Ch 2-Inverse Trigonometric Functions

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

 Mathematics, in general, is fundamentally the science of Mathematics, in general, is fundamentally the science of

self-evident things. — FELIX KLEIN

self-evident things. — FELIX KLEIN __

2.1 Introduction

In Chapter 1, we have studied that the inverse of a

In Chapter 1, we have studied that the inverse of a functionfunction

f

f , denoted by, denoted by f f –1–1_{, exists if}_{, exists if}_f_f_{is one-one and onto. There are}_{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

inverses do not exist. In this chapter, we shall study aboutchapter, 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

Besides, some elementary properties will also be discussed.discussed.

The inverse trigonometric functions play an important

role in calculus for they serve to define many integrals.

The concepts of inverse trigonometric functi

The concepts of inverse trigonometric functions is also used in science and engineering.ons is also used in science and engineering.

2.2

2.2 Basic

Basic

Concepts

In Class XI, we have studied trigonometric functions, which are

In Class XI, we have studied trigonometric functions, which are defined as follows:defined as follows:

sine function, i.e., sine :

sine function, i.e., sine :RR→→[– 1, 1][– 1, 1]

cosine function, i.e., cos :

cosine function, i.e., cos : RR→→[– 1, 1][– 1, 1]

tangent function, i.e., tan :

tangent function, i.e., tan :RR – {– { x x:: x x= (2= (2nn+ 1)+ 1)

2 2

ππ

,

,nn∈∈ ZZ}}→→RR

cotangent function, i.e., cot :

cotangent function, i.e., cot : RR– {– { x x:: x x==nnππ,, nn∈∈ZZ}}→→RR

secant function, i.e., sec :

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

2 2

ππ

,

, nn∈∈ ZZ}} →→RR – (– 1, 1)– (– 1, 1)

cosecant function, i.e., cosec :

cosecant function, i.e., cosec :RR – {– { x x:: x x== nnππ,, nn∈∈ ZZ}} →→ RR – (– 1, 1)– (– 1, 1)

Chapter

2

INVERSE

TRIGONOME

TRIC

FUNCTIONS

Arya Bhatta Arya Bhatta (476-550 A. D.) (476-550 A. D.)

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3

344 MMAATTHHEEMMAATTIICCSS

W

We have also learnt in Chapter 1 e have also learnt in Chapter 1 that if that if f f : X: X→→Y such thatY such that f f (( x x) =) = y yis one-one andis one-one and

onto, then we can define a unique function

onto, then we can define a unique functiongg: Y: Y→→X such thatX such thatgg(( y y) =) = x x, where, where x x∈∈XX

and

and y y== f f (( x x),), y y∈∈Y. Here, the domain of Y. Here, the domain of gg= range of = range of f f and the range of and the range of gg= domain= domain

of

of f f . The function. The function ggis called the inverse of is called the inverse of f f and is denoted byand is denoted by f f –1–1_{. Further,}_{. Further,} _g_g_{is also}_{is also}

one-one and onto and inverse of

one-one and onto and inverse of ggisis f f . Thus,. Thus, gg–1–1_{= (}_{= (}_f_f–1–1₎₎–1–1₌₌ _f_f_{. We also have}_{. We also have}

(

( f f –1–1 _o_o_f_f₎₎₍₍_x_x₎₎₌₌_f_f–1–1 ₍₍_f_f₍₍_x_x₎₎₎₎₌₌_f_f–1–1₍₍_y_y₎₎₌₌_x_x

a

anndd (( f f oo f f –1–1_{) (}_{) (}_y_y₎₎₌₌_f_f₍₍_f_f–1–1₍₍_y_y₎₎₎₎ ₌₌_f_f₍₍_x_x₎₎₌₌_y_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

closed interval [–1, 1]. If we restrict its domain to ,, 2 2 22

−−π

π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

, then it becomes one-one, then it becomes one-one

and onto with range

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

3 3 –– ,, 2 2 22

−

− π

π

ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

,, 22 ,, 22

−π

−π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

,, 3 3 ,, 2 2 22

π

π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

etc., is one-one and its range is [–1, 1]. We can,etc., is one-one and its range is [–1, 1]. We can,

therefore, define the inverse of

therefore, define the inverse of sine function in each sine function in each of these intervals. We denote theof these intervals. We denote the

inverse of sine function by sin

inverse of sine function by sin–1–1 _{(arc sine function). Thus, sin}_{(arc sine function). Thus, sin}–1–1 _{is a function whose}_{is a function whose}

domain is [– 1, 1] and range could be any of the intervals

domain is [– 1, 1] and range could be any of the intervals 33 ,, 2 2 22

−

− π

π −−ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

,, 22 ,, 22

−π

−π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

oror 3 3 ,, 2 2 22

π

π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

, and so on. Corresponding to each such interval, we get a, and so on. Corresponding to each such interval, we get a branchbranch of theof the

function sin

function sin–1–1_{. The branch with range}_{. The branch with range} _,,

2 2 22

−π

−π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

is called theis called the principal value branch, principal value branch,

whereas other intervals as range give different branches of

whereas other intervals as range give different branches of sinsin–1–1_{. When we refer}_{. When we refer}

to the function sin

to the function sin–1–1_{, we take it as the function whose domain is [–1, 1] and range is}_{, we take it as the function whose domain is [–1, 1] and range is}

,, 2 2 22

−π

−π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

. . We We write write sinsin –1 –1 _{: [–1, 1]}_{: [–1, 1]} _→_→ _,, 2 2 22

−π

−π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

From the definition of the inverse functions, it follows that sin (sin

From the definition of the inverse functions, it follows that sin (sin–1–1_x_x₎₎₌₌_x_x

if

if – – 11 ≤≤ x x ≤≤1 and sin1 and sin–1–1 _(sin_(sin_x_x_{) =}_{) =}_x_x_if_if

2

2 x x 22

π

ππ

−

− ≤

≤ ≤≤

. In other words, if . In other words, if y y= sin= sin–1–1_x_x_{, then}_{, then}

sin

sin y y== x x..

Remarks

(i

(i)) WWe kne know fow from rom ChaChaptepter 1, r 1, that that if if y y== f f (( x x) is an invertible function, then) is an invertible function, then x x== f f –1–1₍₍_y_y_)._).

Thus, the graph of sin

Thus, the graph of sin–1–1 _{function can be obtained from the graph of original}_{function can be obtained from the graph of original}

function

function by by interchanginginterchanging x xandand y yaxes, i.e., if (axes, i.e., if (aa,,bb) is a point on the graph of ) is a point on the graph of

sine function, then (

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of sine function. Thus, the graph of the function

of sine function. Thus, the graph of the function y y= sin= sin–1–1_x_x_{can be obtained from}_{can be obtained from}

the graph of

the graph of y y= sin= sin x xby interchangingby interchanging x xandand y yaxes. The graphs of axes. The graphs of y y= sin= sin x xandand

y

y= sin= sin–1–1_x_x_{are as given in Fig 2.1 (i), (ii), (iii). The dark portion of the graph of}_{are as given in Fig 2.1 (i), (ii), (iii). The dark portion of the graph of}

y

y= sin= sin–1–1_x_x_{represent the principal value branch.}_{represent the principal value branch.}

(i

(ii)i) It can be shIt can be shown that thown that the graph oe graph of an inverf an inverse funcse function can btion can be obtaine obtained from thed from thee

corresponding graph of original functi

corresponding graph of original function as a mirror image (i.e., reflection) alongon as a mirror image (i.e., reflection) along

the line

the line y y == x x. This can be visualised by looking the graphs of . This can be visualised by looking the graphs of y y = sin= sin x x andand

y

y= sin= sin–1–1_x_x_{as given in the same axes (Fig 2.1 (iii)).}_{as 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 function

to [0,

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

Fig 2.1 (ii)

Fig 2.1 (ii) Fig 2.1 (iii)Fig 2.1 (iii) Fig 2.1 (i)

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restricted to any of the intervals [–

restricted to any of the intervals [– ππ, , 0], 0], [0,[0,ππ],],[[ππ, 2, 2ππ] etc., is bijective with range as] etc., is bijective with range as

[–1, 1]. We can, therefore, define the inverse of cosine function in each of these

intervals. W

intervals. We denote the e denote the inverse of the inverse of the cosine function by coscosine function by cos–1–1_{(arc cosine function).}_{(arc cosine function).}

Thus, cos

Thus, cos–1–1 _{is a function whose domain is [–1, 1] and range}_{is a function whose domain is [–1, 1] and range}

could be any of the intervals [–

could be any of the intervals [–ππ, 0], [0,, 0], [0, ππ], ], [[ππ, , 22ππ] etc.] etc.

Corresponding to each such interval, we get a branch of the

function cos

function cos–1–1_{. The branch with range [0,}_{. The branch with range [0,}_π_π_{] is}_{] is}_calle_calle_{d the}_{d the}_principal_principal

value branch

value branchof the function cosof the function cos–1–1_{. We write}_{. We write}

cos

cos–1–1 _{: [–1, 1]}_{: [–1, 1]} _→_→ _[0,_[0, _π_π_]._].

The grap

The graph of the function given h of the function given byby y y= cos= cos–1–1_x_x_{can be drawn}_{can be drawn}

in the same way as discussed about the graph of

in the same way as discussed about the graph of y y= sin= sin–1–1_x_x_{. The}_{. The}

graphs of

graphs of y y= cos= cos x xandand y y= cos= cos–1–1_x_x_{are given in Fig 2.2 (i) and (ii).}_{are given in Fig 2.2 (i) and (ii).}

Fig 2.2 (ii) Fig 2.2 (ii)

Let us now discuss cosec

Let us now discuss cosec–1–1_x_x _{and sec}_{and sec}–1–1_x_x _{as follows:}_{as follows:}

Since, cosec

Since, cosec x x== 11

sin

sin x x, the domain of the cosec function is the set {, the domain of the cosec function is the set { x x:: x x∈∈RRandand x

x ≠≠ nnππ,, nn ∈∈ ZZ} and the range is the set {} and the range is the set { y y :: y y ∈∈ RR,, yy ≥≥ 1 1 oror y y ≤≤ –1} i.e., the set–1} i.e., the set

R

R– (–1, 1). It means that– (–1, 1). It means that y y= cosec= cosec x xassumes all real values except –1 <assumes all real values except –1 < y y< 1 and is< 1 and is

not defined for integral multiple of

not defined for integral multiple of ππ. If we restrict the domain of cosec function to. If we restrict the domain of cosec function to

,, 2 2 22

π

π ππ

⎡

₋₋

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

– {0}, then it is one to one and onto – {0}, then it is one to one and onto with with its range as the setits range as the setRR– (– – (– 1, 1). Actually,1, 1). Actually,

cosec function restricted to any of the intervals

cosec function restricted to any of the intervals 33 ,, {{ }}

2 2 22

− π

− π −−ππ

⎡

_{⎤⎤ − − −π}

_−π

⎢

⎥⎥

⎣

⎦⎦

,, 22 2,,2

−π

−π ππ

⎡

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

– {0},– {0}, 3 3 ,, {{ }} 2 2 22

π

π ππ

⎡

_{⎤⎤ − − ππ}

⎢

⎥⎥

⎣

⎦⎦

etc., is bijective and its range is the set of all real numbersetc., is bijective and its range is the set of all real numbers RR– (–1, 1).– (–1, 1).

Fig 2.2 (i) Fig 2.2 (i)

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Thus cosec

Thus cosec–1–1_{can be defined as a function whose domain is}_{can be defined as a function whose domain is}_R_R_{– (–1, 1) and range could}_{– (–1, 1) and range could}

be any of the intervals

be any of the intervals ,, {{00}}

2 2 22

−π

−π ππ

⎡

_{⎤⎤ −−}

⎢

⎥⎥

⎣

⎦⎦

,, 3 3 ,, {{ }} 2 2 22

− π −π

⎡

_{⎤⎤ − − −π}

_−π

⎢

⎥⎥

⎣

⎦⎦

,, 3 3 ,, {{ }} 2 2 22

π

π ππ

⎡

_{⎤⎤ − − ππ}

⎢

⎥⎥

⎣

⎦⎦

etc. Theetc. The

function corresponding to the range

function corresponding to the range ,, {{00}} 2 2 22

−π

−π ππ

⎡

_{⎤⎤ −−}

⎢

⎥⎥

⎣

⎦⎦

is called theis called the principal value branch principal value branch

of cosec

of cosec–1–1_{. We thus have principal branch as}_{. We thus have principal branch as}

cosec cosec–1–1 _{: R}_{: R} _{– (–1, 1)}_{– (–1, 1)} _→_→ ,, {{00}} 2 2 22

−π

−π ππ

⎡

_{⎤⎤ −−}

⎢

⎥⎥

⎣

⎦⎦

The graphs of

The graphs of y y= cosec= cosec x xandand y y= cosec= cosec–1–1_x_x_{are given in Fig 2.3 (i), (ii).}_{are given in Fig 2.3 (i), (ii).}

Also, since sec

Also, since sec x x== 11

cos

cos x x , the domain of , the domain of y y= sec= sec x xis the setis the setRR– {– { x x:: x x= (= (22nn+ 1)+ 1) 22

ππ

,

n

n∈∈ZZ} and range is the set} and range is the set RR – (–1, 1). It means that sec (secant function) assumes– (–1, 1). It means that sec (secant function) assumes

all real values except –1 <

all real values except –1 < y y < 1 and is not defined for odd multiples of < 1 and is not defined for odd multiples of

2 2

ππ

. If we

restrict the domain of

restrict the domain of secant function to [0,secant function to [0,ππ] ] – – {{

2 2

ππ

}, then it is one-one and onto with

Fig 2.3 (i)

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3

its range as the set

its range as the set RR – (–1, 1). – (–1, 1). Actually, Actually, secant function secant function restricted to any of therestricted to any of the

intervals [– intervals [–ππ, 0] – {, 0] – { 2 2

−−ππ

}, }, [[00,, ] –] – 2 2

ππ

⎧

⎧ ⎫⎫

ππ

_⎨

_{⎨ ⎬⎬}

⎩

⎩ ⎭⎭

, [, [ππ, 2, 2ππ] – ] – {{ 3 3 2 2

ππ

} etc., is bijective and its range

is

isRR– {–1, 1}. Thus sec– {–1, 1}. Thus sec–1–1_{can be defined as a function whose domain is}_{can be defined as a function whose domain is} _R_R_{– (–1, 1) and}_{– (–1, 1) and}

range could be any of the intervals [–

range could be any of the intervals [–ππ, 0] – {, 0] – {

2 2

−−ππ

}, [0, }, [0,ππ] ] – – {{ 2 2

ππ

}, }, [[ππ, 2, 2ππ] ] – – {{33 2 2

ππ

} etc. } etc. Corresponding to each

Corresponding to each of these intervals, we get different branchof these intervals, we get different branches of the functies of the function secon sec–1–1_._.

The branch with range [0,

The branch with range [0, ππ] ] – – {{

2 2

ππ

} is called the

} is called the principal value branch principal value branchof theof the

function sec

function sec–1–1_{. We thus have}_{. We thus have}

sec sec–1–1 _:_: _R_R _{– (–1,1)}_{– (–1,1)} _→_→_[0,_[0, _π_π_{] –}_{] –}_{_{ 2 2

ππ

} }

The graphs of the functions

The graphs of the functions y y= sec= sec x xandand y y= sec= sec-1-1_x_x_{are given in Fig 2.4 (i), (ii).}_{are given in Fig 2.4 (i), (ii).}

Finally

Finally, we now , we now discuss tandiscuss tan–1–1_{and cot}_{and cot}–1–1

We know that the domain of the tan function (tangent function) is the set

{ { x x::xx∈∈ RRandand x x≠≠(2(2nn+1)+1) 2 2

ππ

,

, nn∈∈ZZ} and the range is} and the range is RR. It means that tan function. It means that tan function

is not defined for odd multiples of

2 2

ππ

. If we restrict the domain of tangent function to

Fig 2.4 (i)

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,, 2 2 22

−π

−π ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

, then it is one-one and onto with its range as, then it is one-one and onto with its range as RR. Actually. Actually, tangent , tangent functionfunction

restricted to any of the intervals

restricted to any of the intervals 33 ,, 2 2 22

− π −π

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

,, 22 2,,2

−π

−π ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

,, 3 3 ,, 2 2 22

π

π ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

etc., is bijectiveetc., is bijective

and its range is

and its range is RR. Thus tan. Thus tan–1–1 _{can be defined as a function whose domain is}_{can be defined as a function whose domain is} _R_R _and_and

range could be any of the intervals

range could be any of the intervals 33 ,,

2 2 22

− π −π

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

,, 22 ,,22

−π

−π ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

,, 3 3 ,, 2 2 22

π

π ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

and so on. Theseand so on. These

intervals give different branches of

intervals give different branches of the function the function tantan–1–1_{. The branch with range}_{. The branch with range} ,,

2 2 22

−π

−π ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

is called the

is called the principal value branch principal value branchof the function tanof the function tan–1–1_._.

We thus have We thus have tan tan–1–1 _:_: _R_R _→_→ ,, 2 2 22

−π

−π ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

The graphs of the function

The graphs of the function y y==tantan x xandand y y = tan= tan–1–1_x_x_{are given in Fig 2.5 (i), (ii).}_{are given in Fig 2.5 (i), (ii).}

F

Fiigg22..55((ii)) FiFigg 22..55((iiii))

We know that domain of the cot function (cotangent function) is the set

{

{ x x:: x x∈∈RRandand x x≠≠nnππ,,nn∈∈ZZ} and range is} and range is RR. It means that cotangent function is not. It means that cotangent function is not

defined for integral multiples of

defined for integral multiples of ππ. If we restrict the domain of cotangent function to. If we restrict the domain of cotangent function to

(0,

(0,ππ), then it is bijective with and its range as), then it is bijective with and its range asRR. In fact, cotangent function restricted. In fact, cotangent function restricted

to any of the intervals (–

to any of the intervals (–ππ, 0), (0,, 0), (0, ππ), (), (ππ, 2, 2ππ) etc., is bijective and its range is) etc., is bijective and its range isRR. Thus. Thus

cot

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intervals (–

intervals (–ππ, 0), (0,, 0), (0, ππ), ), ((ππ, , 22ππ) etc. These intervals give different branches of the) etc. These intervals give different branches of the

function cot

function cot–1–1_{. The function with range (0,}_{. The function with range (0,}_π_π_{) is called the}_{) is called the}_{principal value branch}_{principal value branch}_of_of

the function cot

the function cot–1–1_{. We thus have}_{. We thus have}

cot

cot–1–1 _:_: _R_R _→_→ _(0,_(0, _π_π₎₎

The graphs of

The graphs of y y==cotcot x xandand y y= cot= cot–1–1_x_x_{are given in Fig 2.6 (i), (ii).}_{are given in Fig 2.6 (i), (ii).}

Fig 2.6 (i)

Fig 2.6 (i) _{Fig 2.6 (ii)}_{Fig 2.6 (ii)}

The following table gives the inverse trigonometric function (principal value

branches) along with their domains and ranges.

sin sin–1–1 _:_: _[_[_–_–₁₁_,_,₁₁_]_] _→_→ ,, 2 2 22

π

π ππ

⎡

₋₋

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

cos cos–1–1 _:_: _[_[_–_–₁₁_,_,₁₁_]_] _→_→ _[0,_[0, _π_π_]_] cosec cosec–1–1 _:_: _R_R _{– (–1,1)}_{– (–1,1)} _→_→ ,, 2 2 22

π

π ππ

⎡

₋₋

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

– {0}– {0} se secc–1–1 _:_: _R_R _{– (–1, 1)}_{– (–1, 1)} _→_→ _[0,_[0,_π_π_{] –}_{] –} _{_{{ }}_} 2 2

ππ

ta tann–1–1 _:_: _R_R _→_→ ,, 2 2 22

−π

−π ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

cot cot–1–1 _:_: _R_R _→_→ _(0,_(0,_π_π₎₎

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

NoteNote

1

1.. ssiinn–1–1_x_x _{should not be confused with (sin}_{should not be confused with (sin}_x_x₎₎–1–1_{. In fact (sin}_{. In fact (sin}_x_x₎₎–1–1₌₌

1

1 sin

sin x

x

andand similarly for other trigonometric functions.

similarly for other trigonometric functions.

2.

2. WheneWhenever no brancver no branch of an inverse trigoh of an inverse trigonometrnometric functioic functions is mentionns is mentioned, weed, we

mean the principal value branch of that

mean the principal value branch of that function.function.

3.

3. The value of an invThe value of an inverse trigerse trigonomeonometric functric functions whtions which lies in the rangich lies in the range of e of

principal branch is called the

principal branch is called the principal value principal valueof that inverse trigonometricof that inverse trigonometric

functions.

W

We now e now consider some examples:consider some examples:

Example 1

Example 1Find the principal value of sinFind the principal value of sin–1–1 11

2 2

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

.. Solution

SolutionLet sinLet sin–1–1 11

2 2

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

== y y. Then, sin. Then, sin y y==

1 1 2 2 ..

We know that the range of the principal value branch of sin

We know that the range of the principal value branch of sin–1–1 _is_is ,,

2 2 22

π

π ππ

⎛

₋₋

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

andand sin sin 4 4

ππ

⎛

⎛ ⎞⎞

⎜

⎜ ⎟⎟

⎝

⎝ ⎠⎠

== 1 1 2

2 . Therefore, principal value of sin. Therefore, principal value of sin

–1 –1 11 2 2

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

isis 44

ππ

Example 2

Example 2Find the principal value of cotFind the principal value of cot–1–1 11

3 3

−−

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

Solution

Solution Let cotLet cot–1–1 11

3 3

−−

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

== y y. Then,. Then,

1 1

ccoott ccoott 3 3 3 3 y y

=

−

= −

⎛

⎛ ⎞⎞

_⎜

_{⎜ ⎟⎟}

ππ

⎝

⎝ ⎠⎠

== cotcot 33

ππ

⎛

_{ππ −−}

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

== 2 2 cot cot 3 3

ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

We know that the range of principal value branch of cot

We know that the range of principal value branch of cot–1–1 _{is (0,}_{is (0,} _π_π_{) and}_{) and}

cot cot 22 3 3

ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

== 1 1 3 3

−−

. Hence, principal value of cot

. Hence, principal value of cot–1–1 11

3 3

−−

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

isis 2 2 3 3

ππ

EXERCISE 2.1

Find the principal values of the following:

1. 1. sinsin–1–1 11 2 2

⎛

₋₋

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

2.2. coscos–1–1 3 3 2 2

⎛

⎞⎞

⎜

⎟⎟

⎜

⎟⎟

⎝

⎠⎠

3.3. coseccosec –1 –1 ₍₂₎₍₂₎ 4. 4. tantan–1–1 ((

−−

3)3) 5.5. coscos–1–1 11 2 2

⎛

₋₋

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

6.6. tantan–1–1 (–1)(–1)

(10)

4 422 MMAATTHHEEMMAATTIICCSS 7. 7. secsec–1–1 22 3 3

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

8.8. cotcot–1–1 (( 3)3) 9.9. coscos–1–1

1 1 2 2

⎛

⎞⎞

−−

⎜

⎟⎟

⎝

⎠⎠

10 10.. coseccosec–1–1 ₍₍ 2 2

−−

)) Find the

Find the values of values of the following:the following:

11.

11. tantan–1–1_{(1) + cos}_{(1) + cos}–1–1

1

2

2 ⎛

⎛

₋₋

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

+ sin+ sin–1–1

1

2

2 ⎛

⎛

₋₋

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

1212.. coscos–1–1

1

2

2 ⎛

⎛ ⎞⎞

⎜

⎜ ⎟⎟

⎝

⎝ ⎠⎠

+ 2 sin+ 2 sin–1–1

1

2

2 ⎛

⎛ ⎞⎞

⎜

⎜ ⎟⎟

⎝

⎝ ⎠⎠

13

13.. If sinIf sin–1–1_x_x₌₌_y_y_{, then}_{, then}

( (AA)) 00 ≤≤ y y≤ π≤ π (B)(B) 2 2 y y 22

π

ππ

−

− ≤

≤ ≤≤

( (CC)) 0 0 << y y<< ππ (D)(D) 2 2 y y 22

π

ππ

−

− <

< <<

14

14.. tantan–1–1 ₃₃

−

_sseecc−−11

( ( ))

−−

₂₂ _{is equal to}_{is equal to}

(A) (A) ππ (B)(B) 3 3

ππ

−−

(C)(C) 3 3

ππ

(D) (D) 22 3 3

ππ

2.3

2.3 Properties of Inverse T

Properties of Inverse T

rigonometric Functions

In this section, we shall prove some important properties of inverse trigonometric

functions. It may be mentioned here that these results are valid within the principal

value branches of the corresponding inverse trigonometric functions and wherever

they are defined. Some results may not be valid for all

they are defined. Some results may not be valid for all values of the domains of inversevalues of the domains of inverse

trigonometric functions. In fact, they will be valid only

trigonometric functions. In fact, they will be valid only for some values of for some values of x xfor whichfor which

inverse trigonometric functions are defined. We will not go into the details of these

values of

values of x xin the domain as this discussion goes beyond the in the domain as this discussion goes beyond the scope of this text book.scope of this text book.

Let us recall that if

Let us recall that if y y= sin= sin–1–1_x_x_{, then}_{, then}_x_x_{= sin}_{= sin}_y_y_{and if}_{and if}_x_x_{= sin}_{= sin}_y_y_{, then}_{, then}_y_y_{= sin}_{= sin}–1–1_x_x_{. This is}_{. This is}

equivalent to

sin (sin

sin (sin–1–1_x_x_{) =}_{) =}_x_x_,_,_x_x _∈_∈_{[– 1, 1] and sin}_{[– 1, 1] and sin}–1–1 _(sin_(sin_x_x_{) =}_{) =}_x_x_,_,_x_x_∈_∈ ,,

2 2 22

π

π ππ

⎡

₋₋

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

Same is true for

Same is true for other five inverse trigonometric other five inverse trigonometric functions as well. We now provefunctions as well. We now prove

some properties of inverse trigonometric functions.

1. 1. ((ii)) ssiinn–1–1 11 x x= cosec= cosec –1 –1_x_x_,_,_x_x _≥_≥_≥_≥_≥_≥_≥_≥_≥_≥ ₁₁_or_or_x_x _≤_≤_≤_≤_≤_≤_≤_≤_≤_≤ _–_–₁₁ ( (iiii)) ccooss–1–1 11 x x = sec= sec –1 –1_x_x_,_,_x_x ≥ ≥ ≥ ≥≥≥ ≥ ≥ ≥ ≥ 1 1 oror x x ≤≤≤≤≤≤≤≤≤≤ – – 11

(11)

( (iiiiii)) ttaann–1–1 11 x x= cot= cot –1 –1_x_x_,_,_x_x _>_>₀₀

To prove the first result, we put cosec

To prove the first result, we put cosec–1–1_x_x ₌₌_y_y_{, i.e.,}_{, i.e.,}_x_x _{= cosec}_{= cosec}_y_y

Therefore

Therefore 11

x

x = sin= sin y y

H Heennccee ssiinn–1–1 11 x x== y y o orr ssiinn–1–1 11 x x = cosec= cosec –1 –1_x_x Similarly

Similarly, we can , we can prove the other parts.prove the other parts.

2. 2. ((ii)) ssiinn–1–1 _(–_(–_x_x₎₎₌₌ _–_–_sin_sin–1–1_x_x_,_,_x_x ∈ ∈ ∈ ∈∈∈ ∈ ∈ ∈ ∈ [– 1, 1][– 1, 1] ( (iiii)) ttaann–1–1 _(–_(–_x_x_{) = – tan}_{) = – tan}–1–1_x_x_,_,_x_x _∈_∈_∈_∈_∈_∈_∈_∈_∈_∈ _R_R ( (iiiiii)) cocosesecc–1–1 _(–_(–_x_x_{) = – cosec}_{) = – cosec}–1–1_x_x_,_,_|_|_x_x_|_| _≥_≥_≥_≥_≥_≥_≥_≥_≥_≥ ₁₁ Let sin

Let sin–1–1_(–_(–_x_x_{) =}_{) =}_y_y_{, i.e., –}_{, i.e., –}_x_x_{= sin}_{= sin}_y_y_{so that}_{so that} _x_x _{= – sin}_{= – sin}_y_y_{, i.e.,}_{, i.e.,}_x_x_{= sin (–}_{= sin (–}_y_y_)._).

H

Heennccee ssiinn–1–1_x_x _{= –}_{= –}_y_y _{= – sin}_{= – sin}–1–1 _(–_(–_x_x₎₎

T

Thheerreeffoorree ssiinn–1–1 _(–_(–_x_x_{) = – sin}_{) = – sin}–1–1_x_x

Similarly

3. 3. ((ii)) ccooss–1–1_(–_(–_x_x₎₎₌₌ _π_π_π_π_π_π_π_π_π_π _{– cos}_{– cos}–1–1_x_x_,_,_x_x _∈_∈_∈_∈_∈_∈_∈_∈_∈_∈ _{[– 1, 1]}_{[– 1, 1]} ( (iiii)) sseecc–1–1 _(–_(–_x_x₎₎₌₌ _π_π_π_π_π_π_π_π_π_π _{– sec}_{– sec}–1–1_x_x_,_,_|_|_x_x_|_| _≥_≥_≥_≥_≥_≥_≥_≥_≥_≥ ₁₁ ( (iiiiii)) ccoott–1–1 _(–_(–_x_x₎₎₌₌ _π_π_π_π_π_π_π_π_π_π _{– cot}_{– cot}–1–1_x_x_,_,_x_x _∈_∈_∈_∈_∈_∈_∈_∈_∈_∈ _R_R Let cos

Let cos–1–1 _(–_(–_x_x_{) =}_{) =}_y_y _{i.e., –}_{i.e., –}_x_x_{= cos}_{= cos}_y_y_{so that}_{so that}_x_x _{= – cos}_{= – cos}_y_y _{= cos (}_{= cos (}_π_π _–_–_y_y₎₎

T

Thheerreeffoorree ccooss–1–1_x_x ₌₌ _π_π _–_–_y_y ₌₌ _π_π _{– cos}_{– cos}–1–1 _(–_(–_x_x₎₎

H

Heennccee ccooss–1–1 _(–_(–_x_x₎₎₌₌ _π_π _{– cos}_{– cos}–1–1_x_x

Similarly

4. 4. ((ii)) ssiinn–1–1_x_x _{+ cos}_{+ cos}–1–1_x_x ₌₌ 2 2 ,, x x ∈∈∈∈∈∈∈∈∈∈ [– 1, 1][– 1, 1] ( (iiii)) ttaann–1–1_x_x _{+ cot}_{+ cot}–1–1_x_x ₌₌ 2 2 ,, x x ∈∈∈∈∈∈∈∈∈∈ RR (i (iiiii)) cocosesecc–1–1_x_x _{+ sec}_{+ sec}–1–1_x_x ₌₌ 2 2 , , || x x|| ≥≥≥≥≥≥≥≥≥≥ 11 Let sin

Let sin–1–1_x_x₌₌_y_y_{. Then}_{. Then}_x_x_{= sin}_{= sin} _y_y _{= cos}_{= cos}

2 2 y y

ππ

⎛

₋₋

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

T Thheerreeffoorree ccooss–1–1_x_x ₌₌ 2 2 y y

ππ

−−

₌₌ _sin_sin–1–1 2 2 x x

ππ

−−

(12)

4 444 MMAATTHHEEMMAATTIICCSS H Heennccee ssiinn–1–1_x_x _{+ cos}_{+ cos}–1–1_x_x ₌₌ 2 2

ππ

Similarly, we can prove the other parts.

5.

5. ((ii)) ttaann–1–1_x_x _{+ tan}_{+ tan}–1–1_y_y _{= tan}_{= tan}–1–1

– – + + 1 1 x x yy y

y ,, xy xy < < 11

(

(iiii)) ttaann–1–1_x_x _{– tan}_{– tan}–1–1_y_y _{= tan}_{= tan}–1–1

+ + – – 1 1 x x yy y

y ,, xy xy > > – – 11

( (iiiiii)) 22ttaann–1–1_x_x _{= tan}_{= tan}–1–1 22 2 2 1 1 –– x x x x , , || x x| | < < 11 Let tan

Let tan–1–1_x_x ₌₌ _θ_θ _{and tan}_{and tan}–1–1_y_y₌₌ _φ_φ_{. Then}_{. Then}_x_x_{= tan}_{= tan} _θ_θ_,_,_y_y _{= tan}_{= tan}_φ_φ

Now Now ttaann ttaann ttaann(( )) 1 1 ttaann ttaann 11 x x yy xy xy

θθ+ φ

+ φ

++

θθ++φ

φ =

=

==

− θ φ −

This gives

This gives θθ ++ φφ = tan= tan–1–1

1 1 x x yy xy xy

++

−−

H

Heennccee ttaann–1–1_x_x _{+ tan}_{+ tan}–1–1_y_y _{= tan}_{= tan}–1–1

1 1 x x yy xy xy

++

−−

In the above result, if we replace

In the above result, if we replace y yby by –– y y, we get the second result and by replacing, we get the second result and by replacing

y

ybyby x x, we get the third result., we get the third result.

6. 6. ((ii)) 22ttaann–1–1_x_x _{= sin}_{= sin}–1–1 2 2 2 2 1 1 ++ x x ,, || x x|| ≤≤≤≤≤≤≤≤≤≤ 11 ( (iiii)) 22ttaann–1–1_x_x _{= cos}_{= cos}–1–1 2 2 2 2 1 1 –– 1 1 ++ x x , , x x ≥≥≥≥≥≥≥≥≥≥ 00 ( (iiiiii)) 2 t2 taann–1–1_x_x _{= tan}_{= tan}–1–1 2 2 2 2 1 1 –– x x x x ,, – 1 – 1 << x x < 1< 1 Let tan

Let tan–1–1_x_x₌₌_y_y_{, then}_{, then}_x_x _{= tan}_{= tan}_y_y_{. Now}_{. Now}

sin sin–1–1 2 2 2 2 1 1 x x x x

++

= sin= sin–1–1 22 2tan 2tan 1 1 ttaann y y y y

++

=

(13)

Also cos Also cos–1–1 2 2 2 2 1 1 1 1 x x x x

−−

++

= cos= cos–1–1 2 2 2 2 1 1 ttaann 1 1 ttaann y y y y

−−

++

= cos= cos–1–1 (cos 2(cos 2 y) = y) =22 y y = 2tan= 2tan–1–1 xx

(i

(iiiii)) Can be worked out similarlyCan be worked out similarly..

W

We now e now consider some examples.consider some examples.

Example 3

Example 3 Show thatShow that

( (ii)) ssiinn–1–1

( (

₂₂

))

2 2 x x 11

−−

xx = 2 sin= 2 sin–1–1 x x,, 1 1 11 2 2 x x 22

−

≤

≤ ≤≤

( (iiii)) ssiinn–1–1

( (

₂₂

))

2 2 x x 11

−−

xx = 2 cos= 2 cos–1–1 x x,, 1 1 1 1 2 2

≤

≤ ≤≤

x x Solution Solution (

(ii)) LLeett x x = sin= sin θθ. Then sin. Then sin–1–1_x_x ₌₌ _θ_θ_{. We have}_{. We have}

sin

sin–1–1

( (

₂₂

))

2

2 x x 11

−−

xx ==ssiinn–1–1

( (

22 ssiinn

θ

θ −

11 ssiin

−

n22

θθ

))

=

=ssiinn–1–1_(2sin_(2sin_θ_θ_cos_cos_θ_θ_{) = sin}_{) = sin}–1–1_(sin2_(sin2_θ_θ_{) = 2}_{) = 2}_θ_θ

=

=2 s2 sinin–1–1_x_x

(

(iiii)) TTaakkee x x= cos= cosθθ, then proceeding as above, we get, sin, then proceeding as above, we get, sin–1–1

( (

₂₂

))

2

2 x x 11

−−

xx = 2 cos= 2 cos–1–1 x x Example 4

Example 4 SShhoow tw thhaatt ttaann–1–1 11 _ttaan_n–1–1 22 _ttaan_n––1133

2

+

1111

==

44 Solution

SolutionBy property 5 (i), we haveBy property 5 (i), we have

L.H.S. = L.H.S. = ttaann––1111 ttaann––11 22 2 2

++

1111 –1 –1 11 1 1 22 15 15 2 2 1111 ttaann ttaann 1 1 22 ₂₀₂₀ 1 1 2 2 1111 −−

++

=

==

−

− ××

= = tantan 1133 4 4 −− = R.H.S. = R.H.S. Example 5

Example 5 ExpressExpress

tan

11

cos

1 1 ssiin

n

x x x x −−

⎛

⎞⎞

⎜

⎟⎟

⎝

−−

⎠⎠

,, 22 x x 22

π

ππ

−

− <

< <<

in the simplest form.in the simplest form.

Solution

Solution We writeWe write

2 2 22 1 1 –1–1 2 2 22

ccooss ssiinn cos

cos ₂₂ ₂₂

ttaann ttaann

1

1 ssiinn _cco_oss _ssiin_n ₂_2ssiin_{n cco}_oss 2 2 22 22 22 x x xx x x xx xx xx xx x x −−

⎡

₋₋

⎤⎤

⎢

⎥⎥

⎛

_{⎞⎞ ==}

⎢

⎥⎥

⎜

₋₋

⎟⎟

⎝

⎠⎠

_⎢

₊

₋₋

_⎥⎥

⎣

⎦⎦

(14)

4 466 MMAATTHHEEMMAATTIICCSS = = –1–1 2 2 ccooss ssiinn ccooss ssiinn

2

2 22 22 22

tan tan

ccooss ssiinn 2 2 22 xx xx xx xx x x xx

⎡

⎛

₊

⎞⎞⎛

⎛

₋₋

⎞⎞

⎤⎤

⎜

⎟⎟⎜

⎜

⎟⎟

⎢

_⎝

_⎠⎠⎝

_⎝

_⎠⎠

⎥⎥

⎢

⎥⎥

⎛

⎞⎞

⎢

₋₋

⎥⎥

⎜

⎟⎟

⎢

_⎝

_⎠⎠

⎥⎥

⎣

⎦⎦

= = –1–1

ccooss ssiinn 2

2 22

tan tan

ccooss ssiinn 2 2 22 x x xx x x xx

⎡

₊₊

⎤⎤

⎢

⎥⎥

⎢

⎥⎥

⎢

−−

⎥⎥

⎣

⎦⎦

–1 –1 1 1 ttaann 2 2 tan tan 1 1 ttaann 2 2 x x x x

⎡

₊₊

⎤⎤

⎢

⎥⎥

==

_⎢

_⎥⎥

⎢

−−

⎥⎥

⎣

⎦⎦

= = ttaann–1–1 ttaann 4 4 22 44 22 x x xx

⎡

⎛

π

₊

⎞⎞

⎤⎤

₌

_{= ++}

ππ

⎜

⎟⎟

⎢

⎥⎥

⎝

⎠⎠

⎣

⎦⎦

Alternatively, Alternatively, – –11 ––11 ––11 2 2 ssiinn ssiinn cos cos 22 22

ttaann ttaann ttaann

2 2 1

1 ssiinn

1

1 ccooss 11 ccooss

2 2 22 x x x x x x x x x x x x

⎡

⎛

π

₋₋

⎞

⎤

⎡

⎛

ππ −−

⎞⎞

⎤⎤

⎜

⎟

⎜

⎟⎟

⎢

⎥

⎢

⎥⎥

⎛

⎞⎞

₌

_⎢

⎝

⎠

_⎥

₌₌

_⎢

⎝

⎠⎠

_⎥⎥

⎜

₋₋

⎟⎟

_⎛

_π

_⎞

_⎛

_{ππ −−}

_⎞⎞

⎢

⎥

⎢

⎥⎥

⎝

⎠⎠

₋

₋₋

⎜

⎟

⎜

⎟⎟

⎢

_⎣

_⎝

_⎠

⎥

⎢

_⎝

_⎠⎠

⎥⎥

⎣

⎦

⎣

⎦⎦

= = –1–1 2 2 2 2 22 2

2ssiinn ccooss 4 4 44 tan tan 2 2 2sin 2sin 4 4 x x xx x x

⎡

⎛

ππ −

−

⎞

⎞ ⎛

⎛

ππ −−

⎞⎞

⎤⎤

⎜

⎟

⎟ ⎜

⎜

⎟⎟

⎢

_⎝

_⎠

_{⎠ ⎝}

_⎝

_⎠⎠

⎥⎥

⎢

⎥⎥

ππ −−

⎛

⎞⎞

⎢

⎥⎥

⎜

⎟⎟

⎢

_⎝

_⎠⎠

⎥⎥

⎣

⎦⎦

=

= ttaann–1–1 ccoott 22

4 4 x x

⎡

⎛

ππ −−

⎞⎞

⎤⎤

⎜

⎟⎟

⎢

_⎝

_⎠⎠

⎥⎥

⎣

⎦⎦

–1 –1 22 ttaann ttaann 2 2 44 x x

⎡

⎛

π

π π −

π −

⎞⎞

⎤⎤

=

_⎢

_⎜

−−

_⎟⎟

_⎥⎥

⎝

⎠⎠

⎣

⎦⎦

= = ttaann–1–1 ttaann 4 4 22 x x

⎡

⎛

ππ

₊₊

⎞⎞

⎤⎤

⎜

⎟⎟

⎢

_⎝

_⎠⎠

⎥⎥

⎣

⎦⎦

44 22 x x

ππ

=

= ++

Example 6

Example 6WriteWrite –1–1

2 2 1 1 cot cot 1 1 x x

⎛

⎞⎞

⎜

⎟⎟

−−

⎝

⎠⎠

, |, | x x| > 1 in the | > 1 in the simplest form.simplest form.

Solution

Solution LetLet x x = sec= sec θθ, then, then _x_x22

₋₋

₁₁₌₌ 22

(15)

Therefore, Therefore, –1–1 2 2 1 1 cot cot 1 1 x x

−−

= cot

= cot–1–1_(cot_(cot_θ_θ_{) =}_{) =}_θ_θ_{= sec}_{= sec}–1–1_x,_x,_{which is the simplest form.}_{which is the simplest form.}

Example 7

Example 7 Prove that tanProve that tan–1–1_x_x ₊₊ –1–1

2 2 2 2 tan tan 1 1 x x x x

−−

= tan= tan–1–1 3 3 2 2 3 3 1 1 33 x x xx x x

⎛

−−

⎞⎞

⎜

⎟⎟

−−

⎝

⎠⎠

,, 1 1 || || 3 3 x x

<<

Solution

Solution LetLet x x= tan= tan θθ. Then. Then θθ = tan= tan–1–1_x_x_{. We have}_{. We have}

R.H.S. = R.H.S. = 3 3 33 – –11 ––11 2 2 22 3 3 33ttaann ttaann ttaann ttaann 1 1 33 11 33ttaann x x xx x x

⎛

−

⎞

⎛

θθ−

−

θθ

⎞⎞

==

⎜

⎟

⎜

⎟⎟

−

θθ

⎝

⎠

⎝

⎠⎠

=

=ttaann–1–1 _(tan3_(tan3_θ_θ_{) =}_{) =}₃₃_θ_θ _{= 3tan}_{= 3tan}–1–1_x_x_{= tan}_{= tan}–1–1 _x_x _{+ 2 tan}_{+ 2 tan}–1–1 _x_x

= =ttaann–1–1_x_x _{+ tan}_{+ tan}–1–1 2 2 2 2 1 1 x x x x

−−

= L.H.S. (Why?)= L.H.S. (Why?) Example 8

Example 8 Find the Find the value of value of cos (scos (secec–1–1_x_x_{+ cosec}_{+ cosec}–1–1_x_x_{), |}_{), |}_x_x_|_| _≥_≥₁₁

Solution

Solution We have cos (secWe have cos (sec–1–1_x_x _{+ cosec}_{+ cosec}–1–1_x_x_{) = cos}_{) = cos}

2 2

ππ

⎛

⎛ ⎞⎞

⎜

⎜ ⎟⎟

⎝

⎝ ⎠⎠

= = 00

EXERCISE 2.2

Prove the following:

1.

1. 3sin3sin–1–1_x_x _{= sin}_{= sin}–1–1 ₍₃₍₃_x_x _{– 4}_{– 4}_x_x33_),_), _–_{– ,,}11 11

2 2 22 x x

∈

∈ ⎢ ⎢

⎡

⎤⎤

_⎥⎥

⎣

⎦⎦

2.

2. 3cos3cos–1–1_x_x _{= cos}_{= cos}–1–1 ₍₄₍₄_x_x33_{– 3}_{– 3}_x_x_),_), 11_{,, 1}₁

2 2 x x

∈

∈ ⎢ ⎢ ⎥⎥

⎡

⎤⎤

⎣

⎦⎦

3.

3. tantan–1–1 22 _ttaan_n 11 77 _ttaan_n 11 11

1 111 2244 22 − − −−

+

==

4.

2 2 ttaan

n

11

1

1 ttaan

n

11

1

1 ttaan

n

11

31

3

1

2

7

1

17

7

−

₊

− −

₌₌

−−

Write the following functions in the simplest form:

5. 5. 2 2 1 1 11 11 tan tan x x x x −−

+

−−

_,_,_x_x ≠ ≠ 00 6.6. 11 ₂₂ 1 1 tan tan 1 1 x x −−

−−

, |, | x x| > 1| > 1 7.

7. tantan 11 11 ccooss

1 1 ccooss x x x x −−

⎛

−−

⎞⎞

⎜

⎟⎟

⎜

₊₊

⎟⎟

⎝

⎠⎠

,, x x<<ππ 8.8. 1

1 ccooss ssiinn

tan tan

ccooss ssiinn

x x xx x x xx −−

⎛

−−

⎞⎞

⎜

₊₊

⎟⎟

⎝

⎠⎠

,, x x<< ππ

(16)

4 488 MMAATTHHEEMMAATTIICCSS 9. 9. 11 2 2 22 tan tan x x a a xx −−

−−

, |, | x x| <| < aa 10 10.. 2 2 33 1 1 3 3 22 3 3 tan tan 3 3 a a xx xx a a axax −−

⎛

−−

⎞⎞

⎜

⎟⎟

−−

⎝

⎠⎠

,,aa> 0;> 0; 33 33 a a aa x x

−−

≤

≤ ≤≤

Find the values of each of the following:

11.

ttaan

n

––11

2 2 cco

oss 2

2ssiin

n

––11

1

2

2 ⎡

⎡

⎛

⎞⎞

⎤⎤

⎜

⎟⎟

⎢

_⎝

_⎠⎠

⎥⎥

⎣

⎦⎦

1212.. cot (tancot (tan

–1 –1_a_a _{+ cot}_{+ cot}–1–1_a_a₎₎ 13 13.. 2 2 – –11 ––11 2 2 22 1 1 22 11

ttaann ssiinn ccooss 2 2 11 11 x x yy x x yy

⎡

−−

⎤⎤

++

⎢

⎥⎥

+

++

⎣

⎦⎦

, |, | x x| < 1,| < 1, y y> 0 and> 0 and xy xy< 1< 1

14

14.. If sinIf sin ssiinn––1111 ccooss–1–1 11

5 5 x x

⎛

₊

⎞⎞

₌₌

⎜

⎟⎟

⎝

⎠⎠

, then find the value of , then find the value of x x

15

15.. If If ttaann––11 11 ttaann––11 11

2 2 22 44 x x xx x x xx

−

+

ππ

+

==

−

++

, then find the value of , then find the value of x x

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

16 16.. ssiinn–1–1 ssiinn22 3 3

ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

1717.. –1 –1 33 ttaann ttaann 4 4

ππ

⎛

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

18

18.. ttaann ssiinn––1133 ccoott–1–1 33 5 5 22

⎛

₊₊

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

19

19.. ccooss 11 ccooss77 iiss eeqquuaal tl too

6 6 −−

⎛

ππ

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

(A) (A) 77 6 6

ππ

(B) (B) 55 6 6

ππ

(C) (C) 3 3

ππ

(D) (D) 6 6

ππ

20 20.. ssiinn ssiinn ((11 11)) 3 3 22 −−

ππ

⎛

₋

₋₋

⎞⎞

⎜

⎟⎟

⎝

⎠⎠

is equal tois equal to (A) (A) 11 2 2 (B)(B) 1 1 3 3 (C)(C) 1 1 4 4 ((DD)) 11 21

21.. ttaann− − 11 33

−

ccoott ((−−11

−−

3))3 is equal tois equal to

(A) (A) ππ (B)(B) 2 2

ππ

−−

((CC)) 00 ((DD)) ₂_{2 3}₃

(17)

Miscellaneous Examples

Example 9

Example 9 Find the value of Find the value of ssiinn ((ssiin11 n33 ))

5 5

−−

ππ

Solution

Solution We know thatWe know that ssiinn ((ssiin−−11 n )) x x

==

xx. Therefore,. Therefore, ssiinn ((ssiin11 n33 )) 33

5 5 55 −−

π

₌₌

ππ

But But 33 ,, 5 5 22 22

π

_∉

_{∉ −−}

⎡

π

π ππ

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

, which is the principal branch of sin, which is the principal branch of sin–1–1 x x

However

However ssiinn ((33 )) ssiinn(( 33 )) ssiinn22 5 5 55 55

π

ππ

=

= ππ −

− ==

andand 22 ,, 5 5 22 22

π

_∈

_{∈ −−}

⎡

π

π ππ

⎤⎤

⎢

⎥⎥

⎣

⎦⎦

Therefore

Therefore ssiinn ((ssiin11 n33 )) ssiinn ((ssiin11 n22 )) 22

5

5 55 55

−

π

₌

−−

π

₌₌

ππ

Example 10

Example 10 Show thatShow that ssiinn 1133 ssiinn 11 88 ccooss 118484

5

5 1177 8855

−

₋

− −

₌₌

−−

Solution

Solution LetLet sinsin 1133 5 5 x x −−

₌₌

_and_and 11 88 sin sin 17 17 y y −−

₌₌

Therefore

Therefore sinsin 33

5 5

x

==

andand sinsin 88 17 17

y y

==

Now

Now ccooss 1 s1 siinn22 11 99 44 25 25 55

x

=

= −

−

xx

=

= −

− ==

(Why?)(Why?)

and

and ccooss 1 s1 siinn22 11 6644 1155 2

28899 1177 y

y

=

= −

−

yy

=

= −

− ==

W

We e hhaavvee ccoos s (( x x–– y y) = cos) = cos x x coscos y y+ sin+ sin x x sinsin y y

=

= 44 1155 33 88 8844

5

×

× +

1177

+ ×

55

× ==

1177 8855

Therefore

Therefore coscos 11 8484

85 85 x x

−

− ==

yy −−

⎛

⎛ ⎞⎞

_⎜

_{⎜ ⎟⎟}

⎝

⎝ ⎠⎠

Hence

Hence _ssiin_n 1133 _ssiin_n 11 88 _cco_oss 118484

5

5 1177 8855

−