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RD Sharma Class 12 Exercise 3.2 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 3.2 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 3.2 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Updated on 21 Jan 2022, 11:52 AM IST

Students preparing for their class 12th board exams have already heard a lot about RD Sharma Solutions. The RD Sharma class 12th exercise 3.2 is one of their unique solutions books with all the information you need to prepare for your board exams. This particular book is based on the third chapter of the class 12 mathematics book, which is Inverse Trigonometric Functions.

This Story also Contains

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise
Inverse Trigonometric Functions Excercise: 3.2
Inverse trigonometric functions exercise 3.2 question 5 (ii)
Answer:

The RD Sharma class 12 chapter 3 exercise 3.2 has 11 questions, including subparts in level 1, which will cover the concept of Domain of functions and Principal values. R D Sharma Solutions Students will get a clear idea of trigonometric functions and concepts when practicing the RD Sharma class 12 solutions Inverse Trigonometric Functions exercise 3

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

Inverse Trigonometric Functions Excercise: 3.2

Inverse trigonometric functions exercise 3.2 question 1

Answer: Domain of f(x) is $\left [ -\sqrt{5},-\sqrt{3} \right ]\cup \left [ \sqrt{3},\sqrt{5} \right ]$
Hint: Domain of $\cos ^{-1}\left ( x \right )$ lies in the interval [-1, 1].
Given: $f(x)=\cos ^{-1}\left ( x^{2}-4 \right )$
The domain of $\cos ^{-1}\left ( x \right )$ is [-1, 1]
$\therefore -1\leq x^{2}-4\leq 1$
$\Rightarrow -1+4\leq x^{2}-4+4\leq 1+4$
$\Rightarrow 3\leq x^{2}\leq 5$
$\Rightarrow \pm \sqrt{3}\leq x\leq \pm \sqrt{5}$
$\Rightarrow x\in \left [ -\sqrt{5},-\sqrt{3} \right ]\cup \left [ \sqrt{3},\sqrt{5} \right ]$
Hence, the domain of f(x) is $\left [ -\sqrt{5},-\sqrt{3} \right ]\cup \left [ \sqrt{3},\sqrt{5} \right ]$

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inverse trigonometric functions exercise 3.2 question 2

Answer:$\left [ \frac{-1}{2},\frac{1}{2} \right ]$
Hint: Domain of $(2x)$ lies in the interval [-1, 1].
Given: $f(x) = 2\cos^{-1}2x +\sin^{-1}x$
Solution:
The domain of $\cos ^{-1}(2x)$ lies in the interval [-1, 1].
$-1\leq 2x\leq 1$
$\frac{-1}{2}\leq x\leq \frac{1}{2}$
Domain of $\cos ^{-1}(2x)$ is $\frac{-1}{2}\leq x\leq \frac{1}{2}$
Domain of $\sin ^{-1}(x)$ is (-1,1)
Hence, domain of $2\cos^{-1}2x +\sin^{-1}x$ lies in the interval $\left [ \frac{-1}{2},\frac{1}{2} \right ]$

Inverse trigonometric functions exercise 3.2 question 3

Answer: [-1, 1]
Hint: Domain of $\cos ^{-1}x$ lies in the interval [-1, 1]
Given:$f(x)=\cos ^{-1}x+\cos x$
Solution:
Domain of (x) lies in the interval [-1, 1].
Domain of $\cos x$ is R as $\cos x$ is defined for all real values.
Hence,N
Domain of $\cos ^{-1}x+\cos x$ will be the intersection of both the functions,R ∩ [-1, 1]=[-1,1]

Inverse trigonometric functions exercise 3.2 question 4 (i)

Answer:

$\frac{5\pi }{6}$
Hint: Any x $x\in \left [ -1,1 \right ],\cos ^{-1}$represents an angle in $\left ( 0,\pi \right )$
Given: $\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )$
Solution:
$\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )$ is an angle in $\left [ 0,\pi \right ]$whose cosine is $\left ( \frac{-\sqrt{3}}{2} \right )$
$=\pi -\frac{\pi }{6}$
$=\frac{6\pi-\pi }{6}$
$\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right ) = \frac{5\pi }{6}$

Inverse trigonometric functions exercise 3.2 question 4 (ii)

Answer:

$\frac{3\pi }{4}$
Hint: The range of the principal value branch of $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\cos ^{-1}\left ( \frac{-1}{\sqrt{2}} \right )$
Solution:
Let $\cos ^{-1}\left ( \frac{-1}{\sqrt{2}} \right ) =y$
$\cos y=\left ( \frac{-1}{\sqrt{2}} \right )$
$y=-\cos \frac{\pi }{4}$
$=\cos \left ( \pi -\frac{\pi }{4} \right )$
$=\cos \left ( \frac{3\pi }{4} \right )$
$y=\frac{3\pi }{4}$
Therefore, the principal value of $\cos ^{-1}\left ( \frac{-1}{\sqrt{2}} \right )$ is $\frac{3\pi }{4}$.

Inverse trigonometric functions exercise 3.2 question 4 (iv)

Answer:$\pi$
Hint: The range of the principal value branch of $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\cos ^{-1}\left ( \tan \frac{3\pi }{4} \right )$
Solution:
$\cos ^{-1}\left ( \tan \frac{3\pi }{4} \right ) = \cos ^{-1}\left ( \tan \left ( \frac{\pi }{2}+\frac{3\pi }{4} \right ) \right )$

$\therefore \cos ^{-1}\left ( \tan \frac{3\pi }{4} \right )=\pi$

Inverse trigonometric functions exercise 3.2 question 5 (i)

Answer:${\frac{2\pi }{3}}$
Hint: The range of the principal value branch of $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\cos ^{-1}\left ( \frac{1}{2} \right )+2\sin ^{-1}\left ( \frac{1}{2} \right )$
Solution:
Let $\cos ^{-1}\left ( \frac{1}{2} \right )=x$
$\cos x=\frac{1}{2}=\cos \left ( {\frac{\pi }{3}}\right )$
$\cos ^{-1}\left ( \frac{1}{2} \right )=\frac{\pi }{3}$ ....(1)
Let $\sin ^{-1}\left ( \frac{1}{2} \right )=y$
$\sin y=\frac{1}{2}=\sin \left ( {\frac{\pi }{6}}\right )$
$\sin ^{-1}\left ( \frac{1}{2} \right )=\frac{\pi }{6}$ .....(2)
From (1) and (2),
$\cos ^{-1}\left ( \frac{1}{2} \right )+2\sin ^{-1}\left ( \frac{1}{2} \right )=\frac{\pi }{3}+2\left ( \frac{\pi }{6} \right )$
$=\frac{\pi }{3}+\frac{\pi }{3}$
$=\frac{2\pi }{3}$
$\therefore$Principal value of $\cos ^{-1}\left ( \frac{1}{2} \right )+2\sin ^{-1}\left ( \frac{1}{2} \right )$$=\frac{2\pi }{3}$

Inverse trigonometric functions exercise 3.2 question 5 (ii)

Answer:${\frac{2\pi }{3}}$
Hint: The range of the principal value branch of $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\cos ^{-1}\left ( \frac{1}{2} \right )-2\sin ^{-1}\left ( \frac{-1}{2} \right )$
Solution:
Let $\cos ^{-1}\left ( \frac{1}{2} \right )=x$
$\cos x=\frac{1}{2}=\cos \left ( {\frac{\pi }{3}}\right )$
$\cos ^{-1}\left ( \frac{1}{2} \right )=\frac{\pi }{3}$ ....(1)
Let $\sin ^{-1}\left ( \frac{-1}{2} \right )=y$
$\sin y=\frac{-1}{2}=-\sin \left ( {\frac{\pi }{6}}\right )$
$\sin ^{-1}\left ( \frac{-1}{2} \right )=\frac{-\pi }{6}$ .....(2)
From (1) and (2),
$\cos ^{-1}\left ( \frac{1}{2} \right )-2\sin ^{-1}\left ( \frac{-1}{2} \right )=\frac{\pi }{3}-2\left ( \frac{-\pi }{6} \right )$
$=\frac{\pi }{3}+\frac{\pi }{3}$
$=\frac{2\pi }{3}$
$\therefore$Principal value of $\cos ^{-1}\left ( \frac{1}{2} \right )-2\sin ^{-1}\left ( \frac{-1}{2} \right )$$=\frac{2\pi }{3}$

Inverse trigonometric functions exercise 3.2 question 5 (iii)

Answer:${\frac{3\pi }{2}}$
Hint: The range of the principal value branch of $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\sin ^{-1}\left ( \frac{-1}{2} \right )+2\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )$
Solution:
Let $\sin ^{-1}\left ( \frac{-1}{2} \right )=x$
$\sin\; \sin x=\frac{-1}{2}=-\sin\; \sin \left ( \frac{\pi }{6} \right )$
$\sin ^{-1}\left ( \frac{-1}{2} \right )=\frac{-\pi }{6}$
Let $\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )=y$
$\cos y=\left ( \frac{-\sqrt{3}}{2} \right )=\cos \left ( \pi -\frac{\pi }{6} \right )$
$\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )=\frac{5\pi }{6}$
Hence, $\sin ^{-1}\left ( \frac{-1}{2} \right )+2\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )=\frac{-\pi }{6}+2\left ( \frac{5\pi }{6} \right )$
$=\frac{-\pi }{6}+\frac{10\pi }{6}$
$=\frac{9\pi }{6}$
$=\frac{3\pi }{2}$
Principal Value of $\sin ^{-1}\left ( \frac{-1}{2} \right )+2\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )$ is ${\frac{3\pi }{2}}$.

Inverse trigonometric functions exercise 3.2 question 5 (iv)

Answer:${\frac{-\pi }{6}}$
Hint: The range of the principal value branch of $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\sin^{-1}\left ( \frac{-\sqrt{3}}{2} \right )+\cos^{-1}\left ( \frac{\sqrt{3}}{2} \right )$
Solution:
$\sin^{-1}\left ( \frac{-\sqrt{3}}{2} \right )+\cos^{-1}\left ( \frac{\sqrt{3}}{2} \right )=\frac{-\pi }{3}+\frac{\pi }{6}$
Since $\sin ^{-1}x$ is an angle in $\left ( \frac{-\pi }{2},\frac{\pi }{2} \right ); \; \cos ^{-1}x$is an angle in $\left ( 0,\pi \right )$
Hence, $\sin ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )+\cos ^{-1}\left ( \frac{\sqrt{3}}{2} \right )=\frac{-\pi }{6}$

Inverse trigonometric functions exercise 3.2 question 4 (iii)

Answer:

$\frac{5\pi }{6}$
Hint: The range of the principal value branch of $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\cos ^{-1}\left ( \sin \left ( \frac{4\pi }{3} \right ) \right )$
Solution:
$\cos ^{-1}\left ( \sin \left ( \frac{4\pi }{3} \right ) \right )=\cos ^{-1}\left ( \sin \left ( \pi +\frac{\pi }{3} \right ) \right )$
$=\cos^{-1} \left ( \frac{-\sqrt{3} }{2} \right )$
For any $x\in \left [ -1,1 \right ],\cos ^{-1}x$ represents an angle in $\left ( 0,\pi \right )$
$\cos^{-1} \left ( \frac{-\sqrt{3} }{2} \right )=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$
Therefore, the principal value of $\cos ^{-1}\left ( \sin \left ( \frac{4\pi }{3} \right ) \right )$ is $\frac{5\pi }{6}$.

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