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Transcript of Class12 Mathematics1 Unit02 NCERT TextBook EnglishEdition
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7/26/2019 Class12 Mathematics1 Unit02 NCERT TextBook EnglishEdition
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Mathematics, in general , is fundamental ly the science of
self-evident things. FELIX KLEI N
2.1 Introduction
In Chapter 1, we have studied that the inverse of a function
f, denoted byf1, exists iffis one-one and onto. There are
many functions which are not one-one, onto or both and
hence we can not talk of their inverses. In Class XI, we
studied that trigonometric functions are not one-one and
onto over their natural domains and ranges and hence their
inverses do not exist. In this chapter, we shall study about
the restrictions on domains and ranges of trigonometric
functions which ensure the existence of their inverses andobserve their behaviour through graphical representations.
Besides, some elementary properties will also be discussed.
The inverse trigonometric functions play an important
role in calculus for they serve to define many integrals.
The concepts of inverse trigonometric functions is also used in science and engineering.
2.2 Basic Concepts
In Class XI, we have studied trigonometric functions, which are defined as follows:
sine function, i.e., sine : R[ 1, 1]
cosine function, i.e., cos : R[ 1, 1]
tangent function, i.e., tan : R {x :x= (2n + 1)2
,nZ} R
cotangent function, i.e., cot : R {x :x= n, nZ} R
secant function, i.e., sec : R {x :x= (2n + 1)2
,nZ} R ( 1, 1)
cosecant function, i.e., cosec : R { x :x= n, nZ} R ( 1, 1)
Chapter 2
INVERSE TRIGONOMETRICFUNCTIONS
Arya Bhatta
(476-550 A. D.)
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34 MATHEMATICS
We have also learnt in Chapter 1 that iff : XY such thatf(x) =yis one-one and
onto, then we can define a unique functiong: YX such thatg(y) =x, wherex X
andy=f(x),yY. Here, the domain ofg = range offand the range ofg= domain
off. The functiongis called the inverse off and is denoted by f1. Further,g is also
one-one and onto and inverse of gis f. Thus,g1 = (f 1)1 =f. We also have
(f 1o f ) (x) =f 1(f (x)) =f 1(y) =x
and (fof1) (y) =f (f 1(y)) =f(x) =y
Since the domain of sine function is the set of all real numbers and range is the
closed interval [1, 1]. If we restrict its domain to ,2 2
, then it becomes one-one
and onto with range [ 1, 1]. Actually, sine function restricted to any of the intervals
3 ,
2 2
, ,2 2
,3
,2 2
etc., is one-one and its range is [1, 1]. We can,
therefore, define the inverse of sine function in each of these intervals. We denote the
inverse of sine function by sin1(arc sine function). Thus, sin1is a function whose
domain is [ 1, 1] and range could be any of the intervals3
,2 2
, ,2 2
or
3,
2 2
, and so on. Corresponding to each such interval, we get a branchof the
function sin1. The branch with range ,2 2
is called theprincipal value branch,
whereas other intervals as range give different branches of sin1. When we refer
to the function sin1, we take it as the function whose domain is [1, 1] and range is
,2 2
. We write sin1: [1, 1] ,2 2
From the definition of the inverse functions, it follows that sin (sin1x) = x
if 1 x1 and sin1
(sinx) =xif 2 2x
. In other words, ify = sin1
x, then
siny=x.
Remarks
(i) We know from Chapter 1, that ify=f(x) is an invertible function, then x=f1(y).
Thus, the graph of sin1function can be obtained from the graph of original
function by interchangingxandyaxes, i.e., if (a, b) is a point on the graph ofsine function, then (b,a) becomes the corresponding point on the graph of inverse
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INVERSE TRIGONOMETRIC FUNCTIONS 35
of sine function. Thus, the graph of the functiony = sin1xcan be obtained fromthe graph ofy = sinxby interchangingxandyaxes. The graphs ofy = sinxand
y = sin1x are as given in Fig 2.1 (i), (ii), (iii). The dark portion of the graph of
y = sin1x represent the principal value branch.
(ii) It can be shown that the graph of an inverse function can be obtained from thecorresponding graph of original function as a mirror image (i.e., reflection) alongthe line y=x. This can be visualised by looking the graphs of y = sin xand
y = sin1xas given in the same axes (Fig 2.1 (iii)).
Like sine function, the cosine function is a function whose domain is the set of all
real numbers and range is the set [1, 1]. If we restrict the domain of cosine functionto [0, ], then it becomes one-one and onto with range [1, 1]. Actually, cosine function
Fig 2.1 (ii) Fig 2.1 (iii)
Fig 2.1 (i)
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36 MATHEMATICS
restricted to any of the intervals [ , 0], [0,],[, 2] etc., is bijective with range as
[1, 1]. We can, therefore, define the inverse of cosine function in each of these
intervals. We denote the inverse of the cosine function by cos1(arc cosine function).
Thus, cos1 is a function whose domain is [1, 1] and range
could be any of the intervals [, 0], [0, ], [, 2] etc.
Corresponding to each such interval, we get a branch of the
function cos1. The branch with range [0, ] is called theprincipal
value branch of the function cos1. We write
cos1: [1, 1] [0, ].
The graph of the function given byy= cos1xcan be drawn
in the same way as discussed about the graph ofy= sin1x. The
graphs ofy= cosxandy= cos1xare given in Fig 2.2 (i) and (ii).
Fig 2.2 (ii)
Let us now discuss cosec1xand sec1xas follows:
Since, cosecx=1
sinx, the domain of the cosec function is the set {x:x Rand
x n, n Z} and the range is the set {y : yR, y 1 or y 1} i.e., the setR (1, 1). It means thaty= cosecxassumes all real values except 1
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INVERSE TRIGONOMETRIC FUNCTIONS 37
Thus cosec1can be defined as a function whose domain is R (1, 1) and range could
be any of the intervals , {0}2 2
,3
, { }2 2
,3
, { }2 2
etc. The
function corresponding to the range , {0}2 2
is called theprincipal value branch
of cosec1. We thus have principal branch as
cosec1 : R (1, 1) , {0}
2 2
The graphs ofy= cosecxandy= cosec1xare given in Fig 2.3 (i), (ii).
Also, since secx= 1cos
, the domain ofy= secxis the set R {x:x= (2n+ 1)2
,
nZ} and range is the set R (1, 1). It means that sec (secant function) assumes
all real values except 1 < y< 1 and is not defined for odd multiples of2
. If we
restrict the domain of secant function to [0, ] {2
}, then it is one-one and onto with
Fig 2.3 (i) Fig 2.3 (ii)
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38 MATHEMATICS
its range as the set R (1, 1). Actually, secant function restricted to any of the
intervals [, 0] {2
}, [0, ]
2
, [, 2] {
3
2
} etc., is bijective and its range
is R {1, 1}. Thus sec1can be defined as a function whose domain is R (1, 1) and
range could be any of the intervals [ , 0] {2
}, [0, ] {
2
}, [, 2] {
3
2
} etc.
Corresponding to each of these intervals, we get different branches of the function sec1.
The branch with range [0, ] { 2
} is called the principal value branch of thefunction sec1. We thus have
sec1: R (1,1) [0, ] {2
}
The graphs of the functionsy= secxandy= sec-1xare given in Fig 2.4 (i), (ii).
Finally, we now discuss tan1and cot1
We know that the domain of the tan function (tangent function) is the set
{x :x Randx(2n+1)2
, nZ} and the range is R. It means that tan function
is not defined for odd multiples of2
. If we restrict the domain of tangent function to
Fig 2.4 (i) Fig 2.4 (ii)
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INVERSE TRIGONOMETRIC FUNCTIONS 39
,2 2
, then it is one-one and onto with its range as R. Actually, tangent function
restricted to any of the intervals3
,2 2
, ,2 2
,3
,2 2
etc., is bijective
and its range is R. Thus tan1can be defined as a function whose domain is Rand
range could be any of the intervals3
,2 2
, ,2 2
,3
,2 2
and so on. These
intervals give different branches of the function tan1. The branch with range ,2 2
is called the principal value branchof the function tan1.
We thus have
tan1:R ,2 2
The graphs of the function y =tanxandy= tan1xare given in Fig 2.5 (i), (ii).
Fig 2.5 (i) Fig 2.5 (ii)
We know that domain of the cot function (cotangent function) is the set
{x:xRandxn, nZ} and range is R. It means that cotangent function is not
defined for integral multiples of . If we restrict the domain of cotangent function to
(0, ), then it is bijective with and its range as R. In fact, cotangent function restricted
to any of the intervals (, 0), (0, ), (, 2) etc., is bijective and its range is R. Thus
cot1can be defined as a function whose domain is the Rand range as any of the
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40 MATHEMATICS
intervals (, 0), (0, ), (, 2) etc. These intervals give different branches of the
function cot1. The function with range (0, ) is called theprincipal value branchof
the function cot1. We thus have
cot1: R(0, )
The graphs ofy =cotxandy= cot1xare given in Fig 2.6 (i), (ii).
Fig 2.6 (i) Fig 2.6 (ii)
The following table gives the inverse trigonometric function (principal value
branches) along with their domains and ranges.
sin1 : [1, 1] ,2 2
cos1 : [1, 1] [0, ]
cosec1 : R (1,1) ,2 2
{0}
sec1 : R (1, 1) [0, ] { }2
tan1 : R ,2 2
cot1 : R (0, )
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INVERSE TRIGONOMETRIC FUNCTIONS 41
Note
1. sin1x should not be confused with (sinx)1. In fact (sin x)1 =1
sin and
similarly for other trigonometric functions.
2. Whenever no branch of an inverse trigonometric functions is mentioned, wemean the principal value branch of that function.
3. The value of an inverse trigonometric functions which lies in the range ofprincipal branch is called the principal value of that inverse trigonometricfunctions.
We now consider some examples:
Example 1Find the principal value of sin11
2
.
SolutionLet sin11
2
=y. Then, siny=1
2.
We know that the range of the principal value branch of sin1is ,2 2
and
sin4
=1
2. Therefore, principal value of sin1
1
2
is4
Example 2Find the principal value of cot11
3
SolutionLet cot11
3
=y. Then,
1cot cot
33y
= =
= cot3
=2
cot3
We know that the range of principal value branch of cot1 is (0, ) and
cot2
3
=1
3
. Hence, principal value of cot1
1
3
is2
3
EXERCISE 2.1
Find the principal values of the following:
1. sin11
2
2. cos13
2
3. cosec1 (2)
4. tan1 ( 3) 5. cos11
2
6. tan1 (1)
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42 MATHEMATICS
7. sec12
3
8. cot1 ( 3) 9. cos11
2
10. cosec1 ( 2 )Find the values of the following:
11. tan1(1) + cos11
2
+ sin1
1
2
12. cos1
1
2
+ 2 sin
11
2
13. If sin1
x=y, then
(A) 0 y (B)2 2
y
(C) 0 < y< (D)2 2
y
< 0
To prove the first result, we put cosec1x=y, i.e.,x= cosecy
Therefore1
= siny
Hence sin11
=y
or sin1 1 = cosec1x
Similarly, we can prove the other parts.
2. (i) sin1(x) = sin1x, x [ 1, 1]
(ii) tan1(x) = tan1x, x R
(iii) cosec1 (x) = cosec1x, | x| 1
Let sin1 (x) =y, i.e., x= sinyso thatx= siny, i.e.,x= sin (y).
Hence sin1x= y= sin1(x)
Therefore sin1(x) = sin1x
Similarly, we can prove the other parts.
3. (i) cos1
(x) = cos1
x, x [ 1, 1](ii) sec1 (x) = sec1 x, | x| 1
(iii) cot1 (x) = cot1 x, x R
Let cos1(x) = y i.e., x= cosyso thatx= cosy= cos ( y)
Therefore cos1x= y= cos1(x)
Hence cos1 (x) = cos1xSimilarly, we can prove the other parts.
4. (i) sin1x+ cos1x =2
, x [ 1, 1]
(ii) tan1 x+ cot1 x=
2
, xR
(iii) cosec1 x+ sec1 x =2
, | x| 1
Let sin1x=y. Thenx= siny= cos2
y
Therefore cos1x=2
y
= 1sin2
x
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44 MATHEMATICS
Hence sin1x+ cos1x=2
Similarly, we can prove the other parts.
5. (i) tan1x+ tan1y= tan1
+
1
x y
xy, xy< 1
(ii) tan1x tan1y= tan1+
1
x y
xy, xy> 1
(iii) 2tan1x= tan12
21
x
x, | x| < 1
Let tan1x= and tan1y= . Thenx= tan ,y= tan
Nowtan tan
tan( )1 tan tan 1
y
xy
+ ++ = =
This gives + = tan11
y
xy
+
Hence tan1x+ tan1y= tan11
y
xy
+
In the above result, if we replaceyby y, we get the second result and by replacing
ybyx, we get the third result.
6. (i) 2tan1x= sin12
2
1 +
x
x, |x| 1
(ii) 2tan1x= cos12
2
1
1 +
x
x,x 0
(iii) 2 tan1x= tan12
21
xx
, 1 < x< 1
Let tan1x=y, thenx= tany. Now
sin1 22
1
x
+= sin1 2
2tan
1 tan
y
y+
= sin1(sin 2y) = 2y = 2tan1x
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INVERSE TRIGONOMETRIC FUNCTIONS 45
Also cos1
2
2
1
1
x
x
+= cos1
2
2
1 tan
1 tan
y
y
+= cos1(cos 2y) = 2y = 2tan1x
(iii) Can be worked out similarly.
We now consider some examples.
Example 3Show that
(i) sin1
( )2
2 1x x= 2 sin1x,
1 1
2 2
x
(ii) sin1 ( )22 1 x = 2 cos1x,1
12
Solution
(i) Letx= sin . Then sin1x= . We have
sin1 ( )22 1 x = sin1 ( )22sin 1 sin = sin1 (2sincos) = sin1 (sin2) = 2
= 2 sin1x
(ii) Takex= cos , then proceeding as above, we get, sin1
( )2
2 1x x= 2 cos1x
Example 4Show that tan11 11 2 3tan tan
2 11 4+ =
SolutionBy property 5 (i), we have
L.H.S. =1 11 2tan tan
2 11+ 1 1
1 2
152 11tan tan
1 2 201
2 11
+
= =
=1 3
tan4
= R.H.S.
Example 5Express 1 costan1 sin
xx
,
2 2x < < in the simplest form.
SolutionWe write
2 2
1 1
2 2
cos sincos 2 2tan tan
1 sincos sin 2sin cos
2 2 2 2
x xx
x x xx
= +
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46 MATHEMATICS
=1
2
cos sin cos sin2 2 2 2
tan
cos sin2 2
x x x x
x x
+
=1
cos sin2 2tan
cos sin2 2
x x
x x
+
1
1 tan2tan
1 tan2
x
x
+ =
=1
tan tan4 2 4 2
x + = +
Alternatively,
1 1 1
2sin sin
cos 2 2tan tan tan
21 sin1 cos 1 cos
2 2
xx
x
xxx
= =
=1
2
2 22sin cos
4 4tan
22sin
4
x
x
=1 2tan cot
4
x
1 2tan tan2 4
x =
= 1tan tan4 2 +
4 2= +
Example 6Write1
2
1cot
1x
, |x | > 1 in the simplest form.
SolutionLetx= sec , then 2 1x = 2sec 1 tan =
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INVERSE TRIGONOMETRIC FUNCTIONS 47
Therefore,1
2
1cot
1x = cot1(cot ) = = sec1x, which is the simplest form.
Example 7Prove that tan1x+1
2
2tan
1= tan1
3
2
3
1 3
x x
x
,
1| |
3x 1
7.1 1 cos
tan1 cos
+
,x< 8.1 cos sin
tancos sin
x
x
+
,x<
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48 MATHEMATICS
9.1
2 2tan
x
a x
, |x | < a
10.
2 31
3 2
3tan
3
a x x
a ax
,a> 0;
3 3
a ax
Find the values of each of the following:
11.
1 1 1
tan 2cos 2sin 2
12. cot (tan1
a+ cot1
a)
13.
21 1
2 2
1 2 1tan sin cos
2 1 1
x y
y
+
+ + , |x | < 1,y> 0 andxy< 1
14. If sin1 11sin cos 1
5x
+ =
, then find the value ofx
15. If1 11 1
tan tan2 2 4
x x
x
+ + =
+, then find the value ofx
Find the values of each of the expressions in Exercises 16 to 18.
16.1 2sin sin
3
17. 13
tan tan4
18.1 13 3
tan sin cot5 2
+
19.1 7cos cos is equal to
6
(A)7
6
(B)
5
6
(C)
3
(D)
6
20. 1 1sin sin ( )3 2
is equal to
(A)1
2(B)
1
3(C)
1
4(D) 1
21.1 1tan 3 cot ( 3) is equal to
(A) (B)2
(C) 0 (D) 2 3
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INVERSE TRIGONOMETRIC FUNCTIONS 49
Miscell aneous Examples
Example 9Find the value of1 3
sin (sin )5
SolutionWe know that 1sin (sin ) x = . Therefore,1 3 3
sin (sin )5 5
=
But3
,
5 2 2
, which is the principal branch of sin1x
However3 3 2
sin ( ) sin( ) sin5 5 5
= = and
2,
5 2 2
Therefore1 13 2 2sin (sin ) sin (sin )
5 5 5
= =
Example 10Show that1 1 13 8 84
sin sin cos5 17 85
=
SolutionLet 13
sin5
x = and 18
sin17
y =
Therefore3
sin5
x= and8
sin17
y=
Now2 9 4cos 1 sin 1
25 5x x= = = (Why?)
and2 64 15
cos 1 sin 1289 17
y y= = =
We have cos (xy) = cosxcosy+ sinxsiny
=4 15 3 8 84
5 17 5 17 85 + =
Therefore1 84cos
85x y
=
Hence 1 1 13 8 84
sin sin cos5 17 85
=
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50 MATHEMATICS
Example 11Show that1 1 112 4 63
sin cos tan13 5 16
+ + =
SolutionLet 1 1 112 4 63
sin , cos , tan13 5 16
x y z = = =
Then12 4 63
sin , cos , tan13 5 16
x y z= = =
Therefore5 3 12 3
cos , sin , tan and tan13 5 5 4
x y x y= = = =
We havetan tan
tan( )1 tan tan
yx y
y
++ =
12 3635 4
12 3 161
5 4
+= =
Hence tan( ) tany z+ =
i.e., tan (x+ y) = tan (z) or tan (x +y) = tan ( z)
Therefore x+ y= z or x+y= zSince x,yandzare positive,x+y z (Why?)
Hence x+ y + z = or1 1 112 4 63sin cos tan
13 5 16+ + =
Example 12Simplify1 cos sintan
cos sin
a x b x
b x a x
+
, ifa
btanx> 1
SolutionWe have,
1 cos sintancos sin
a x b x
b x a x
+
=1
cos sin
costancos sin
cos
a x b x
b xb x a x
b x
+
=1
tantan
1 tan
a
ba
b
+
=1 1tan tan (tan )
a
b = 1tan
a
b
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INVERSE TRIGONOMETRIC FUNCTIONS 51
Example 13Solve tan12x+ tan13x=4
SolutionWe have tan12x+ tan13x=4
or1 2 3
tan1 2 3
x
x
+
=4
i.e. 1 25tan1 6
x =
4
Therefore 25
1 6
x
= tan 1
4
=
or 6x2+ 5x 1 = 0 i.e., (6x 1) (x+ 1) = 0
which gives x=1
6orx= 1.
Sincex= 1 does not satisfy the equation, as the L.H.S. of the equation becomes
negative,1
6x= is the only solution of the given equation.
Miscellaneous Exercise on Chapter 2
Find the value of the following:
1.1 13cos cos
6
2.1 7tan tan
6
Prove that
3.1 13 242sin tan
5 7= 4.
1 1 18 3 77sin sin tan17 5 36
+ =
5.1 1 14 12 33cos cos cos
5 13 65+ = 6. 1 1 112 3 56cos sin sin
13 5 65+ =
7.1 1 163 5 3
tan sin cos16 13 5
= +
8.1 1 1 11 1 1 1tan tan tan tan
5 7 3 8 4
+ + + =
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52 MATHEMATICS
Prove that
9.1 11 1
tan cos2 1
x
= +,x[0, 1]
10.1 1 sin 1 sin
cot21 sin 1 sin
x x x
x x
+ + = +
, 0,4
x
11. 1 11 1 1tan cos4 21 1
x x xx x
+ = + + , 1 1
2 [Hint: Putx= cos 2]
12.1 19 9 1 9 2 2
sin sin8 4 3 4 3
=
Solve the following equations:
13. 2tan1(cosx) = tan1(2 cosecx) 14.1 11 1
tan tan ,( 0)1 2
xx x
x
= >
+
15. sin (tan1x), |x | < 1 is equal to
(A)21
x
(B)
2
1
1(C)
2
1
1+(D)
21+
16. sin1 (1 x) 2 sin1x=2
, thenxis equal to
(A) 0,1
2(B) 1,
1
2(C) 0 (D)
1
2
17.1 1tan tan
x y
y x y
+ is equal to
(A)2
(B)
3
(C)
4
(D)
3
4
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INVERSE TRIGONOMETRIC FUNCTIONS 53
Summary
The domains and ranges (principal value branches) of inverse trigonometric
functions are given in the following table:
Functions Domain Range
(Principal Value Branches)
y= sin1x [1, 1] ,2 2
y= cos1x [1, 1] [0, ]
y= cosec1x R (1,1) ,2 2
{0}
y= sec1x R (1, 1) [0, ] { }2
y= tan1x R ,2 2
y= cot1x R (0, )
sin1x should not be confused with (sinx)1. In fact (sin x)1=
1
sinx and
similarly for other trigonometric functions.
The value of an inverse trigonometric functions which lies in its principal
value branch is called the principal value of that inverse trigonometric
functions.
For suitable values of domain, we have
y= sin1xx= siny x = siny y = sin1x
sin (sin1x) =x sin
1(sinx) =x
sin1 1 = cosec1x cos1 (x) = cos1x
cos11
= sec1x cot1(x) = cot1x
tan1
1= cot1x sec
1 (x) = sec1x
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54 MATHEMATICS
sin1(x) = sin1x tan
1(x) = tan1x
tan1x+ cot1x=2
cosec1 (x) = cosec1x
sin1x+ cos1x =
2
cosec
1x+ sec1x=2
tan1x+ tan1y= tan1
1
y
y
+
2tan1x = tan1
2
21
x
x
tan1x tan1y= tan1
1
y
y
+
2tan1x= sin1 2
2
1+= cos1
2
2
1
1
+
Histori cal Note
The study of trigonometry was first started in India. The ancient Indian
Mathematicians, Aryabhatta (476A.D.), Brahmagupta (598 A.D.), Bhaskara I
(600 A.D.) and Bhaskara II (1114 A.D.) got important results of trigonometry. Allthis knowledge went from India to Arabia and then from there to Europe. The
Greeks had also started the study of trigonometry but their approach was so
clumsy that when the Indian approach became known, it was immediately adopted
throughout the world.
In India, the predecessor of the modern trigonometric functions, known as
the sine of an angle, and the introduction of the sine function represents one of
the main contribution of the siddhantas (Sanskrit astronomical works) to
mathematics.
Bhaskara I (about 600 A.D.) gave formulae to find the values of sine functions
for angles more than 90. A sixteenth century Malayalam work Yuktibhasa
contains a proof for the expansion of sin (A + B). Exact expression for sines or
cosines of 18, 36, 54, 72, etc., were given by Bhaskara II.
The symbols sin1x, cos1x, etc., for arc sinx, arc cosx, etc., were suggested
by the astronomer Sir John F.W. Hersehel (1813) The name of Thales
(about 600 B.C.) is invariably associated with height and distance problems. He
is credited with the determination of the height of a great pyramid in Egypt by
measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known
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INVERSE TRIGONOMETRIC FUNCTIONS 55
height, and comparing the ratios:
H
S
h
s= = tan (suns altitude)
Thales is also said to have calculated the distance of a ship at sea through
the proportionality of sides of similar triangles. Problems on height and distance
using the similarity property are also found in ancient Indian works.