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Chapter

2

INVERSE TRIGONOME TRIGONOMETRIC TRIC FUNCTIONS Mathematics, in general, is fundamentally the science of

self-evident things. — FELIX KLEIN

2.1 Introduction In Chapter 1, we have studied that the inverse of a function f , denoted by f –1, exists if f f is 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 and observe their behaviour through graphical representations. Besides, some elementary properties will also be discussed. Arya Bhatta The inverse trigonometric functions play an important (476-550 A. D.) role in calculus for they serve to define many integrals. The concepts of inverse trigonometric functions functi ons 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]

π

, n ∈ Z} → R 2 cotangent function, i.e., cot : R – { x : x = nπ, n ∈ Z} → R tangent function, i.e., tan : R – { x : x = (2n + 1)

secant function, i.e., sec : R – { x : x = (2n + 1)

π

, n ∈ Z} → R – (– 1, 1) 2 cosecant function, i.e., cosec : R – { x : x = nπ, n ∈ Z} → R – (– 1, 1)

34

M AT H E M AT I C S

We have also learnt in Chapter 1 that if f f : X→Y such that f ( x x) = y is one-one and onto, then we can define a unique function g : Y→X such that g ( y) = x, where x ∈ X x), y ∈ Y. Here, the domain of g = range of f f and the range of g = domain and y = f ( x f . The function g is called the inverse of f f and is denoted by f –1. Further, g is also of f one-one and onto and inverse of g is f . Thus, g –1 = ( f –1)–1 = f . We also have f –1 o f ) ( x) = f –1 ( f f ( x x)) = f –1( y y) = x ( f f o f –1) ( y) = f ( f f –1( y y)) = f ( x x) = y ( f

and

Since the domain of sine function is the set of all real numbers and range is the

⎡ − π , π ⎤ , then it becomes one-one ⎣ 2 2 ⎥⎦

closed interval [–1, 1]. If we restrict its domain to ⎢

and onto with range [– 1, 1]. Actually, sine function restricted to any of the intervals

⎡ −3π , – π ⎤ , ⎡ −π , π ⎤ , ⎡ π , 3π ⎤ etc., ⎢⎣ 2 2 ⎥⎦ ⎢⎣ 2 2 ⎥⎦ ⎢⎣ 2 2 ⎥⎦

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 sin –1 (arc sine function). Thus, sin –1 is a function whose domain is [– 1, 1] and range could be any of the intervals

⎡ −3π , − π ⎤ , ⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦ ⎢⎣ 2 2 ⎥⎦

or

⎡ π , 3π ⎤ ⎢⎣ 2 2 ⎥⎦ , and so on. Corresponding to each such interval, we get a branch of the ⎡ −π , π ⎤ function sin–1. The branch with range ⎢ is called the principal value branch, ⎣ 2 2 ⎥⎦ whereas other intervals as range give different branches of sin–1. When we refer to the function sin–1, we take it as the function whose domain is [–1, 1] and range is

⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦ . We write

sin–1 : [–1, 1] →

⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦

From the definition of the inverse functions, it follows that sin (sin –1 x) = x if – 1 ≤ x ≤ 1 and sin–1 (sin x) = x if −

π 2

≤ x ≤

π 2

. In other words, if y = sin–1 x, then

sin y = x. Remarks (i)) We know (i know from from Cha Chapte pterr 1, that if y = f ( x x) is an invertible function, then x = f –1 ( y y). –1 Thus, the graph of sin function can be obtained from the graph of original function by interchanging x and y axes, i.e., if ( a, b) is a point on the graph of sine function, then (b, a) becomes the corresponding point on the t he graph of inverse

INVERSE TRIGONOMETRIC FUNCTIONS

35

of sine function. Thus, the graph of the function y = sin–1 x can be obtained from y = sin x by interchanging x and y axes. The graphs of y y = sin x and the graph of y –1 y = sin x are as given in Fig 2.1 (i), (ii), (iii). The dark portion of the graph of y = sin–1 x represent the principal value branch. (ii) It can be shown (ii) shown that the the graph of of an inverse inverse function function can be be obtained obtained from the the corresponding graph of original function functi on as a mirror image (i.e., reflection) along y = sin x and the line y = x. This can be visualised by looking the graphs of y y = sin–1 x as given in the same axes (Fig 2.1 (iii)).

Fig 2.1 (i)

Fig 2.1 (ii)

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 function to [0, π], then it becomes one-one and onto with range [–1, 1]. Actually, cosine function

36

M AT H E M AT I C S

restricted to any of the intervals [– π, 0], [0,π], [π, 2π] etc., is bijective with range as [–1, 1]. We can, therefore, define the inverse of cosine function in each of these intervals. We We denote the inverse of the cosine function by cos–1 (arc cosine function). Thus, cos–1 is a function whose domain is [–1, 1] and range could be any of the intervals [– π, 0], [0, π], [ π, 2 π] etc. Corresponding to each such interval, we get a branch of the function cos–1. The branch with range [0, π] is calle called d the principal –1 value branch of the function cos . We write cos–1 : [–1, 1] → [0, π]. The graph graph of the function given by y = cos–1 x can be drawn in the same way as discussed about the graph of y = sin–1 x. The graphs of y y = cos x and y = cos–1 x are given in Fig 2.2 (i) and (ii).

Fig 2.2 (i)

Fig 2.2 (ii)

Let us now discuss cosec –1 x and sec–1 x as follows: 1 Since, cosec x = , the domain of the cosec function is the set { x : x ∈ R and sin x x ≠ nπ, n ∈ Z} and the range is the set { y : y ∈ R, y ≥ 1 or y ≤ –1} i.e., the set R – (–1, 1). It means that y = cosec x assumes all real values except –1 0 x To prove the first result, we put cosec –1 x = y, i.e., x = cosec y

Therefore H ence

1 x

= sin y

sin–1

x

= y

1

= cosec–1 x x Similarly,, we can prove the other parts. Similarly or

sin–1

1

2. (i) sin–1 (– x) = – sin–1 x, x ∈ [– 1, 1] (ii) tan–1 (– x) = – tan–1 x, x ∈ R (iii) co cose secc–1 (– x) = – cosec–1 x, | x | ≥ 1

Let sin–1 (– x) = y, i.e., – x = sin y so that x = – sin y, i.e., x = sin (– y). H ence Therefore

sin–1 x = – y = – sin–1 (– x) sin–1 (– x) = – sin–1 x

Similarly,, we can prove the other parts. Similarly 3. (i) cos–1 (– x) = π – cos–1 x, x ∈ [– 1, 1] (ii) sec–1 (– x) = π – sec–1 x, | x | ≥ 1 (iii) cot–1 (– x) = π – cot–1 x, x ∈ R

Let cos–1 (– x) = y i.e., – x = cos y so that x = – cos y = cos (π – y) cos–1 x = π – y = π – cos–1 (– x) He nce cos–1 (– x) = π – cos–1 x Similarly,, we can prove the other parts. Similarly Therefore

4. (i) sin–1 x + cos–1 x = (ii) tan–1 x + cot –1 x =

2 2

(iii (i ii)) co cose secc–1 x + sec –1 x =

, x ∈ [– 1, 1] , x ∈ R

2

, | x | ≥ 1

Let sin–1 x = y. Then x = sin y = cos Therefore

cos–1 x =

π 2

− y

=

⎛ π − y ⎞ ⎜2 ⎟ ⎝ ⎠ π − sin –1 x 2

43

44

M AT H E M AT I C S

sin–1 x + cos–1 x =

Hence

π 2

Similarly, we can prove the other parts.

x + y

5. (i) tan–1 x + tan –1 y = tan–1

x – y

(ii) tan–1 x – tan–1 y = tan–1

1 + y

2 x

(iii) 2tan–1 x = tan–1

1–

, xy < 1

1 – y

, xy > – 1

, | x | < 1

x 2

Let tan–1 x = θ and tan–1 y = φ. Then x = tan θ, y = tan φ tan θ + tan φ

Now

tan(θ + φ) =

This gives

θ + φ = tan–1

Hence

tan–1 x + tan–1 y = tan–1

1− tan θ tan φ

=

x + y

1− xy

x + y

1− xy x + y

1− xy

In the above result, if we replace y by – y, we get the second result and by replacing y by x, we get the third result. 6. (i) 2tan–1 x = sin–1

–1

(ii) 2tan x = cos

–1

(iii) 2 ta tan–1 x = tan–1

2 1 + x

2

1 – x

2

1+

2

, | x | ≤ 1

2 x 1 – x 2

, x ≥ 0

, – 1 < x < 1

Let tan–1 x = y, then x = tan y. Now sin–1

2 x 1 + x

2

= sin–1

2tan y 2 1 + tan y

= sin–1 (sin 2 y) = 2 y = 2tan–1 x

INVERSE TRIGONOMETRIC FUNCTIONS

Also cos–1

1 − x 2 1 + x

= cos–1

2

1− tan 2 y

= cos–1 (cos 2 y) = 2 y = 2tan–1 x

1 + tan y 2

(iii (i ii)) Can be worked out similarly. similarly. We now consider some examples. Example 3 Show that

(i) sin–1 ( 2 x 1 − x 2 ) = 2 sin–1 x, (ii) sin–1 ( 2 x 1 − x 2 ) = 2 cos–1 x,

1

−

2 1 2

≤ x ≤

1 2

≤ x ≤ 1

Solution

(i) Let x = sin θ. Then sin–1 x = θ. We have

(

sin–1 ( 2 x 1 − x 2 ) = sin–1 2 sin θ 1 − sin 2 θ

)

= sin–1 (2sinθ cosθ) = sin–1 (sin2θ) = 2θ = 2 ssin in–1 x (ii) Take x = cos θ, then proceeding as above, we get, sin–1 ( 2 x 1 − x 2 ) = 2 cos–1 x that tan–1 Example 4 Show th

1 2

+ tan –1

2 11

= tan –1

3 4

Solution By property 5 (i), we have

1 L.H.S. = tan

–1

1 2

+ tan –1

Example 5 Express tan

−1

2 11

= tan –1

+

2

2 11 1 2

1−

= tan −1

×

15

− = tan

1

20

3 4

= R.H.S.

2 11

⎛ cos x ⎞ , π π ⎜⎝ 1 − sin x ⎟⎠ − 2 1 in the simplest form. 2 x − 1 ⎠ 1

Solution Let x = sec θ, then

x

2

−1 =

sec 2 θ − 1 = tan θ

INVERSE TRIGONOMETRIC FUNCTIONS

Therefore, cot

1

–1

x

2

−1

= cot–1 (cot θ) = θ = sec–1 x, which is the simplest form.

Example 7 Prove that tan x + tan –1

–1

2 x 1 − x

2

⎛ 3 x − x3 ⎞ ⎜ 2 ⎟, ⎝ 1 − 3 x ⎠

–1

= tan

| x | <

1 3

Solution Let x = tan θ. Then θ = tan–1 x. We have

R.H.S. = tan

–1

3 ⎛ 3 x − x3 ⎞ –1 ⎛ 3 tan θ − tan θ ⎞ = tan ⎜ ⎜ ⎟ 2 ⎟ 2 x 1 − 3 ⎝ ⎠ ⎝ 1 − 3 tan θ ⎠

= tan–1 (tan3θ) = 3θ = 3tan–1 x = tan–1 x + 2 tan–1 x = tan–1 x + tan–1

2 x

= L.H.S. (Why?) 2 1 − x Example 8 Find the value of cos (s (sec ec–1 x + cosec–1 x), | x | ≥ 1

⎛ π⎞ ⎜ 2 ⎟= 0 ⎝ ⎠

Solution We have cos (sec–1 x + cosec–1 x) = cos

EXERCISE 2.2 Prove the following:

⎡ 1 1⎤ 1. 3sin–1 x = sin–1 (3 x – 4 x3), x ∈ ⎢ – , ⎥ ⎣ 2 2⎦ ⎡1 ⎤ 2. 3cos–1 x = cos–1 (4 x3 – 3 x), x ∈ ⎢ , 1⎥ ⎣2 ⎦ 3. tan –1 4.

2 11

2 tan −1

+ tan −1 1 2

7 24

= tan −1

1 2

1

31

7

17

+ tan −1 = tan −1

Write the following functions in the simplest form: 1 + x

−1

5.

tan

7.

tan −

x 1

⎛ ⎜⎜ ⎝

2

− 1 , x ≠ 0

⎞ ⎟ , x < π 1 + cos x ⎟⎠ 1 − cos x

47

1

−1

6.

tan

8.

tan −

x 1

2

−1

x | > 1 , | x

⎛ cos x − sin x ⎞ ⎜ cos x + sin x ⎟ , x < π ⎝ ⎠

48

9.

10 .

M AT H E M AT I C S

x

tan −1 a

tan

2

− x2

, | x | < a

−a ⎞ ≤ x ≤ , > 0; a ⎜ 3 ⎟ 2 3 − 3 a ax ⎝ ⎠

−1 ⎛ 3a

2

x − x3

a

3

Find the values of each of the following: 11.

13 .

tan

tan

⎡ ⎛ 2 sin –1 1 ⎞ ⎤ 2 c o s ⎜ ⎟⎥ ⎢ ⎝ 2 ⎠⎦ ⎣

–1

1⎡

⎢sin

1 + x

2⎣

⎛ ⎝

2

+ cos

− y2 ⎤ 2 ⎥ , | x | < 1, y > 0 and xy < 1 1+ y ⎦

–1 1

+ cos–1 x ⎞⎟ =1 , then find the value of x x 5 ⎠ x +1 π –1 x − 1 + tan –1 = , then find the value of x If tan x x − 2 x+2 4

14 . If sin ⎜ sin 15 .

2 x

–1

12 . cot (tan–1a + cot–1a)

–1

1

Find the values of each of the expressions in Exercises 16 to 18. 16 .

18 . 19 .

sin

–1

⎛ ⎝

6

–1

tan

–1

⎛ tan 3π ⎞ ⎜ ⎟ 4 ⎠ ⎝

3

(B)

6

(C)

π 3

(D)

π 6

⎛ π − sin −1 (− 1 ) ⎞ ⎟ is equal to 2 ⎠ ⎝3

sin ⎜

(A) 21 .

17 .

3 + cot –1 ⎞⎟ 5 2⎠ ⎛ 7 π ⎞ is equal to cos −1 ⎜ cos to ⎟ 6 ⎠ ⎝ 7π 5π tan ⎜ sin

(A) 20 .

⎛ sin 2π ⎞ ⎜ ⎟ 3 ⎠ ⎝

1 2

(B)

1

(C)

3

1 4

(D) 1

tan −1 3 − cot −1 (− 3 ) is equal to (A) π

(B)

−

π 2

(C ) 0

(D) 2 3

INVERSE TRIGONOMETRIC FUNCTIONS

Miscellaneous Examples 3π

− Example 9 Find the value of sin (sin 1

5

) −1

Solution We know that sin −1 (sin x) = x . Therefore, sin (sin

But However

Therefore

3π 5

3π

)=

3π 5

π π ∉ ⎡⎢ − , ⎤⎥ , which is the principal branch of sin–1 x 5 ⎣ 2 2⎦ 2π ⎡ π π ⎤ 3π 3π 2π ∈ − , sin ( ) = sin(π − ) = sin and 5 ⎢⎣ 2 2 ⎥⎦ 5 5 5 sin −1 (sin

3π

) = sin −1 (sin

5

3

− Example 10 Show that sin

1

Solution Let sin −1

3

= x

5

3

and sin −1

)=

2π 5

= cos−1

17

8

84 85

= y

17

8

Therefore

sin x =

Now

cos x = 1 − si sin 2 x

=

1−

and

2 cos y = 1 − si sin y

=

1−

We have

x– y) = cos x cos y + sin x sin y cos ( x

5

and sin y =

5

8

− sin −1

5

2π

4

=

5

×

15

= cos−1

84

Hence

sin −

3

8

5

17

289

17

+ ×

=

64

5

17

⎛ 84 ⎞ ⎜⎝ 85 ⎟⎠

− sin −1

25

8

x − y = cos − 1

9

3

Therefore

1

17

85

=

4

(Why?)

5

=

15 17

84 85

49

50

M AT H E M AT I C S

1 12

− Example 11 Show that sin

Solution Let sin −1

12 13

co cos−

= x,

Then

sin x =

12

Therefore

cos x =

5

13

13

4

63

5

16

+ cos−1 + tan −1 1

4 5

, cco os y =

= y , tan −1 4 5

,

63 16

=π

=z 63

tan z =

16

3 12 3 , sin y = , tan x = and tan y = 13 5 5 4 12 tan x + tan y

We have

tan( x + y ) =

Hence

tan( x + y) = − tan z

1 − tan x tan y

=

+

3

5 4 12 3

1−

5

×

=−

63 16

4

tan ( x x + y) = tan (– z) or tan ( x x + y) = tan (π – z)

i.e., Therefore

x + y = – z or x + y = π – z

Since

x, y and z are positive, x + y ≠ – z (Why?)

Hence

x + y + z = π or sin

Example 12 Simplify tan

–1

–1 12

13

4

63

5

16

+ cos–1 + tan –1

⎡ a cos x − b sin x ⎤ a ⎢ b cos x + a sin x ⎥ , if b ⎣ ⎦

=π

tan x > –1 –1

Solution We have,

⎡ a cos x − b sin x ⎤ tan –1 ⎢ ⎥ ⎣ b cos x + a sin x ⎦

=

⎡ a cos x − b sin x ⎤ ⎢ ⎥ b cos x tan –1 ⎢ ⎥ ⎢ b cos x + a sin x ⎥ b cos x ⎣ ⎦

–1 = tan

a b

=

⎡ a − tan x ⎤ ⎢ ⎥ tan –1 ⎢ b ⎥ a ⎢1 + tan x ⎥ ⎣ b ⎦ a

− tan –1 (tan x) = tan –1 − x b

INVERSE TRIGONOMETRIC FUNCTIONS

π

Example 13 Solve tan–1 2 x + tan–1 3 x =

or

–1

⎛ 2 x + 3x ⎞ ⎜ 1 − 2 x × 3x ⎟ ⎝ ⎠

=

⎛ 5 x ⎞ ⎜ 1 − 6 x 2 ⎟ ⎝ ⎠

=

tan

i.e.

–1

5 x

4

π

Solution We have tan–1 2 x + tan–1 3 x =

tan

51

4

π 4

π 4

π

=1 2 1 − 6 x 4 6 x2 + 5 x – 1 = 0 i. i.e., e., (6 x – 1) ( x + 1) = 0

Therefore or which gives

= tan

x =

1

or x = – 1. 6 Since x = – 1 does not satisfy the equation, as the L.H.S. of the equation becomes

negative, x =

1 6

is the only solution of the given equation.

Miscellaneous Exercise on Chapter 2 Find the value of the following: 1.

–1

cos

⎛ cos 13π ⎞ ⎜ ⎟ 6 ⎠ ⎝

–1

2.

tan

4.

sin

6.

cos

⎛ tan 7π ⎞ ⎜ ⎟ 6 ⎠ ⎝

Prove that –1

3.

2 sin

5.

cos

7.

tan

–1

8.

tan

–1

–1

3 5

4 5

+ cos–1

63 16 1 5

= tan –1

24 7

12 13

= sin –1

= cos–1

5 13

+ cos –1

33 65 3 5

1

1

1

π

7

3

8

4

+ tan −1 + tan −1 + tan −1 =

–1

8 17

–1 12

13

+ sin –1 = tan –1

3

77

5

36

3

56

5

65

+ sin –1 = sin –1

52

M AT H E M AT I C S

Prove that 9.

tan

–1

10 .

cot

–1

11.

tan

–1

12 .

9π 8

−

1

=

x

2

cos

⎛ ⎜⎜ ⎝

1 + sin x

⎛ ⎜⎜ ⎝

1 + x

–1

+ 1 + sin x −

9 4

− 1 + x + sin −

1

1 3

=

⎛ 1 − x ⎞ ⎜⎝ 1 + x ⎟⎠ , x ∈ [0, 1] ⎞ x ⎛ π⎞ ⎟⎟ = , x ∈ ⎜⎝ 0, 4 ⎟⎠ 1 − sin x ⎠ 2 1 − sin x

⎞ π ⎟= − 1 − x ⎟⎠ 4 1− x

9 4

sin −

1

1 2

–1

cos x ,

−

1 2

≤ x ≤ 1 [Hint: Put x = cos 2θ]

2 2 3

Solve the following equations: 13 . 2tan–1 (cos x) = tan–1 (2 cosec x)

14 .

tan

− x 1 –1 = tan x, ( x > 0) 1 + x 2

–1 1

15 . sin (tan–1 x), | x x | < 1 is equal to x

(A)

1 − x 2

16 . sin–1 (1 – x) – 2 sin –1 x =

(A) 0,

17 .

1 2

1 − x 2

π 2

(B) 1,

1 2

is equal to

π

π

2

(B)

3

1

(C)

1 + x 2

x

(D)

1 + x 2

, then x is equal to

⎛ x ⎞ − tan −1 x − y ⎟ x+ y ⎝ y⎠

tan −1 ⎜

(A)

1

(B)

(C ) 0

(C)

π 4

(D)

(D)

1 2

−3π 4

INVERSE TRIGONOMETRIC FUNCTIONS

Summary

The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table: Fu n cti on s

Domain

R a n ge (Principal Value Branches)

y = sin–1 x

[–1, 1]

⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦

y = cos–1 x

[–1, 1]

[0, π]

y = cosec–1 x

R – (–1,1)

⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦ – {0}

y = sec–1 x

R – (–1, 1)

[0, π] – { } 2

y = tan–1 x

R

⎛− π,π⎞ ⎜ 2 2⎟ ⎝ ⎠

y = cot–1 x

R

(0, π)

π

–1 –1 –1 sin x should not be confused with (sin x) . In fact (sin x) =

1 sin x

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 = sin–1 x ⇒ x = sin y

–1 cos

tan–1

1 x

1 x

1 x

x = sin y ⇒ y = sin–1 x

sin–1 (sin x) = x

= cosec–1 x

cos–1 (– x) = π – cos–1 x

= sec–1 x

cot–1 (– x) = π – cot–1 x

sec–1 (– x) = π – sec–1 x

sin (sin–1 x) = x

–1 sin

= cot–1 x

53

54

M AT H E M AT I C S

sin–1 (– x) = – sin–1 x

tan–1 x + cot–1 x =

sin–1 x + cos–1 x =

π 2

π 2

x + y –1 –1 –1 tan + tan = tan x y 1 − xy

tan–1 (– x) = – tan–1 x

cosec–1 (– x) = – cosec–1 x

cosec–1 x + sec–1 x =

–1 –1 2tan x = tan

π 2

2 x 1 − x 2

x − y –1 –1 –1 x y tan – tan = tan 1 + xy

–1

–1

2tan x = sin

2 x 1 + x 2

= cos

–1

1 − x 2 1 + x 2

Historical 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 importa nt results of trigonometry. tr igonometry. All this 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 sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, 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

INVERSE TRIGONOMETRIC FUNCTIONS

55

height, and comparing the ratios:

H S

=

h s

= tan (sun’s 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.

— —

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2

INVERSE TRIGONOME TRIGONOMETRIC TRIC FUNCTIONS Mathematics, in general, is fundamentally the science of

self-evident things. — FELIX KLEIN

2.1 Introduction In Chapter 1, we have studied that the inverse of a function f , denoted by f –1, exists if f f is 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 and observe their behaviour through graphical representations. Besides, some elementary properties will also be discussed. Arya Bhatta The inverse trigonometric functions play an important (476-550 A. D.) role in calculus for they serve to define many integrals. The concepts of inverse trigonometric functions functi ons 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]

π

, n ∈ Z} → R 2 cotangent function, i.e., cot : R – { x : x = nπ, n ∈ Z} → R tangent function, i.e., tan : R – { x : x = (2n + 1)

secant function, i.e., sec : R – { x : x = (2n + 1)

π

, n ∈ Z} → R – (– 1, 1) 2 cosecant function, i.e., cosec : R – { x : x = nπ, n ∈ Z} → R – (– 1, 1)

34

M AT H E M AT I C S

We have also learnt in Chapter 1 that if f f : X→Y such that f ( x x) = y is one-one and onto, then we can define a unique function g : Y→X such that g ( y) = x, where x ∈ X x), y ∈ Y. Here, the domain of g = range of f f and the range of g = domain and y = f ( x f . The function g is called the inverse of f f and is denoted by f –1. Further, g is also of f one-one and onto and inverse of g is f . Thus, g –1 = ( f –1)–1 = f . We also have f –1 o f ) ( x) = f –1 ( f f ( x x)) = f –1( y y) = x ( f f o f –1) ( y) = f ( f f –1( y y)) = f ( x x) = y ( f

and

Since the domain of sine function is the set of all real numbers and range is the

⎡ − π , π ⎤ , then it becomes one-one ⎣ 2 2 ⎥⎦

closed interval [–1, 1]. If we restrict its domain to ⎢

and onto with range [– 1, 1]. Actually, sine function restricted to any of the intervals

⎡ −3π , – π ⎤ , ⎡ −π , π ⎤ , ⎡ π , 3π ⎤ etc., ⎢⎣ 2 2 ⎥⎦ ⎢⎣ 2 2 ⎥⎦ ⎢⎣ 2 2 ⎥⎦

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 sin –1 (arc sine function). Thus, sin –1 is a function whose domain is [– 1, 1] and range could be any of the intervals

⎡ −3π , − π ⎤ , ⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦ ⎢⎣ 2 2 ⎥⎦

or

⎡ π , 3π ⎤ ⎢⎣ 2 2 ⎥⎦ , and so on. Corresponding to each such interval, we get a branch of the ⎡ −π , π ⎤ function sin–1. The branch with range ⎢ is called the principal value branch, ⎣ 2 2 ⎥⎦ whereas other intervals as range give different branches of sin–1. When we refer to the function sin–1, we take it as the function whose domain is [–1, 1] and range is

⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦ . We write

sin–1 : [–1, 1] →

⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦

From the definition of the inverse functions, it follows that sin (sin –1 x) = x if – 1 ≤ x ≤ 1 and sin–1 (sin x) = x if −

π 2

≤ x ≤

π 2

. In other words, if y = sin–1 x, then

sin y = x. Remarks (i)) We know (i know from from Cha Chapte pterr 1, that if y = f ( x x) is an invertible function, then x = f –1 ( y y). –1 Thus, the graph of sin function can be obtained from the graph of original function by interchanging x and y axes, i.e., if ( a, b) is a point on the graph of sine function, then (b, a) becomes the corresponding point on the t he graph of inverse

INVERSE TRIGONOMETRIC FUNCTIONS

35

of sine function. Thus, the graph of the function y = sin–1 x can be obtained from y = sin x by interchanging x and y axes. The graphs of y y = sin x and the graph of y –1 y = sin x are as given in Fig 2.1 (i), (ii), (iii). The dark portion of the graph of y = sin–1 x represent the principal value branch. (ii) It can be shown (ii) shown that the the graph of of an inverse inverse function function can be be obtained obtained from the the corresponding graph of original function functi on as a mirror image (i.e., reflection) along y = sin x and the line y = x. This can be visualised by looking the graphs of y y = sin–1 x as given in the same axes (Fig 2.1 (iii)).

Fig 2.1 (i)

Fig 2.1 (ii)

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 function to [0, π], then it becomes one-one and onto with range [–1, 1]. Actually, cosine function

36

M AT H E M AT I C S

restricted to any of the intervals [– π, 0], [0,π], [π, 2π] etc., is bijective with range as [–1, 1]. We can, therefore, define the inverse of cosine function in each of these intervals. We We denote the inverse of the cosine function by cos–1 (arc cosine function). Thus, cos–1 is a function whose domain is [–1, 1] and range could be any of the intervals [– π, 0], [0, π], [ π, 2 π] etc. Corresponding to each such interval, we get a branch of the function cos–1. The branch with range [0, π] is calle called d the principal –1 value branch of the function cos . We write cos–1 : [–1, 1] → [0, π]. The graph graph of the function given by y = cos–1 x can be drawn in the same way as discussed about the graph of y = sin–1 x. The graphs of y y = cos x and y = cos–1 x are given in Fig 2.2 (i) and (ii).

Fig 2.2 (i)

Fig 2.2 (ii)

Let us now discuss cosec –1 x and sec–1 x as follows: 1 Since, cosec x = , the domain of the cosec function is the set { x : x ∈ R and sin x x ≠ nπ, n ∈ Z} and the range is the set { y : y ∈ R, y ≥ 1 or y ≤ –1} i.e., the set R – (–1, 1). It means that y = cosec x assumes all real values except –1 0 x To prove the first result, we put cosec –1 x = y, i.e., x = cosec y

Therefore H ence

1 x

= sin y

sin–1

x

= y

1

= cosec–1 x x Similarly,, we can prove the other parts. Similarly or

sin–1

1

2. (i) sin–1 (– x) = – sin–1 x, x ∈ [– 1, 1] (ii) tan–1 (– x) = – tan–1 x, x ∈ R (iii) co cose secc–1 (– x) = – cosec–1 x, | x | ≥ 1

Let sin–1 (– x) = y, i.e., – x = sin y so that x = – sin y, i.e., x = sin (– y). H ence Therefore

sin–1 x = – y = – sin–1 (– x) sin–1 (– x) = – sin–1 x

Similarly,, we can prove the other parts. Similarly 3. (i) cos–1 (– x) = π – cos–1 x, x ∈ [– 1, 1] (ii) sec–1 (– x) = π – sec–1 x, | x | ≥ 1 (iii) cot–1 (– x) = π – cot–1 x, x ∈ R

Let cos–1 (– x) = y i.e., – x = cos y so that x = – cos y = cos (π – y) cos–1 x = π – y = π – cos–1 (– x) He nce cos–1 (– x) = π – cos–1 x Similarly,, we can prove the other parts. Similarly Therefore

4. (i) sin–1 x + cos–1 x = (ii) tan–1 x + cot –1 x =

2 2

(iii (i ii)) co cose secc–1 x + sec –1 x =

, x ∈ [– 1, 1] , x ∈ R

2

, | x | ≥ 1

Let sin–1 x = y. Then x = sin y = cos Therefore

cos–1 x =

π 2

− y

=

⎛ π − y ⎞ ⎜2 ⎟ ⎝ ⎠ π − sin –1 x 2

43

44

M AT H E M AT I C S

sin–1 x + cos–1 x =

Hence

π 2

Similarly, we can prove the other parts.

x + y

5. (i) tan–1 x + tan –1 y = tan–1

x – y

(ii) tan–1 x – tan–1 y = tan–1

1 + y

2 x

(iii) 2tan–1 x = tan–1

1–

, xy < 1

1 – y

, xy > – 1

, | x | < 1

x 2

Let tan–1 x = θ and tan–1 y = φ. Then x = tan θ, y = tan φ tan θ + tan φ

Now

tan(θ + φ) =

This gives

θ + φ = tan–1

Hence

tan–1 x + tan–1 y = tan–1

1− tan θ tan φ

=

x + y

1− xy

x + y

1− xy x + y

1− xy

In the above result, if we replace y by – y, we get the second result and by replacing y by x, we get the third result. 6. (i) 2tan–1 x = sin–1

–1

(ii) 2tan x = cos

–1

(iii) 2 ta tan–1 x = tan–1

2 1 + x

2

1 – x

2

1+

2

, | x | ≤ 1

2 x 1 – x 2

, x ≥ 0

, – 1 < x < 1

Let tan–1 x = y, then x = tan y. Now sin–1

2 x 1 + x

2

= sin–1

2tan y 2 1 + tan y

= sin–1 (sin 2 y) = 2 y = 2tan–1 x

INVERSE TRIGONOMETRIC FUNCTIONS

Also cos–1

1 − x 2 1 + x

= cos–1

2

1− tan 2 y

= cos–1 (cos 2 y) = 2 y = 2tan–1 x

1 + tan y 2

(iii (i ii)) Can be worked out similarly. similarly. We now consider some examples. Example 3 Show that

(i) sin–1 ( 2 x 1 − x 2 ) = 2 sin–1 x, (ii) sin–1 ( 2 x 1 − x 2 ) = 2 cos–1 x,

1

−

2 1 2

≤ x ≤

1 2

≤ x ≤ 1

Solution

(i) Let x = sin θ. Then sin–1 x = θ. We have

(

sin–1 ( 2 x 1 − x 2 ) = sin–1 2 sin θ 1 − sin 2 θ

)

= sin–1 (2sinθ cosθ) = sin–1 (sin2θ) = 2θ = 2 ssin in–1 x (ii) Take x = cos θ, then proceeding as above, we get, sin–1 ( 2 x 1 − x 2 ) = 2 cos–1 x that tan–1 Example 4 Show th

1 2

+ tan –1

2 11

= tan –1

3 4

Solution By property 5 (i), we have

1 L.H.S. = tan

–1

1 2

+ tan –1

Example 5 Express tan

−1

2 11

= tan –1

+

2

2 11 1 2

1−

= tan −1

×

15

− = tan

1

20

3 4

= R.H.S.

2 11

⎛ cos x ⎞ , π π ⎜⎝ 1 − sin x ⎟⎠ − 2 1 in the simplest form. 2 x − 1 ⎠ 1

Solution Let x = sec θ, then

x

2

−1 =

sec 2 θ − 1 = tan θ

INVERSE TRIGONOMETRIC FUNCTIONS

Therefore, cot

1

–1

x

2

−1

= cot–1 (cot θ) = θ = sec–1 x, which is the simplest form.

Example 7 Prove that tan x + tan –1

–1

2 x 1 − x

2

⎛ 3 x − x3 ⎞ ⎜ 2 ⎟, ⎝ 1 − 3 x ⎠

–1

= tan

| x | <

1 3

Solution Let x = tan θ. Then θ = tan–1 x. We have

R.H.S. = tan

–1

3 ⎛ 3 x − x3 ⎞ –1 ⎛ 3 tan θ − tan θ ⎞ = tan ⎜ ⎜ ⎟ 2 ⎟ 2 x 1 − 3 ⎝ ⎠ ⎝ 1 − 3 tan θ ⎠

= tan–1 (tan3θ) = 3θ = 3tan–1 x = tan–1 x + 2 tan–1 x = tan–1 x + tan–1

2 x

= L.H.S. (Why?) 2 1 − x Example 8 Find the value of cos (s (sec ec–1 x + cosec–1 x), | x | ≥ 1

⎛ π⎞ ⎜ 2 ⎟= 0 ⎝ ⎠

Solution We have cos (sec–1 x + cosec–1 x) = cos

EXERCISE 2.2 Prove the following:

⎡ 1 1⎤ 1. 3sin–1 x = sin–1 (3 x – 4 x3), x ∈ ⎢ – , ⎥ ⎣ 2 2⎦ ⎡1 ⎤ 2. 3cos–1 x = cos–1 (4 x3 – 3 x), x ∈ ⎢ , 1⎥ ⎣2 ⎦ 3. tan –1 4.

2 11

2 tan −1

+ tan −1 1 2

7 24

= tan −1

1 2

1

31

7

17

+ tan −1 = tan −1

Write the following functions in the simplest form: 1 + x

−1

5.

tan

7.

tan −

x 1

⎛ ⎜⎜ ⎝

2

− 1 , x ≠ 0

⎞ ⎟ , x < π 1 + cos x ⎟⎠ 1 − cos x

47

1

−1

6.

tan

8.

tan −

x 1

2

−1

x | > 1 , | x

⎛ cos x − sin x ⎞ ⎜ cos x + sin x ⎟ , x < π ⎝ ⎠

48

9.

10 .

M AT H E M AT I C S

x

tan −1 a

tan

2

− x2

, | x | < a

−a ⎞ ≤ x ≤ , > 0; a ⎜ 3 ⎟ 2 3 − 3 a ax ⎝ ⎠

−1 ⎛ 3a

2

x − x3

a

3

Find the values of each of the following: 11.

13 .

tan

tan

⎡ ⎛ 2 sin –1 1 ⎞ ⎤ 2 c o s ⎜ ⎟⎥ ⎢ ⎝ 2 ⎠⎦ ⎣

–1

1⎡

⎢sin

1 + x

2⎣

⎛ ⎝

2

+ cos

− y2 ⎤ 2 ⎥ , | x | < 1, y > 0 and xy < 1 1+ y ⎦

–1 1

+ cos–1 x ⎞⎟ =1 , then find the value of x x 5 ⎠ x +1 π –1 x − 1 + tan –1 = , then find the value of x If tan x x − 2 x+2 4

14 . If sin ⎜ sin 15 .

2 x

–1

12 . cot (tan–1a + cot–1a)

–1

1

Find the values of each of the expressions in Exercises 16 to 18. 16 .

18 . 19 .

sin

–1

⎛ ⎝

6

–1

tan

–1

⎛ tan 3π ⎞ ⎜ ⎟ 4 ⎠ ⎝

3

(B)

6

(C)

π 3

(D)

π 6

⎛ π − sin −1 (− 1 ) ⎞ ⎟ is equal to 2 ⎠ ⎝3

sin ⎜

(A) 21 .

17 .

3 + cot –1 ⎞⎟ 5 2⎠ ⎛ 7 π ⎞ is equal to cos −1 ⎜ cos to ⎟ 6 ⎠ ⎝ 7π 5π tan ⎜ sin

(A) 20 .

⎛ sin 2π ⎞ ⎜ ⎟ 3 ⎠ ⎝

1 2

(B)

1

(C)

3

1 4

(D) 1

tan −1 3 − cot −1 (− 3 ) is equal to (A) π

(B)

−

π 2

(C ) 0

(D) 2 3

INVERSE TRIGONOMETRIC FUNCTIONS

Miscellaneous Examples 3π

− Example 9 Find the value of sin (sin 1

5

) −1

Solution We know that sin −1 (sin x) = x . Therefore, sin (sin

But However

Therefore

3π 5

3π

)=

3π 5

π π ∉ ⎡⎢ − , ⎤⎥ , which is the principal branch of sin–1 x 5 ⎣ 2 2⎦ 2π ⎡ π π ⎤ 3π 3π 2π ∈ − , sin ( ) = sin(π − ) = sin and 5 ⎢⎣ 2 2 ⎥⎦ 5 5 5 sin −1 (sin

3π

) = sin −1 (sin

5

3

− Example 10 Show that sin

1

Solution Let sin −1

3

= x

5

3

and sin −1

)=

2π 5

= cos−1

17

8

84 85

= y

17

8

Therefore

sin x =

Now

cos x = 1 − si sin 2 x

=

1−

and

2 cos y = 1 − si sin y

=

1−

We have

x– y) = cos x cos y + sin x sin y cos ( x

5

and sin y =

5

8

− sin −1

5

2π

4

=

5

×

15

= cos−1

84

Hence

sin −

3

8

5

17

289

17

+ ×

=

64

5

17

⎛ 84 ⎞ ⎜⎝ 85 ⎟⎠

− sin −1

25

8

x − y = cos − 1

9

3

Therefore

1

17

85

=

4

(Why?)

5

=

15 17

84 85

49

50

M AT H E M AT I C S

1 12

− Example 11 Show that sin

Solution Let sin −1

12 13

co cos−

= x,

Then

sin x =

12

Therefore

cos x =

5

13

13

4

63

5

16

+ cos−1 + tan −1 1

4 5

, cco os y =

= y , tan −1 4 5

,

63 16

=π

=z 63

tan z =

16

3 12 3 , sin y = , tan x = and tan y = 13 5 5 4 12 tan x + tan y

We have

tan( x + y ) =

Hence

tan( x + y) = − tan z

1 − tan x tan y

=

+

3

5 4 12 3

1−

5

×

=−

63 16

4

tan ( x x + y) = tan (– z) or tan ( x x + y) = tan (π – z)

i.e., Therefore

x + y = – z or x + y = π – z

Since

x, y and z are positive, x + y ≠ – z (Why?)

Hence

x + y + z = π or sin

Example 12 Simplify tan

–1

–1 12

13

4

63

5

16

+ cos–1 + tan –1

⎡ a cos x − b sin x ⎤ a ⎢ b cos x + a sin x ⎥ , if b ⎣ ⎦

=π

tan x > –1 –1

Solution We have,

⎡ a cos x − b sin x ⎤ tan –1 ⎢ ⎥ ⎣ b cos x + a sin x ⎦

=

⎡ a cos x − b sin x ⎤ ⎢ ⎥ b cos x tan –1 ⎢ ⎥ ⎢ b cos x + a sin x ⎥ b cos x ⎣ ⎦

–1 = tan

a b

=

⎡ a − tan x ⎤ ⎢ ⎥ tan –1 ⎢ b ⎥ a ⎢1 + tan x ⎥ ⎣ b ⎦ a

− tan –1 (tan x) = tan –1 − x b

INVERSE TRIGONOMETRIC FUNCTIONS

π

Example 13 Solve tan–1 2 x + tan–1 3 x =

or

–1

⎛ 2 x + 3x ⎞ ⎜ 1 − 2 x × 3x ⎟ ⎝ ⎠

=

⎛ 5 x ⎞ ⎜ 1 − 6 x 2 ⎟ ⎝ ⎠

=

tan

i.e.

–1

5 x

4

π

Solution We have tan–1 2 x + tan–1 3 x =

tan

51

4

π 4

π 4

π

=1 2 1 − 6 x 4 6 x2 + 5 x – 1 = 0 i. i.e., e., (6 x – 1) ( x + 1) = 0

Therefore or which gives

= tan

x =

1

or x = – 1. 6 Since x = – 1 does not satisfy the equation, as the L.H.S. of the equation becomes

negative, x =

1 6

is the only solution of the given equation.

Miscellaneous Exercise on Chapter 2 Find the value of the following: 1.

–1

cos

⎛ cos 13π ⎞ ⎜ ⎟ 6 ⎠ ⎝

–1

2.

tan

4.

sin

6.

cos

⎛ tan 7π ⎞ ⎜ ⎟ 6 ⎠ ⎝

Prove that –1

3.

2 sin

5.

cos

7.

tan

–1

8.

tan

–1

–1

3 5

4 5

+ cos–1

63 16 1 5

= tan –1

24 7

12 13

= sin –1

= cos–1

5 13

+ cos –1

33 65 3 5

1

1

1

π

7

3

8

4

+ tan −1 + tan −1 + tan −1 =

–1

8 17

–1 12

13

+ sin –1 = tan –1

3

77

5

36

3

56

5

65

+ sin –1 = sin –1

52

M AT H E M AT I C S

Prove that 9.

tan

–1

10 .

cot

–1

11.

tan

–1

12 .

9π 8

−

1

=

x

2

cos

⎛ ⎜⎜ ⎝

1 + sin x

⎛ ⎜⎜ ⎝

1 + x

–1

+ 1 + sin x −

9 4

− 1 + x + sin −

1

1 3

=

⎛ 1 − x ⎞ ⎜⎝ 1 + x ⎟⎠ , x ∈ [0, 1] ⎞ x ⎛ π⎞ ⎟⎟ = , x ∈ ⎜⎝ 0, 4 ⎟⎠ 1 − sin x ⎠ 2 1 − sin x

⎞ π ⎟= − 1 − x ⎟⎠ 4 1− x

9 4

sin −

1

1 2

–1

cos x ,

−

1 2

≤ x ≤ 1 [Hint: Put x = cos 2θ]

2 2 3

Solve the following equations: 13 . 2tan–1 (cos x) = tan–1 (2 cosec x)

14 .

tan

− x 1 –1 = tan x, ( x > 0) 1 + x 2

–1 1

15 . sin (tan–1 x), | x x | < 1 is equal to x

(A)

1 − x 2

16 . sin–1 (1 – x) – 2 sin –1 x =

(A) 0,

17 .

1 2

1 − x 2

π 2

(B) 1,

1 2

is equal to

π

π

2

(B)

3

1

(C)

1 + x 2

x

(D)

1 + x 2

, then x is equal to

⎛ x ⎞ − tan −1 x − y ⎟ x+ y ⎝ y⎠

tan −1 ⎜

(A)

1

(B)

(C ) 0

(C)

π 4

(D)

(D)

1 2

−3π 4

INVERSE TRIGONOMETRIC FUNCTIONS

Summary

The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table: Fu n cti on s

Domain

R a n ge (Principal Value Branches)

y = sin–1 x

[–1, 1]

⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦

y = cos–1 x

[–1, 1]

[0, π]

y = cosec–1 x

R – (–1,1)

⎡ −π , π ⎤ ⎢⎣ 2 2 ⎥⎦ – {0}

y = sec–1 x

R – (–1, 1)

[0, π] – { } 2

y = tan–1 x

R

⎛− π,π⎞ ⎜ 2 2⎟ ⎝ ⎠

y = cot–1 x

R

(0, π)

π

–1 –1 –1 sin x should not be confused with (sin x) . In fact (sin x) =

1 sin x

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 = sin–1 x ⇒ x = sin y

–1 cos

tan–1

1 x

1 x

1 x

x = sin y ⇒ y = sin–1 x

sin–1 (sin x) = x

= cosec–1 x

cos–1 (– x) = π – cos–1 x

= sec–1 x

cot–1 (– x) = π – cot–1 x

sec–1 (– x) = π – sec–1 x

sin (sin–1 x) = x

–1 sin

= cot–1 x

53

54

M AT H E M AT I C S

sin–1 (– x) = – sin–1 x

tan–1 x + cot–1 x =

sin–1 x + cos–1 x =

π 2

π 2

x + y –1 –1 –1 tan + tan = tan x y 1 − xy

tan–1 (– x) = – tan–1 x

cosec–1 (– x) = – cosec–1 x

cosec–1 x + sec–1 x =

–1 –1 2tan x = tan

π 2

2 x 1 − x 2

x − y –1 –1 –1 x y tan – tan = tan 1 + xy

–1

–1

2tan x = sin

2 x 1 + x 2

= cos

–1

1 − x 2 1 + x 2

Historical 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 importa nt results of trigonometry. tr igonometry. All this 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 sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, 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

INVERSE TRIGONOMETRIC FUNCTIONS

55

height, and comparing the ratios:

H S

=

h s

= tan (sun’s 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.

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