Exercise 4.1 - Chapter 4 Inverse Trigonometric Functions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Inverse Trigonometric Functions
Ex $4.1$
Question 1.
Find all the values of $\mathrm{x}$ such that
(i) $-10 \pi \leq x \leq 10 \pi$ and $\sin x=0$
(ii) $-8 \pi \leq x \leq 8 \pi$ and $\sin x=-1$
Solution:
(i) $\sin x=0$
$\Rightarrow \mathrm{x}=\mathrm{n} \pi$
where $\mathrm{n}=0, \pm 1, \pm 2, \pm 3, \ldots \ldots, \pm 10$
(ii) $\sin x=-1$
$\Rightarrow \mathrm{x}=(4 \mathrm{n}-1) \frac{\pi}{2}, \mathrm{n}=0, \pm 1, \pm 2, \pm 3,4$
Question 2.
Find the period and amplitude of
(i) $y=\sin 7 x$
(ii) $y=-\sin \left(\frac{1}{3} \mathrm{x}\right)$
(iii) $y=4 \sin (-2 x)$
Solution:
(i) $y=\sin 7 x$
Period of the function $\sin \mathrm{x}$ is $2 \pi$
Period of the function $\sin 7 x$ is $\frac{2 \pi}{7}$
The amplitude of $\sin 7 x$ is 1 .
(ii) $y=-\sin \frac{1}{3} x$
Period of $\sin x$ is $2 \pi$
So, period of $\sin \frac{1}{3} \mathrm{x}$ is $6 \pi$ and the amplitude is 1 .
(iii) $y=4 \sin (-2 x)=-4 \sin 2 x$
Period of $\sin x$ is $2 \pi$
$\pi$ Period of $\sin 2 x$ is $\pi$ and the amplitude is 4 .
Question 3.
Sketch the graph of $y=\sin \left(\frac{1}{3} x\right)$ for $0 \leq x<6 \pi$.
Solution:
The period of $\sin \left(\frac{1}{3} x\right)$ is $6 \pi$ and the amplitude is 1 .

The graph is

Question 4.
Find the value of
(i) $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$
(ii) $\sin ^{-1}\left(\sin \left(\frac{5 \pi}{4}\right)\right)$
Solution:
(i)
$
\text { (i) } \begin{aligned}
\sin \frac{2 \pi}{3} & =\sin \left(\pi-\frac{\pi}{3}\right)=\sin \left(\frac{\pi}{3}\right) \\
\therefore \quad \sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) & =\sin ^{-1}\left(\sin \frac{\pi}{3}\right)=\frac{\pi}{3}
\end{aligned}
$
(ii)
$
\therefore \quad \sin ^{-1}\left(\sin \frac{5 \pi}{4}\right)=\sin ^{-1}\left(\sin \left(\frac{-\pi}{4}\right)\right)=\frac{-\pi}{4}
$

For that value of $x$ does $\sin x=\sin ^{-1} x$ ?
Solution:
$\sin \mathrm{x}=\sin ^{-1} \mathrm{x}$ is possible only when $\mathrm{x}=0(\because \mathrm{x} \in \mathrm{R})$
Question 6.
Find the domain of the following
(i) $f(x)=\sin ^{-1}\left(\frac{x^2+1}{2 x}\right)$
(ii) $g(x)=2 \sin ^{-1}(2 x-1)-\frac{\pi}{4}$
Solution:
(i) $f(x)=\sin ^{-1}\left(\frac{x^2+1}{2 x}\right)$
The range of $\sin -1 \mathrm{x}$ is $-1$ to 1
$-1 \leq \frac{x^2+1}{2 x} \leq 1$
$\Rightarrow \frac{x^2+1}{2 x} \geq-1$ or $\frac{x^2+1}{2 x} \leq 1$
$\Rightarrow \mathrm{x}^2+1 \geq-2 \mathrm{x}$ or $\mathrm{x}^2+1 \leq 2 \mathrm{x}$
$\Rightarrow \mathrm{x}^2+1+2 \mathrm{x} \geq 0$ or $\mathrm{x}^2+1-2 \mathrm{x} \leq 0$
$\Rightarrow(\mathrm{x}+1)^2 \geq 0$ or $(\mathrm{x}-1)^2 \leq 0$ which is not possible
$\Rightarrow-1 \leq \mathrm{x} \leq 1$ or
(ii) $g(x)=2 \sin ^{-1}(2 x-1)-\frac{\pi}{4}$
$-1 \leq(2 \mathrm{x}-1) \leq 1$
$0 \leq 2 \mathrm{x} \leq 2$
$0 \leq \mathrm{x} \leq 1$
$\mathrm{x} \in[0,1]$

Question 7.
Find the value of $\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} \sin \frac{\pi}{9}\right)$
Solution:
$
\begin{aligned}
\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} \sin \frac{\pi}{9} & =\sin \left(\frac{5 \pi}{9}+\frac{\pi}{9}\right)=\sin \frac{6 \pi}{9}=\sin \frac{2 \pi}{3} \\
& =\sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3} \\
\sin ^{-1}\left[\sin \frac{\pi}{3}\right] & =\frac{\pi}{3}
\end{aligned}
$