Inverse Trigonometric Functions

Teachers: Anonymous
Source: Internet

In class 11 maths, students have already learnt trigonometric function. The inverse trigonometric function is the inverse of a trigonometric function.

In this chapter of maths, students will learn concepts related to the definition of inverse trigonometric functions, domain, range, and principal value, graph, and properties of inverse trigonometric functions.

Students have already learnt that trigonometric functions are not one to one and onto in their natural ranges. All these questions & solutions are explained in step-by-step method for class 12 students. These are important while practice the problems and in depth understanding of this section.

ক্লাস 11 গণিতে, শিক্ষার্থীরা ইতিমধ্যে ত্রিকোণমিতিক অপেক্ষকসমূহ শিখেছে। বিপরীত ত্রিকোণমিতিক অপেক্ষকসমূহ হল একটি ত্রিকোণমিতিক অপেক্ষকসমূহের বিপরীত।

গণিতের এই অধ্যায়ে, শিক্ষার্থীরা বিপরীত ত্রিকোণমিতিক অপেক্ষকসমূহ সংজ্ঞা সম্পর্কিত ধারণাগুলি শিখবে। শিক্ষার্থীরা ইতিমধ্যেই শিখেছে যে ত্রিকোণমিতিক অপেক্ষকসমূহগুলি তাদের প্রাকৃতিক সীমার মধ্যে এক থেকে এক নয়।

এই সমস্ত প্রশ্ন এবং সমাধান 12 শ্রেনীর শিক্ষার্থীদের জন্য ধাপে ধাপে পদ্ধতিতে ব্যাখ্যা করা হয়েছে। এই বিভাগের সমস্যা এবং গভীরভাবে বোঝার অনুশীলন করার সময় এগুলি গুরুত্বপূর্ণI

Question:1

Find the principal values of

Answer:

Let $x = sin^{-1}left ( frac{-1}{2} right )$

$implies sin x = frac{-1}{2}= -sin(frac{pi}{6}) = sin(-frac{pi}{6})$
We know, principle value range of $sin^{-1}$ is $[-frac{pi}{2}, frac{pi}{2}]$

$therefore$ The principal value of $sin^{-1}left ( frac{-1}{2} right )$ is $-frac{pi}{6},$

Question:2

Find the principal values of

Answer:
So, let us assume that $cos^{-1}left(frac{sqrt3}{2} right ) = x$ then,

Taking inverse both sides we get
$cos x = (frac{sqrt{3}}{2})$ , or $cos (frac{pi}{6}) = (frac{sqrt{3}}{2})$
and as we know that the principal values of $cos^{-1}$ is from [0, $pi$ ],
Hence $cos x = (frac{sqrt{3}}{2})$ when x = $frac{pi}{6}$ .
Therefore, the principal value for $cos^{-1}left(frac{sqrt3}{2} right )$ is $frac{pi}{6}$ .

Question:3

Find the principal values of

Answer:
Let us assume that $textup{cosec}^{-1}(2) = x$ ,
then we have; $Cosec x = 2$ , or $Cosec( frac{pi}{6}) = 2$ .
And we know the range of principal values is $[frac{-pi}{2},frac{pi}{2}] - left { 0 right }.$
Therefore the principal value of $textup{cosec}^{-1}(2)$ is $frac{pi}{6}$ .

Question:4

Find the principal values of

Answer:
Let us assume that $tan^{-1}(-sqrt3) = x$ ,
then we have $tan x = (-sqrt 3)$ or

-tan (frac{pi}{3}) = tan left ( frac{-pi}{3} right ).

and as we know that the principal value of $tan^{-1}$ is $left ( frac{-pi}{2}, frac{pi}{2} right )$ .
Hence the only principal value of $tan^{-1}(-sqrt3)$ when $x = frac{-pi}{3}$ .

Question:5

Find the principal values of

Answer:
Let us assume that $cos^{-1}left(-frac{1}{2} right ) =y$ then,
Easily we have $cos y = left ( frac{-1}{2} right )$ or
we can write it as

-cos left ( frac{pi}{3} right ) = cos left ( pi - frac{pi}{3} right ) = cos left ( frac{2pi}{3} right ).

as we know that the range of the principal values of $cos^{-1}$ is $left [ 0,pi right ]$ .
Hence $frac{2pi}{3}$ lies in the range it is a principal solution.

Question:6

Find the principal values of

Answer:
Given $tan^{-1}(-1)$ so we can assume it to be equal to ‘z’ $tan^{-1}(-1) =z$ ,

-tan (frac{pi}{4}) = tan(frac{-pi}{4})= -1

And as we know the range of principal values of $tan^{-1}$ from $left ( frac{-pi}{2}, frac{pi}{2} right )$ .
As only one value z = $-frac{pi}{4}$ lies hence we have only one principal value that is $-frac{pi}{4}$ .

Question:7

Find the principal values of

Answer:
Let us assume that $sec^{-1}left (frac{2}{sqrt3}right) = z$ then,
we can also write it as; $sec z = left (frac{2}{sqrt3}right)$ .Or $sec (frac{pi}{6}) = left (frac{2}{sqrt3}right)$
and the principal values lies between $left [ 0, pi right ] - left { frac{pi}{2} right }$ .
Hence we get only one principal value of $sec^{-1}left (frac{2}{sqrt3}right)$ i.e., $frac{pi}{6}$ .

Question:8

Find the principal values of

Answer:
Let us assume that $cot^{-1}(sqrt3) = x$ ,
then we can write in other way $cot x = (sqrt3)$ or $cot (frac{pi}{6}) = (sqrt3)$ .
Hence when $x=frac{pi}{6}$ we have $cot (frac{pi}{6}) = (sqrt3)$
and the range of principal values of $cot^{-1}$ lies in $left ( 0, pi right )$ .
Then the principal value of $cot^{-1}(sqrt3)$ is $frac{pi}{6}$

Question:9

Find the principal values of

Answer:
Let us assume $cos^{-1}left(-frac{1}{sqrt2} right ) = x$ ;
Then we have $cos x = left ( frac{-1}{sqrt 2} right )$ or $-cos (frac{pi}{4}) = left ( frac{-1}{sqrt 2} right )$ , $cos (pi - frac{pi}{4}) = cos (frac{3pi}{4})$ .
And we know the range of principal values of $cos^{-1}$ is $[0,pi]$ .
So, the only principal value which satisfies $cos^{-1}left(-frac{1}{sqrt2} right ) = x$ is $frac{3pi}{4}$ .

Question:10

Find the principal values of

Answer:
Let us assume the value of $textup{cosec}^{-1}(-sqrt2) = y$ ,
then we have $cosec y = (-sqrt 2)$ or $-cosec (frac{pi}{4}) = (-sqrt 2) = cosec (frac{-pi}{4})$
and the range of the principal values of $textup{cosec}^{-1}$ lies between $left [ frac{-pi}{2},frac{pi}{2} right ] - left { 0 right }$
hence the principal value of $textup{cosec}^{-1}(-sqrt2)$ is $frac{-pi}{4}$ .

Question:11

Find the principal values of

tan^{-1}(1) + cos^{-1}left(-frac{1}{2} right ) + sin^{-1}left(-frac{1}{2} right )

Answer:
To find the values first we declare each term to some constant $tan^{-1}(1) = x$ ,
So we have $tan x = 1$ or $tan (frac{pi}{4}) = 1$
Therefore $x = frac{pi}{4}$ &

So, we have $cos y = left ( frac{-1}{2} right ) = -cos left ( frac{pi}{3} right ) = cos(pi - frac{pi}{3}) = cos left ( frac{2pi}{3} right )$ .
Therefore $y = frac{2pi}{3}$ , $sin^{-1}(frac{-1}{2}) = z$ ,
So we have $sin z = frac{-1}{2}$ or $-sin (frac{pi}{6}) =sin (frac{-pi}{6}) = frac{-1}{2}$
Therefore $z = -frac{pi}{6}$
Hence we can calculate the sum

$=frac{3pi + 8pi -2pi}{12} = frac{9pi}{12}=frac{3pi}{4}$

Question:12

Find the principal values of

Answer:
Here we have $cos^{-1}left(frac{1}{2} right ) + 2sin^{-1}left(frac{1}{2} right )$
let us assume that the value of $cos^{-1}left ( frac{1}{2} right ) = x, :and:sin^{-1}left(frac{1}{2} right ) = y$
then we have to find out the value of x +2y.
Calculation of x

Rightarrow cos^{-1}left ( frac{1}{2} right ) = x

$Rightarrow cos frac{pi}{3} = frac{1}{2}$
Hence $x = frac{pi}{3}$ .
Calculation of y

Rightarrow sin^{-1}left(frac{1}{2} right ) = y

$Rightarrow sin frac{pi}{6} = frac{1}{2}$ .
Hence $y = frac{pi}{6}$ .
The required sum will be = $frac{pi}{3}+2(frac{pi}{6}) = frac{2pi}{3}$

Question:13

Prove the following

3sin^{-1}x = sin^{-1}(3x - 4x^3),;;xinleft[-frac{1}{2},frac{1}{2} right ]

Answer:
Given to prove $3sin^{-1}x = sin^{-1}(3x - 4x^3)$ where $x:epsilon left[-frac{1}{2},frac{1}{2} right ]$ .
Take $theta= sin ^{-1}x$ or $x = sin theta$
Take R.H.S value

= $sin^{-1}(3sin theta - 4sin^3 theta)$ = $sin^{-1}(sin 3theta)$ = $3theta$ = $3sin^{-1}x$ = L.H.S

Question:14

Prove that $3cos^{-1} x = cos^{-1}(4x^3 - 3x), ;;xinleft[frac{1}{2},1 right ]$

Answer:
Given to prove $3cos^{-1} x = cos^{-1}(4x^3 - 3x), ;;xinleft[frac{1}{2},1 right ]$ .
Take $cos^{-1}x = theta$ or $cos theta = x$ ;
Then we have R.H.S.

= $cos^{-1}(4cos^3 theta - 3costheta)$ $left [ because 4cos^3 theta - 3costheta = cos3 theta right ]$ = $cos^{-1}(cos3theta)$ = $3theta$ = $3cos^{-1}x$ = L.H.S
Hence Proved.

Question:15

Prove that $tan^{-1}frac{2}{11} + tan^{-1}frac{7}{24} = tan^{-1}frac{1}{2}$

Answer:
Given to prove $tan^{-1}frac{2}{11} + tan^{-1}frac{7}{24} = tan^{-1}frac{1}{2}$
We have L.H.S

=tan^{-1}frac{frac{2}{11} + frac{7}{24} }{1 - left ( frac{2}{11}timesfrac{7}{24} right ) }

left [ because tan^{-1}x + tan^{-1}y = tan^{-1} frac{x +y}{1 - xy} right ]

=tan^{-1}frac{11times 24 }{frac{11times24 -14}{11times 24} }

=tan^{-1}left ( frac{125}{250}right ) = tan^{-1}left ( frac{1}{2} right )

= R.H.S
Hence proved.

Question:16

Prove that $2tan^{-1} frac{1}{2} + tan^{-1}frac{1}{7} = tan^{-1}frac{31}{17}$

Answer:
Given to prove $2tan^{-1} frac{1}{2} + tan^{-1}frac{1}{7} = tan^{-1}frac{31}{17}$
Then taking L.H.S.
We have $2tan^{-1} frac{1}{2} + tan^{-1}frac{1}{7}$

=tan^{-1} frac{2.frac{1}{2}}{1 - left ( frac{1}{2} right )^2} + tan^{-1} frac{1}{7}

because 2tan^{-1} x = tan^{-1} frac{2x}{1- x^2}

=tan^{-1} frac{1}{(frac{3}{4})} + tan^{-1} frac{1}{7}

=tan^{-1} frac{4}{3} + tan^{-1} frac{1}{7}

=tan^{-1} frac{frac{4}{3} + frac{1}{7}}{1 - frac{4}{3}.frac{1}{7}}

=tan^{-1} left ( frac{frac{28+3}{21}}{frac{21-4}{21}} right )

= R.H.S.
Hence proved.

Question:17

Write the following functions in the simplest form

tan^{-1}frac{sqrt{1 + x^2}- 1}{x},;;xneq 0

Answer:
We have $tan^{-1}frac{sqrt{1 + x^2}- 1}{x}$
Take

tan^{-1} frac {sqrt{1+x^2} - 1}{x} = tan^{-1}frac{sqrt{1+tan^2 Theta - 1}}{tan Theta}

=tan^{-1}(frac{sec Theta-1}{tan Theta}) = tan^{-1}left ( frac{1-cos Theta}{sin Theta} right )

=tan^{-1}left ( frac {2sin^2left ( frac{Theta}{2} right )}{2sinfrac{Theta}{2}cosfrac{Theta}{2}} right )

=tan^{-1}left ( tanfrac{Theta}{2} right ) = frac{Theta}{2} =frac{1}{2}tan^{-1}x

$=frac{1}{2}tan^{-1}x$ is the simplified form.

Question:18

Write the simplest form of function $tan^{-1} frac{1}{sqrt{x^2 -1}},;; |x| > 1″></p> <p>Answer:<br>Given that <img decoding=$ or $Theta = cosec ^{-1}x$

=tan^{-1} frac{1}{sqrt{cosec^2 Theta -1}}

$=tan^{-1}(tan Theta)$ = $Theta$ = $cosec^{-1}x$

[because cosec^{-1}x + sec^{-1}x = frac{pi}{2}]

Question:19

Write the following functions in the simplest form

Answer:
Given that $tan^{-1}left(sqrt{frac{1-cos x}{1 + cos x}} right ),;; 0< x < pi$
We have in inside the root the term : $frac{1-cos x}{1 + cos x}$
Put $1-cos x = 2sin^2frac{x}{2}$ and $1+cos x = 2cos^2frac{x}{2}$ ,
Then we have,