Exercise 4.2 - Chapter 4 Inverse Trigonometric Functions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$E \times 4.2$
Question 1.
Find all the values of $x$ such that
(i) $-6 \pi \leq x \leq 6 \pi$ and $\cos x=0$
(ii) $-5 \pi \leq x \leq 5 \pi$ and $\cos x=1$
Solution:
(i) $\cos x=0$
$\Rightarrow \mathrm{x}=(2 \mathrm{n}+1) \pm \frac{\pi}{2}$
$\mathrm{n}=0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$
(ii) $\cos x=1=\cos 0$
$\Rightarrow \mathrm{x}=2 \mathrm{n} \pi \pm 0$
$\mathrm{n}=0, \pm 1, \pm 2$

Question 2.
State the reason for $\cos ^{-1}\left[\cos \left(-\frac{\pi}{6}\right)\right] \neq-\frac{\pi}{6}$
Solution:
We know $\cos (-\pi)=\cos \pi$
so
$
\cos ^{-1}\left[\cos \left(-\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[\cos \frac{\pi}{6}\right]=\frac{\pi}{6} \neq-\frac{\pi}{6}
$
Question 3.
Is $\cos ^{-1}(-x)=\pi-\cos ^{-1} \mathrm{x}$ true? Justify your answer.
Solution:
Let $\pi=\cos ^{-1}(-x)$
$
\begin{aligned}
& \Rightarrow \cos \pi=-\mathrm{x} \\
& \Rightarrow-\cos \pi=\mathrm{x} \\
& \Rightarrow \cos (\pi-\pi)=\mathrm{x} \\
& \Rightarrow \pi-\pi=\cos ^{-1} \mathrm{x} \\
& \Rightarrow \pi-\cos ^{-1} \mathrm{x}=\pi \\
& \Rightarrow \pi-\cos ^{-1} \mathrm{x}=\cos ^{-1}(-\mathrm{x})
\end{aligned}
$
Question 4.
Find the principal value of $\cos ^{-1}\left(\frac{1}{2}\right)$
Solution:
$
\begin{aligned}
& \text { Let } \cos ^{-1}\left(\frac{1}{2}\right)=\pi \\
& \Rightarrow \cos \pi=\frac{1}{2}=\cos \frac{\pi}{3} \\
& \Rightarrow \pi=\frac{\pi}{3} \\
& \Rightarrow \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}
\end{aligned}
$

Question 5.
Find the value of
(i) $2 \cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)$
(ii) $\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}(-1)$
(iii) $\cos ^{-1}\left(\cos \frac{\pi}{7} \cos \frac{\pi}{17}-\sin \frac{\pi}{7} \sin \frac{\pi}{17}\right)$
Solution:
(i)
$
\begin{aligned}
& \cos ^{-1} \frac{1}{2}=\frac{\pi}{3} \\
& \sin ^{-1} \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$
So $2 \cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)=2\left(\frac{\pi}{3}\right)+\frac{\pi}{6}=\frac{2 \pi}{3}+\frac{\pi}{6}=\frac{4 \pi+\pi}{6}=\frac{5 \pi}{6}$
II Method:
We know $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}$
So $2 \cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)=\cos ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}+\sin ^{-1} \frac{1}{2}$
$
=\cos ^{-1}\left(\frac{1}{2}\right)+\frac{\pi}{2}=\frac{\pi}{3}+\frac{\pi}{2}=\frac{2 \pi+3 \pi}{6}=\frac{5 \pi}{6}
$
(ii)
$
\begin{aligned}
\cos ^{-1}\left(-\frac{1}{2}\right) & =\frac{\pi}{3} \\
\sin ^{-1}(-1) & =-\frac{\pi}{2}
\end{aligned}
$
So $\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}(-1)=\frac{\pi}{3}+\left(-\frac{\pi}{2}\right)=\frac{\pi}{3}-\frac{\pi}{2}=\frac{2 \pi-3 \pi}{6}=-\frac{\pi}{6}$
(iii)
Let $\frac{\pi}{7}=\mathrm{A}$ and $\frac{\pi}{17}=\mathrm{B}$
Now $\cos A \cos B-\sin A \sin B=\cos (A+B)$
$
\mathrm{A}+\mathrm{B}=\frac{\pi}{7}+\frac{\pi}{17}=\frac{\pi(17+7)}{119}=\frac{24 \pi}{119}
$
So $\cos ^{-1}\left[\cos \frac{\pi}{7} \cos \frac{\pi}{17}-\sin \frac{\pi}{7} \sin \frac{\pi}{17}\right]=\cos ^{-1}\left[\cos \left(\frac{24 \pi}{119}\right)\right]=\frac{24 \pi}{119}$

Question 6.
Find the domain of
(i) $f(x)=\sin ^{-1}\left(\frac{|x|-2}{3}\right)+\cos ^{-1}\left(\frac{1-|x|}{4}\right)$
(ii) $g(x)=\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}$
Solution:
(i) $f(x)=\sin ^{-1}\left(\frac{|x|-2}{3}\right)+\cos ^{-1}\left(\frac{1-|x|}{4}\right)=\mathrm{U}(\mathrm{x})+\mathrm{V}(\mathrm{x})$ say
$\mathrm{U}(\mathrm{x})$ :
$-1<\frac{|x|-2}{3}<1$
$-3<|\mathrm{x}|-2<3$
$-1<|\mathrm{x}| \leq 5$
$\mathrm{~V}(\mathrm{x})$
$-1 \leq \frac{1-|x|}{4} \leq 1$
$-4 \leq 1-|\mathrm{x}| \leq 4$
$-5 \leq-|\mathrm{x}| \leq 3$
$-3 \leq|\mathrm{x}| \leq 5$
from $\mathrm{U}(\mathrm{x})$ and $\mathrm{V}(\mathrm{x})$
$\Rightarrow|\mathrm{x}| \leq 5$
$\Rightarrow-5 \leq|\mathrm{x}| \leq 5$
(ii) $\mathrm{g}(\mathrm{x})=\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}$
$-1 \leq \mathrm{x} \leq 1$

Question 7.
For what value of $x$, the inequality $\frac{\pi}{2}<\cos ^{-1}(3 x-1)<\pi$ holds?
Solution:
$
\begin{aligned}
& \frac{\pi}{2}<\cos ^{-1}(3 \mathrm{x}-1)<\pi \\
& \Rightarrow \cos \frac{\pi}{2}<3 \mathrm{x}-1<\cos \pi \\
& \Rightarrow 0<3 \mathrm{x}-1<-1 \\
& \Rightarrow 1<3 \mathrm{x}<0 \\
& \Rightarrow \frac{1}{3}<\mathrm{x}<0
\end{aligned}
$
This inequality holds only if $x<0$ or $x>\frac{1}{3}$

Question 8.
Find the value of
(i) $\cos \left(\cos ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)\right)$
(ii) $\cos ^{-1}\left(\cos \left(\frac{4 \pi}{3}\right)\right)+\cos ^{-1}\left(\cos \left(\frac{5 \pi}{4}\right)\right)$
Solution:
(i) We know $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}$
$
\begin{aligned}
& \therefore \quad \cos ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2} \\
& \Rightarrow \cos \left[\cos ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)\right]=\cos \frac{\pi}{2}=0
\end{aligned}
$
(ii)
$
\begin{aligned}
\cos \frac{4 \pi}{3} & =\cos \left(\pi+\frac{\pi}{3}\right)=-\cos \frac{\pi}{3} \\
\cos ^{-1} \cos \frac{4 \pi}{3} & =-\frac{\pi}{3}
\end{aligned}
$
and
$
\cos \frac{5 \pi}{4}=\cos \left(\pi+\frac{\pi}{4}\right)=-\cos \frac{\pi}{4}
$
$
\begin{aligned}
& \therefore \quad \cos ^{-1} \cos \frac{5 \pi}{4}=-\frac{\pi}{4} \\
& \text { so } \cos ^{-1}\left(\cos \frac{4 \pi}{3}\right)+\cos ^{-1} \cos \left(\frac{5 \pi}{4}\right) \\
& =-\frac{\pi}{3}-\frac{\pi}{4}=-\frac{7 \pi}{12} \\
&
\end{aligned}
$