RD Sharma Class 12 Exercise 3.1 Inverse Trigonometric Functions Solutions Maths

RD Sharma Class 12 Exercise 3.1 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 11:30 AM IST

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RD Sharma Class 12 Solutions Inverse Trigonometric Functions Exercise 3.1 has 14 questions, including subparts of Level 1 difficulty. However, these questions are fundamental concept-based and can be completed relatively quickly if you have gone through the theorems.

Inverse Trigonometric Functions Excercise: 3.1

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Function Exercise 3.1 Question 1 (i)

Answer:
$\frac{-\pi }{3}$
Hint:
$f\! \left ( x \right )=y$
$f\! ^{-1}\left ( y \right )=x$
Given:
$Find\: principal\: value\: o\! f \sin ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )$
Solution:
$Let,\sin ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )= y$
$\Rightarrow \sin y= \frac{-\sqrt{3}}{2}$
$\Rightarrow \sin y= -\sin \left ( \frac{\pi }{3} \right )= \sin \left ( \frac{-\pi }{3} \right )$
$Range\: o\! f\: principal \: value\: o\! f\: \sin ^{-1} is\left ( \frac{-\pi }{2},\frac{\pi }{2} \right )\: and \: \sin \left ( \frac{-\pi }{3} \right )= \frac{-\sqrt{3}}{2}$
$There\! f\! ore, principal \: value\: o\! f\: \sin ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )is\: \frac{-\pi }{3}$

Inverse Trigonometric Function Exercise 3.1 Question 1 (ii)

Answer:
$\frac{-\pi }{6}$
Hint:
$\cos \left ( 90^{\circ}+x \right )= -\sin x$
Given:
$Find\: principal\: value\: o\! f \sin^{-1}\left ( \cos \frac{2\pi }{3} \right )$
Solution:
$Let, \sin^{-1}\left ( \cos \frac{2\pi }{3} \right )= y$
$\Rightarrow \sin y= \left ( \cos \frac{2\pi }{3} \right )$
$\Rightarrow \sin y=\cos \frac{2\pi }{3}= \cos \left ( \frac{\pi }{2}+\frac{\pi }{6} \right )= -\sin\left ( \frac{\pi }{6} \right )$
$W\! e\, know\, principal\, value\, o\! f \sin^{-1} is\, \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and -\sin \sin \left ( \frac{\pi }{6} \right )= \cos \cos \left ( \frac{2\pi }{3} \right )$
$Therefore, principal\, value\, o\! f \, \sin^{-1}\left [ \cos \left ( \frac{2\pi }{3} \right ) \right ] is\, \frac{-\pi }{6}.$

Inverse Trigonometric Function Exercise 3.1 Question 1 (iii)

Answer:
$\frac{\pi }{12}$
Hint:
$Separate\, and\,\, reorganize \,\, the\, values$
Given:
$Find \: principal\: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right )$
Solution:
$\sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right )$ $= \sin^{-1}\left ( \frac{\sqrt{3}}{2\sqrt{2}} -\frac{1}{2\sqrt{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} -\frac{1}{2}\times \frac{1}{\sqrt{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2}\times \sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}}-\frac{1}{\sqrt{2}}\times \sqrt{1-\left ( \frac{\sqrt{3}}{2} \right )^{2}} \right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right )-\sin^{-1}\left ( \frac{1}{\sqrt{2}} \right )$
$= \frac{\pi }{3}-\frac{\pi }{4}$
$=\frac{\pi }{12}$
$There\! fore, principal \: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right ) is\: \frac{\pi }{2}.$

Inverse Trigonometric Function Exercise 3.1 Question 1 (iv)

Answer:
$\frac{7\pi }{12}$
Hint:
$Separate\: and\: reorganize\: the\: values$
Given:
$Find\: principal\: value\: o\! f \sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right )$
Solution:
$\sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right )$ $= \sin^{-1}\left ( \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \times \sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}}+\frac{1}{\sqrt{2}}\times \sqrt{1-\left ( \frac{\sqrt{3}}{2} \right )^{2}}\right )$
$= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right )+\sin^{-1}\left ( \frac{1}{\sqrt{2}} \right )$
$= \frac{\pi }{3}+\frac{\pi }{4}$
$= \frac{7\pi }{2}$
$Therefore, principal\: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right ) is = \frac{7\pi }{2}.$

Inverse Trigonometric Function Exercise 3.1 Question 1 (v)

Answer:
$\frac{-\pi }{4}$
Hint:
$\cos x= - \cos \left ( 180^{\circ}-x \right )$
Given:
$Find\: principal\: value\: o\! f \sin^{-1}\left ( \cos \left ( \frac{3\pi }{4} \right ) \right )$
Solution:
$Let, \sin^{-1}\left ( \cos \left ( \frac{3\pi }{4} \right ) \right )= y$
$\Rightarrow\sin y = \cos \left ( \frac{3\pi }{4} \right )$
$\Rightarrow \sin y= \cos \frac{3\pi }{4}= \cos\left ( \pi -\frac{\pi }{4} \right )= - \cos\left ( \frac{\pi }{4} \right )$
$Principal\: value \: o\! f \cos^{-1}\: is\: \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]\: and\: -\cos \cos\left ( \frac{\pi }{4} \right ) = \cos \cos\left ( \frac{3\pi }{4} \right )$
$There\! f\! ore, principal\: value\: o\! f \sin^{-1}\left ( \cos \left ( \frac{3\pi }{4} \right ) \right ) is \frac{-\pi }{4}.$

Inverse Trigonometric Function Exercise 3.1 Question 1 (vi)

Answer:

Hint:

Given:
$Find\: principal\: value\: o\! f \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right )$
Solution:
$Let , y= \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right )$

$=1$
$= \sin \left ( \frac{\pi }{2} \right )$
$Range\: Principal\: value\: o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \: \sin \left ( \frac{\pi }{2} \right )= \tan \left ( \frac{5\pi }{4} \right )$
$There\! f\! ore, principal\: value\: o\! f \sin^{-1}\left ( \tan \left ( \frac{5\pi }{4} \right ) \right ) is \: \frac{\pi }{2}.$

Inverse Trigonometric Function Exercise 3.1 Question 2 (i)

Answer:
$\frac{-\pi }{3}$
Hint:
$2\sin^{-1}x= \sin^{-1}\left ( 2x\sqrt{1-x^{2}} \right )$
Given:
$Find\, principal \, value\, o\! f \sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}}$
Solution:
$\sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}}$
$= \sin^{-1}\frac{1}{2}-\sin^{-1}\left ( 2\times \frac{1}{\sqrt{2}}\sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}} \right )$
$= \sin^{-1}\frac{1}{2}-\sin^{-1}1$
$= \frac{\pi }{6}-\frac{\pi }{2}$
$= \frac{-\pi }{3}$
$There\! fore, principal\, value\, o\! f \sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}} \, is = \frac{-\pi }{3}.$

Inverse Trigonometric Function Exercise 3.1 Question 2 (ii)

Answer:
$\frac{\pi }{6}$
Hint:
$\sin^{-1}x= y$
$\sin y= x$
Given:
$Find\: principal\, value \, o\! f \sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \}$
Solution:
$\sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \}$
$= \sin^{-1}\left [ \cos \left ( \frac{\pi }{3} \right ) \right ] \; \; \; \; \; \; \; \; \left [ \because \left ( \frac{\sqrt{3}}{2} \right ) = \frac{\pi }{3}\right ]$
$= \left ( \frac{1}{2} \right )\; \; \; \; \; \; \; \; \left [ \because \cos \frac{\pi }{3}= \frac{1}{2} \right ]$
$Let , \left ( \frac{1}{2} \right )= y$
$\sin \sin y =\frac{1}{2}$
$Range\, o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \, \sin \sin \left ( \frac{\pi }{6} \right ) =\frac{1}{2}$
$Therefore, principal \, value\, o\! f \, \sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \} is \, \frac{\pi }{6}.$

Inverse Trigonometric Function Exercise 3.1 Question 3 (i)

Answer:
$\left [ -1,1 \right ]$
Hint:
$The\: domain\: o\! f \sin^{-1}x \: is \left [ -1,1 \right ]$
Given:
$\sin^{-1}x^{2}$
Explanation:
$Let \: f\! \left ( x \right )= \left ( x^{2} \right )$
$\! \! \! \! \! \! \! \! \! \! \! \! The\, domain \, o\! f \sin^{-1} is \left [ -1,1 \right ] which\, implies\, that \, the\, value\, o\! f \, x\, lies\, between -1 and\, 1$
$i.e; -1\leq x\leq 1$
$but\: here\: we\: have\: the\: f\! unction \: x^{2}$
$so,-1\leq x^{2}\leq 1\; \; \; \; \; \; \; [As\, x^{2}\: is\: always\: positive]$
$\Rightarrow 0\leq x^{2}\leq 1$
$\Rightarrow\left | x \right | \leq 1$
$\Rightarrow x\leq 1\: and x\geq -1\: \: \: \: \: \: \: \: \: \: \: \left [ \because a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right ) \right ]$
$Hence, the\: domain\: o\! f \sin^{-1}x^{2} \: is \left [ -1,1 \right ].$

Inverse Trigonometric Function Exercise 3.1 Question 3 (ii)

Answer:
$\left [ -1,1 \right ]$
Hint:
$The\: domain \: o\! f \sin^{-1}x \: is \left [ -1,1 \right ] and\: the\: domain\: o\! f \sin x\: is \: R .$
Given:
$f\! \left ( x \right )= \sin^{-1}x+\sin x$
Explanation:
$W\! hen \, we\, f\! ind\, the\, domain\, o\! f \, the \, f\! unction\, which \, is \, f\! urther\, the\, sum \, o\! f \, two\, di\! f\! f\! erent \, f\! unctions\, then\, we\, take\, the\, intersection \, o\! f \, thei\, r respective\, domains.$ $\left [-1,1 \right ]\cap R$
$= \left [ -1,1 \right ]\cap \left ( -\infty ,\infty \right )$
$= \left [ -1,1 \right ]$
Hence, the domain of $\sin^{-1}x+\sin x$ is $\left [ -1,1 \right ]$ .

Inverse Trigonometric Function Exercise 3.1 Question 3 (iii)

Answer:
$\left [ -\sqrt{2},1 \right ]\cup \left [ 1,\sqrt{2} \right ]$
Hint:
$The\: domain \: o\! f \sin^{-1}x \: is \left [ -1,1 \right ].$
Given:
$f\! \left ( x \right )= \sin^{-1}\sqrt{x^{2}-1}$
Explanation:
$-1\leq \sqrt{x^{2}-1}\leq 1$
$x^{2}-1\leq 1 \: [Squaring\: both \: sides]$
$x^{2}-1+1\leq 1+1$
$x^{2}\leq 2$
$-\sqrt{2}\leq x\leq \sqrt{2} \cdot \cdot \cdot \cdot (i)$
$Also, x^{2}-1\geq 0$
$x^{2}\geq 1$
$\Rightarrow x\leq -1 \: and \: x\geq 1 \cdot \cdot \cdot \cdot (ii)$
$From (i) and (ii), the\: value\: o\! f \: x \: lies\: between$
$1\leq x\leq\sqrt{2} \: and \: -\sqrt{2}\leq x\leq -1$
$The \: domain \: o\! f \sin^{-1}\sqrt{x^{2}-1} \: is \left [ -\sqrt{2,}-1 \right ]\cup \left [ 1,\sqrt{2} \right ]$

Inverse Trigonometric Function Exercise 3.1 Question 3 (iv

Answer:
$\left [ \frac{-1}{2} ,\frac{1}2{}\right ]$
Hint:
$The \, domain\, o\! f \sin^{-1}x \, is \left [ -1,1 \right ].$
Given:
$f\! \left ( x \right )= \sin^{-1}x+\sin^{-1}2x$
Explanation:
$Now, we\, f\! ind\, the\, domain\, o\! f \sin^{-1}2x$
$-1\leq 2x\leq 1$
$\frac{-1}{2}\leq x\leq \frac{1}{2}$
$The\, domain\, o\! f \, f\! \left ( x \right ) is\, the\, inter\! section \, o\! f\, domains\, o\! f \sin^{-1}x\, and \, \sin^{-1}2x.$
$\left [ \frac{-1}2,\frac{1}2{}{} \right ]\cap \left [ -1,1 \right ]$
$= \left [ \frac{-1}2,\frac{1}2{}{} \right ]$

Inverse Trigonometric Function Exercise 3.1 Question 4

Answer:
$4$
Given:
$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z+\sin^{-1}t= 2\pi \: then\, f\! ind\: x^{2}+y^{2}+z^{2}+t^{2}$
Hint:
$We \: know \: that\: the\: range\: o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$
$\sin^{-1}x\leq \frac{\pi }{2},\sin^{-1}y\leq \frac{\pi }{2},\sin^{-1}z\leq \frac{\pi }{2},\sin^{-1}t\leq \frac{\pi }{2}$
Solution:
$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z+\sin^{-1}t\leq 2\pi \: and\: it \: is\: given\: that \: L.H.S \: is\: 2\pi.$
$So\: it\: is\: possible\: only\: when$
$\sin^{-1}x= \frac{\pi }{2},\sin^{-1}y= \frac{\pi }{2},\sin^{-1}z= \frac{\pi }{2},\sin^{-1}t= \frac{\pi }{2}$
$x= 1,y= 1,z= 1,t= 1$
$\therefore x^{2}+y^{2}+z^{2}+t^{2}= 4$

Inverse Trigonometric Function Exercise 3.1 Question 5

Answer:
$3$
Given:
$\left ( \sin^{-1}x \right )^{2}+\left (\sin^{-1}y \right )^{2}+\left ( \sin^{-1}z \right )^{2}= \frac{3}{4}\pi ^{2} \,\, then\, find \,\, x^{2}+y^{2}+z^{2}.$
Hint:
$We\, know\, that \, the\, range\, o\! f \, \sin^{-1} is \left [ \frac{-\pi }{2} ,\frac{\pi }{2}\right ]$
$\sin^{-1}x\leq \frac{\pi }{2},\sin^{-1}y \leq \frac{\pi}{2}, \sin^{-1}z\leq \frac{\pi}{2}$
$\left ( \sin^{-1}x \right )^{2}\leq \frac{\pi ^{2}}{4},\left (\sin^{-1}y \right )^{2}\leq \frac{\pi ^{2}}{4},\left ( \sin^{-1}z \right )^{2}\leq \frac{\pi ^{2}}{4}$
$\left ( \sin^{-1}x \right )^{2}+\left (\sin^{-1}y \right )^{2}+\left ( \sin^{-1}z \right )^{2}\leq \frac{3}{4}\pi ^{2} \, \, and \, it\, is\, given\, that\, L.H.S\, is \,\, \frac{3}{4}\pi ^{2}.$
$So\: it\: is\: possible\: only\: when$
$\sin^{-1}x= \frac{\pi }{2},\sin^{-1}y= \frac{\pi}{2}, \sin^{-1}z= \frac{\pi}{2}$
$x= 1,y= 1,z= 1$ $\therefore x^{2}+y^{2}+z^{2}= 3$

The concepts used in this exercise are:

Evaluating the Inverse of Sin and Cos functions
Finding the domain of inverse functions
Finding values of variables assigned to inverse equations

Students facing problems in understanding the RD Sharma textbook can use RD Sharma Class 12th Exercise 3.1 solutions to study for their exams.

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Both trigonometry and Inverse trigonometric functions are vast chapters that contain hundreds of sums. However, if you divide the sums and practice daily, you can easily cover this chapter with maximum output.

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4. Explain the Inverse of fundamental trigonometric values?

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