Inverse Trignometric Functions
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Transcript of Inverse Trignometric Functions
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Chapter 2INVERSE TRIGONOMETRIC
FUNCTIONS
2.1 Overview
2.1.1 Inverse function
Inverse of a function f exists, if the function is one-one and onto, i.e, bijective.
Since trigonometric functions are many-one over their domains, we restrict their
domains and co-domains in order to make them one-one and onto and then find
their inverse. The domains and ranges (principal value branches) of inverse
trigonometric functions are given below:
Functions Domain Range (Principal value
branches)
y = sin1x [1,1]
,2 2
y = cos1x [1,1] [0,]
y = cosec1x R (1,1)
, {0}2 2
y = sec1x R (1,1) [0,]
2
y = tan1x R
,2 2
y = cot1x R (0,)Notes:
(i) The symbol sin1x should not be confused with (sinx)1. Infact sin1x is an
angle, the value of whose sine isx, similarly for other trigonometric functions.
(ii) The smallest numerical value, either positive or negative, of is called theprincipal value of the function.
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INVERSE TRIGONOMETRIC FUNCTIONS 19
(iii) Whenever no branch of an inverse trigonometric function is mentioned, we mean
the principal value branch. The value of the inverse trigonometic function which
lies in the range of principal branch is its principal value.
2.1.2 Graph of an inverse trigonometric functionThe graph of an inverse trigonometric function can be obtained from the graph of
original function by interchangingx-axis andy-axis, i.e, if (a, b) is a point on the graph
of trigonometric function, then (b, a) becomes the corresponding point on the graph of
its inverse trigonometric function.
It can be shown that the graph of an inverse function can be obtained from the
corresponding graph of original function as a mirror image (i.e., reflection) along the
liney =x.
2.1.3 Properties of inverse trigonometric functions
1. sin1 (sinx) =x :
,2 2
x
cos1(cosx) =x : [0, ]x tan1(tanx) =x :
,
2 2x
cot1(cotx) =x : ( )0, x
sec1(secx) =x :
[0, ] 2
x
cosec1(cosec x) =x :
, {0}2 2
x
2. sin (sin1x) =x : x[1,1]cos(cos1x) =x : x[1,1]
tan(tan1x) =x : xRcot(cot1x) =x : xRsec(sec1x) =x : xR (1,1)cosec(cosec1x) =x : xR (1,1)
3.1 11sin cosec x
x
: xR (1,1)
1 11cos sec xx
: xR (1,1)
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20 MATHEMATICS
1 11tan cot xx
: x > 0
= + cot1x : x < 0
4. sin1 (x) = sin1x : x[1,1]cos1 (x) = cos1x : x[1,1]tan1 (x) = tan1x : xRcot1 (x) = cot1x : xRsec1 (x) = sec1x : xR(1,1)cosec1 (x) = cosec1x : xR(1,1)
5. sin1x + cos1x =
2 : x[1,1]
tan1x + cot1x =
2 : xR
sec1x + cosec1x =
2 : x R[1,1]
6. tan1x + tan1y = tan1 1
y
xy
: xy< 1
tan1x tan1y = tan1; 1
1
x yxy
xy
> +
7. 2tan1x = sin1 22
1 : 1 x 1
2tan1x = cos12
21 1
x
: x 0
2tan1x = tan1 22
1
x: 1
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INVERSE TRIGONOMETRIC FUNCTIONS 21
Solution If cos13
2
= , then cos =3
2.
Since we are considering principal branch, [0, ]. Also, since3
2> 0, being in
the first quadrant, hence cos13
2
=
6.
Example 2 Evaluate tan1
sin2
.
Solution tan1
sin2
= tan1
sin
2
= tan1(1) =
4 .
Example 3 Find the value of cos113
cos6
.
Solution cos113
cos6
= cos1 cos(2 )
6
+
=1 cos cos
6
=6
.
Example 4 Find the value of tan19
tan8
.
Solution tan19
tan 8
= tan1 tan 8
+
=1tan tan
8
=
8
Example 5 Evaluate tan (tan1( 4)).
Solution Since tan (tan1x) = x, x R, tan (tan1( 4) = 4.
Example 6 Evaluate: tan1 3 sec1 (2) .
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22 MATHEMATICS
Solution tan1
3 sec1
( 2) = tan1
3 [ sec1
2]
=1 1 2cos
3 2 3 3 3
+ = + =
.
Example 7 Evaluate:1 1 3sin cos sin
2
.
Solution1 1 13 sin cos sin sin cos
2 3
=
1 1 sin
2 6
.
Example 8 Prove that tan(cot1x) = cot (tan1x). State with reason whether the
equality is valid for all values ofx.
Solution Let cot1x = . Then cot =x
or,
tan =2
1 tan 2
x =
So1 1 1 tan(cot ) tan cot cot cot cot(tan )
2 2x x x = = = =
The equality is valid for all values ofx since tan1x and cot1x are true forx R.
Example 9 Find the value of sec1tan
2
y
.
SolutionLet1tan =
2
y, where
,
2 2
. So, tan =2
y,
which gives
24
sec=2
y.
Therefore,
2
1 4sec tan =sec =2 2
yy +
.
Example 10 Find value of tan (cos1x) and hence evaluate tan1 8cos
17
.
Solution Let cos
1
x = , then cos =x, where [0,]
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INVERSE TRIGONOMETRIC FUNCTIONS 23
Therefore, tan(cos1x) =
2 21cos 1 tan = .
cos
x
x=
Hence
2
1
81
178 15tan cos =
817 8
17
=
.
Example 11 Find the value of1 5sin 2cot
12
Solution Let cot15
12
=y . Then coty =5
12
.
Now1 5sin 2cot
12
= sin 2y
= 2siny cosy =12 5
213 13
since cot 0, so , 2
y y <
120
169
Example 12 Evaluate1 11 4cos sin sec
4 3
Solution1 11 4
cos sin sec4 3
=
1 11 3
cos sin cos4 4
+
=1 1 1 11 3 1 3cos sin cos cos sin sin sin cos
4 4 4 4
=
2 23 1 1 3
1 1 4 4 4 4
=
3 15 1 7 3 15 7
4 4 4 4 16
.
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24 MATHEMATICS
Long Answer (L.A.)
Example 13 Prove that 2sin13
5 tan1
17
31=
4
Solution Let sin13
5= , thensin =
3
5, where ,
2 2
Thus tan =3
4, which gives = tan1
3
4.
Therefore, 2sin13
5 tan1
17
31
= 2 tan117
31= 2 tan1
3
4 tan1
17
31
=1 1
32.
174tan tan9 31
1
16
= tan1124 17tan
7 31
=1
24 17
7 31tan24 17
1 .7 31
+
=4
Example 14 Prove that
cot17 + cot18 + cot118 = cot13
Solution We have
cot17 + cot18 + cot118
= tan11
7+ tan1
1
8+ tan1
1
18(since cot1x = tan1
1
x, ifx > 0)
=1 1
1 1
17 8tan tan1 1 18
17 8
+ +
(sincex. y =1 1
.7 8
< 1)
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INVERSE TRIGONOMETRIC FUNCTIONS 25
=1 13 1tan tan
11 18+ =
1
3 1
11 18tan3 1
111 18
+
(sincexy < 1)
=1 65tan
195=
1 1tan3
= cot1 3
Example 15Which is greater, tan 1 or tan1 1?
Solution From Fig. 2.1, we note that tan x is an increasing function in the interval
,2 2
, since 1 >4
tan 1 > tan
4
. This gives
tan 1 > 1
tan 1 > 1 >4
tan 1 > 1 > tan1 (1).
Example 16Find the value of
1 12sin 2 tan cos(tan 3)3
+
.
SolutionLet tan12
3=x and tan1 3 =y so that tanx =
2
3and tany = 3 .
Therefore,
1 12sin 2 tan cos(tan 3)3
+
= sin (2x) + cosy
= 2 2
2 tan 1
1 tan 1 tan
x
x y+
+ + = ( )2
22.
134
1 1 39
++ +
=12 1 37
13 2 26+ = .
/2p/2 p/4/4 p/2/2X
tan xan x
O
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26 MATHEMATICS
Example 17Solve forx
1 11 1tan tan , 01 2
xx x
x
= > +
Solution From given equation, we have 1 11
2tan tan1
x
x
= +
1 1 12 tan 1 tan tan x =
12 3tan
4x
=
1tan6
x
=
1
3x =
Example 18 Find the values ofx which satisfy the equation
sin1x + sin1 (1 x) = cos1x.
Solution From the given equation, we havesin (sin1x + sin1 (1 x)) = sin (cos1x)
sin (sin1x) cos (sin1 (1 x)) + cos (sin1x) sin (sin1 (1 x) ) = sin (cos1x)
2 2 21 (1 ) (1 ) 1 1x x x x+ =
2 22 1 (1 1) 0x x x x x+ =
( )2 22 1 0x x x x =
x = 0 or 2x x2
= 1 x2
x = 0 or x =1
2.
Example 19 Solve the equation sin16x + sin1 6 3x =2
Solution From the given equation, we havesin1 6x =1sin 6 3
2
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INVERSE TRIGONOMETRIC FUNCTIONS 27
sin (sin1 6x) = sin1sin 6 3
2
6x = cos (sin1 6 3x)
6x = 21 108x . Squaring, we get
36x2 = 1 108x2
144x2 = 1 x= 1
12
Note that x = 1
12is the only root of the equation asx =
1
12does not satisfy it.
Example 20 Show that
2 tan11 sin cos
tan .tan tan2 4 2 cos sin
= +
SolutionL.H.S. =1
2 2
2 tan .tan
2 4 2tan
1 tan tan2 4 2
1 1
22since 2 tan tan
1xx =
=1
2
2
1 tan22tan
21 tan
2tan
1 tan 21 tan2
1 tan2
+
+
=
2
1
2 22
2 tan . 1 tan2 2
tan
1 tan tan 1 tan2 2 2
+
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28 MATHEMATICS
=
2
1
2 2 2
2 tan 1 tan2 2
tan
1 tan 1 tan 2 tan 1 tan2 2 2 2
+ + +
=
2
2 2
1
2
2 2
2 tan 1 tan2 2
1 tan 1 tan
2 2tan1 tan 2 tan
2 2
1 tan 1 tan2 2
+ +
+
+ +
=1 sin costan
cos sin
+
= R.H.S.
Objective type questions
Choose the correct answer from the given four options in each of the Examples 21 to 41.
Example 21 Which of the following corresponds to the principal value branch of tan1?
(A) ,2 2
(B) ,2 2
(C) ,2 2
{0} (D) (0, )
Solution (A) is the correct answer.
Example 22The principal value branch of sec1 is
(A) { }, 02 2
(B) [ ]0,
2
(C) (0, ) (D) ,2 2
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INVERSE TRIGONOMETRIC FUNCTIONS 29
Solution (B) is the correct answer.
Example 23 One branch of cos1 other than the principal value branch corresponds to
(A)3
,2 2
(B) [ ]3
, 22
(C) (0, ) (D) [2, 3]
Solution (D) is the correct answer.
Example 24The value of1 43
sin cos 5
is
(A)3
5
(B)
7
5
(C)
10
(D)
10
Solution (D) is the correct answer.1 140 3 3sin cos sin cos 8
5 5
+ = +
=1 13 3
sin cos sin sin5 2 5
=
=1sin sin
10 10
=
.
Example 25The principal value of the expression cos1 [cos ( 680)] is
(A)2
9
(B)
2
9
(C)
34
9
(D)
9
Solution (A) is the correct answer. cos1 (cos (680)) = cos1 [cos (720 40)]
= cos1 [cos ( 40)] = cos1 [cos (40)] = 40 =2
9
.
Example 26The value of cot (sin1x) is
(A)21+
(B) 21
x
x+
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30 MATHEMATICS
(C)1
x(D)
21 x
x
.
Solution (D) is the correct answer. Let sin1x = , then sin =x
cosec =1
cosec2 = 21
x
1 + cot2 = 21
cot =2
1 x.
Example 27If tan1x =10
for some xR, then the value of cot1x is
(A)5
(B)
2
5
(C)
3
5
(D)
4
5
Solution (B) is the correct answer. We know tan1x + cot1x =2
. Therefore
cot1x =2
10
cot1x =2
10
=
2
5
.
Example 28The domain of sin1 2x is
(A) [0, 1] (B) [ 1, 1]
(C)1 1
,2 2
(D) [2, 2]
Solution(C) is the correct answer. Let sin12x = so that 2x = sin .
Now 1 sin 1, i.e., 1 2x 1 which gives1 1
2 2x .
Example 29The principal value of sin13
2
is
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INVERSE TRIGONOMETRIC FUNCTIONS 31
(A) 23 (B)
3 (C) 4
3 (D) 5
3 .
Solution (B) is the correct answer.
1 1 13sin sin sin sin sin
2 3 3 3
= = = .
Example 30The greatest and least values of (sin1x)2 + (cos1x)2 are respectively
(A)2 25
and4 8
(B) and
2 2
(C)2 2
and4 4
(D)
2
and04
.
Solution (A) is the correct answer. We have
(sin1x)2 + (cos1x)2 = (sin1x + cos1x)2 2 sin1x cos1x
=2 1 12sin sin
4 2x
= ( )2
21 1
sin 2 sin4
x x
+
= ( )2
21 12 sin sin
2 8x x
+
=
2 2
12 sin4 16
x +
.
Thus, the least value is
2 2
2 i.e.16 8
and the Greatest value is
2 2
22 4 16
+
,
i.e.25
4
.
Example 31Let = sin1(sin ( 600), then value of is
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32 MATHEMATICS
(A)3 (B)
2 (C) 2
3 (D) 2
3 .
Solution(A) is the correct answer.
1 1 10sin sin 600 sin sin
180 3
=
=1 2sin sin 4
3
=1 2sin sin
3
=1 1sin sin sin sin
3 3 3
= =
.
Example 32The domain of the functiony = sin1(x2) is
(A) [0, 1] (B) (0, 1)
(C) [1, 1] (D)
Solution(C) is the correct answer. y = sin1 (x2) siny = x2
i.e. 1 x2 1 (since 1 siny 1) 1 x2 1
0 x2 1
1 . . 1 1x i e x
Example 33The domain of y = cos1(x2 4) is
(A) [3, 5] (B) [0, ]
(C) 5, 3 5, 3 (D) 5, 3 3, 5
Solution(D) is the correct answer. y = cos1 (x2 4 ) cosy =x2 4i.e. 1 x2 4 1 (since 1 cosy 1)
3 x2 5
3 5x
5, 3 3, 5x
Example 34The domain of the function defined byf(x) = sin1x + cosxis
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INVERSE TRIGONOMETRIC FUNCTIONS 33
(A) [1, 1] (B) [1, + 1]
(C) ( ) , (D)
Solution (A) is the correct answer. The domain of cos is Rand the domain of sin1 is
[1, 1].Therefore, the domain of cosx + sin1x is R [ ]1,1 , i.e., [1, 1].
Example 35The value of sin (2 sin1 (.6)) is
(A) .48 (B) .96 (C) 1.2 (D) sin 1.2
Solution (B) is the correct answer. Let sin1 (.6) = , i.e., sin = .6.
Now sin (2) = 2 sin cos = 2 (.6) (.8) = .96.
Example 36If sin1x + sin1y =2
, then value of cos1x + cos1y is
(A)2
(B) (C) 0 (D)
2
3
Solution (A) is the correct answer. Given that sin1x + sin1y =2 .
Therefore,1 1 cos cos
2 2 2x y
+ =
cos1x + cos1y =2
.
Example 37The value of tan1 13 1cos tan
5 4
+
is
(A)19
8(B)
8
19(C)
19
12(D)
3
4
Solution (A) is the correct answer. tan1 13 1cos tan
5 4
+
= tan1 14 1tan tan
3 4
+
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34 MATHEMATICS
= tan tan 11
4 119 193 4 tan tan
4 1 8 81
3 4
+ = =
.
Example 38The value of the expression sin [cot1 (cos (tan1 1))] is
(A) 0 (B) 1 (C)1
3(D)
2
3.
Solution(D) is the correct answer.
sin [cot1 (cos4
)] = sin [cot1
1
2]=
1 2 2sin sin
3 3
=
Example 39The equation tan1x cot1x = tan11
3
has
(A) no solution (B) unique solution
(C) infinite number of solutions (D) two solutions
Solution (B) is the correct answer.We have
tan1x cot1x =6
and tan1x + cot1x =
2
Adding them, we get 2tan1x =2
3
tan1x =3
i.e., 3x = .
Example 40If 2 sin1x + cos1x , then
(A) ,2 2
= = (B) 0, = =
(C)3
,2 2
= = (D) 0, 2 = =
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INVERSE TRIGONOMETRIC FUNCTIONS 35
Solution (B) is the correct answer. We have2
sin1x 2
2
+
2
sin1x +
2
2
+
2
0 sin1x + (sin1x + cos1x)
0 2sin1x + cos1x
Example 41The value of tan2 (sec12) + cot2 (cosec13) is
(A) 5 (B) 11 (C) 13 (D) 15
Solution (B) is the correct answer.
tan2 (sec12) + cot2 (cosec13) = sec2 (sec12) 1 + cosec2 (cosec13) 1
= 22 1 + 32 2 = 11.
2.3 EXERCISE
Short Answer (S.A.)
1. Find the value of1 15 13
tan tan cos cos6 6
+
.
2. Evaluate1 3
cos cos2 6
.
3. Prove that1cot 2cot 3 7
4
.
4. Find the value of1 1 11 1
tan cot tan sin23 3
.
5. Find the value of tan12
tan3
.
6. Show that 2tan1
(3) =
2
+
1 4tan3
.
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36 MATHEMATICS
7. Find the real solutions of the equation
( )1 1 2
tan 1 sin 12
x x x x+ + + + = .
8. Find the value of the expression sin ( )1 11
2tan cos tan 2 23
+
.
9. If 2 tan1 (cos ) = tan1 (2 cosec ), then show that =
4,
where n is any integer.
10. Show that1 11 1cos 2tan sin 4tan
7 3
=
.
11. Solve the following equation ( )1 13
cos tan sin cot4
x =
.
Long Answer (L.A.)
12. Prove that
2 2
1 1 2
2 2
1 1 1tan cos
4 21 1
x xx
x x
13. Find the simplified form of1 3 4cos cos sin
5 5x
, wherex
3,
4 4
.
14. Prove that1 1 18 3 77sin sin sin
17 5 85
.
15. Show that1 1 15 3 63sin cos tan
13 5 16 .
16. Prove that1 1 11 2 1
tan tan sin4 9 5
+ = .
17. Find the value of1 11 14 tan tan
5 239.
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INVERSE TRIGONOMETRIC FUNCTIONS 37
18. Show that11 3 4 7
tan sin2 4 3
and justify why the other value
4 7
3+
is ignored?
19. Ifa1, a
2, a
3,...,a
nis an arithmetic progression with common difference d, then
evaluate the following expression.
1 1 1 1
1 2 2 3 3 4 1
tan tan tan tan ... tan1 1 1 1 n n
d d d d
a a a a a a a a
+ + + + + + + +
.
Objective Type Questions
Choose the correct answers from the given four options in each of the Exercises from
20 to 37 (M.C.Q.).
20. Which of the following is the principal value branch of cos1x?
(A)
,2 2
(B) (0, )
(C) [0, ] (D) (0, )
2
21. Which of the following is the principal value branch of cosec1x?
(A)
,2 2
(B) [0, ]
2
(C)
,2 2
(D)
,2 2
{0}
22. If 3tan1
x + cot1
x = , thenx equals
(A) 0 (B) 1 (C) 1 (D)1
2.
23. The value of sin133
cos5
is
(A)3
5(B)
7
5
(C)
10
(D)
10
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38 MATHEMATICS
24. The domain of the function cos1 (2x 1) is
(A) [0, 1] (B) [1, 1]
(C) ( 1, 1) (D) [0, ]
25. The domain of the function defined byf(x) = sin1 1x is
(A) [1, 2] (B) [1, 1]
(C) [0, 1] (D) none of these
26. If cos1 12
sin cos 05
x + =
, thenx is equal to
(A)1
5(B)
2
5(C) 0 (D) 1
27. The value of sin (2 tan1 (.75)) is equal to
(A) .75 (B) 1.5 (C) .96 (D) sin 1.5
28. The value of1 3cos cos
2
is equal to
(A)2
(B)
3
2
(C)
5
2
(D)
7
2
29. The value of the expression 2 sec1 2 + sin11
2
is
(A)
6(B)
5
6(C)
7
6(D) 1
30. If tan1x + tan1y =4
5, then cot1x + cot1y equals
(A)
5(B)
2
5(C)
3
5
(D)
31. If sin1
21 1
2 2 2
2 1 2cos tan
1 1 1
a a x
a a x
, where a,x ]0, 1, then
the value ofx is
(A) 0 (B)
2
a(C) a (D) 2
2
1
a
a
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INVERSE TRIGONOMETRIC FUNCTIONS 39
32. The value of cot1 7cos
25
is
(A)25
24(B)
25
7(C)
24
25(D)
7
24
33. The value of the expression tan11 2
cos2 5
is
(A) 2 5 (B) 5 2
(C)5 2
2
(D) 5 2
1cosHint :tan
2 1 cos
=
+
34. If |x | 1, then 2 tan1x + sin1 22
1
x
x
is equal to
(A) 4 tan1x (B) 0 (C)2
(D)
35. If cos1 + cos1 + cos1 = 3, then ( + ) + (+ ) + ( + )equals
(A) 0 (B) 1 (C) 6 (D) 12
36. The number of real solutions of the equation
11 cos2 2 cos (cos )in ,2x x
+ = is
(A) 0 (B) 1 (C) 2 (D) Infinite
37. If cos1x > sin1x, then
(A)1
12
< (B)1
02
x 0
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40 MATHEMATICS
Fill in the blanks in each of the Exercises 38 to 48.
38. The principal value of cos11
2
is__________.
39. The value of sin13
sin5
is__________.
40. If cos (tan1x + cot1 3 ) = 0, then value ofx is__________.
41. The set of values of sec1
1
2
is__________.
42. The principal value of tan1 3 is__________.
43. The value of cos114
cos3
is__________.
44. The value of cos (sin1x + cos1x), |x| 1 is______ .
45. The value of expression tan
1 1sin cos
2
x x +
,whenx =3
2is_________.
46. Ify = 2 tan1x + sin1 22
1 x
for allx, then____
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INVERSE TRIGONOMETRIC FUNCTIONS 41
54. The minimum value ofn for which tan1 ,4
n n>
N , is valid is 5.
55. The principal value of sin11 1
cos sin2
is3
.