again if we differentiate equation (i) w.r.t. ‘x’, we get d2 y dx
2
=–
3x (1 x 2 )3 / 2
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MATHS
(ii)
d2 y dx
2
–
dy = dx
increasing
(a)
d2 y 1 if x , 0 then <0 dx 2 2
(b)
concavity downwards
1 if x , 1 2
3 1 x
dy >0 dx
2
1 1 if x , and 2 2
1 1 if x , 2 2 d2 y dx
2
=
3x (1 x 2 )3 / 2
1 if x , 0 2
concavity downwards
d2 y 1 if x 0, then >0 dx 2 2
if x 0, 1 2
concavity upwards
Similarly if – 1 x < –
1 1 x < , y = 2 – 3cos–1 x. 2 2
if
(iii)
1 if x , 1 2
<0
dy 1 d2 y then < 0 and > 0. dx 2 dx 2
the graph of y = cos–1 (4x3 – 3x) is as
Self practice problems: 4 5 16 + sin–1 + sin–1 5 13 65
(22)
Evaluate sin–1
(23)
If tan–14 + tan–1 5 = cot–1 , then find ‘’
(24)
Prove that 2 cos–1
(25)
Solve the equation tan–1 (2x) + tan–1 (3x) =
(26)
Solve the equation sin–1x + sin–1 2x =
(27)
Define y = sin–1 (3x – 4x3) in terms of sin–1x and also draw its graph.
(28)
3x x 3 Define y = tan–1 1 3x 2
3 13
+ cot–1
16 1 7 + cos–1 = 63 2 25
4
2 3
in terms of tan–1 x and also draw its graph.
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MATHS Answers.
(22)
2
(27)
1 ; 3 sin x 1 ; y = sin–1 (3x – 4x3) = 3 sin x 3 sin1 x ;
=–
(23)
19 9
(25)
x=
1 6
(26)
x=
1 2
1 1 x 2 2 1 x 1 2 1 1 x 2
graph of y = sin–1 (3x – 4x3)
(28)
y = tan–1
1 ; 3 tan x 3x x 3 1 ; 1 3 x 2 = 3 tan x 1 3 tan x ;
3x x 3 Fig.: Graph of y = tan–1 2 1 3x
1 3
x
x 1 3
1 3 1 3
x
Property 7 : Miscellaneous identities
(i)
sin 2 x 1 x 2 1
1 2 sin x 1 = 2 sin x 2 sin 1 x
if
|x| 1
2 1 x 1 2 if 1 x 1 2 if
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MATHS
graph of y = sin1 2 x 1 x 2
(ii)
2 cos 1 x if 0 x 1 = 1 2 2 cos x if 1 x 0
cos 1 (2 x 2 1)
graph of y = cos 1 (2 x 2 1)
(iii)
1
tan
2 tan 1x 1 = 2 tan x 2 tan 1x
2x
1 x2
graph of y = tan1
(iv)
(v)
cos1
1 x2 1 x 2
if
x 1
if
x 1
2x 1 x2
2 tan 1 x = 2 tan 1 x 2 tan 1 x
2x sin1 1 x2
graph of y = sin1
if | x | 1
if | x | 1
if
x 1
if
x 1
2x 1 x2
2 tan 1 x if x 0 = 1 2 tan x if x 0
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MATHS
graph of y = cos 1
1 x2 1 x 2
(vi)
If tan1 x + tan1 y + tan1 z = , then x + y + z = xyz
(vii)
If tan1 x + tan1 y + tan1 z =
(viii)
tan1 1 + tan1 2 + tan1 3 =
(ix)
tan1 1 + tan1
, then xy + yz + zx = 1 2
1 1 + tan1 = 3 2 2
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