class12 mathematics1 unit02 ncert textbook englishedition

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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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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,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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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


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


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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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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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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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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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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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


23/23


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.

class12 mathematics1 unit02 ncert textbook englishedition

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