Exercise 4.2-Additional Problems - Chapter 4 - Inverse Trigonometric Functions - 12th Maths Guide Samacheer Kalvi Solutions

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Additional Problems
Question 1.
Find the principal value of: $
\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)
$
Solution:
$
\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)
$
The range of the principal value of $\cos ^{-1} x$ is $[0, \pi]$
Since
$
\begin{aligned}
& \cos \left(\pi-\frac{\pi}{6}\right)=-\cos \frac{\pi}{6}=-\frac{\sqrt{3}}{2} \text { and } \frac{5 \pi}{6} \in[0, \pi] \\
& \therefore \quad \cos \frac{5 \pi}{6}=\frac{-\sqrt{3}}{2} \\
& \frac{5 \pi}{6}=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \\
&
\end{aligned}
$

Question 2.
Find the value of $\cos ^{-1}\left(\cos \frac{4 \pi}{3}\right)$
Solution:

$\begin{aligned}
\cos ^{-1}\left(\cos \frac{4 \pi}{3}\right) & =\cos ^{-1}\left(\cos \left(\pi+\frac{\pi}{3}\right)\right)=\cos ^{-1}\left(-\cos \frac{\pi}{3}\right) \\
& =\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)=\frac{2 \pi}{3} \\
\frac{2 \pi}{3} & \in[0, \pi]
\end{aligned}$
Question 3.
$
\cos \left(\cos ^{-1}\left(\frac{-2}{3}\right)-\sin ^{-1}\left(\frac{2}{3}\right)\right)
$
Solution:
Consider
$
\begin{aligned}
\cos ^{-1}\left(\frac{-2}{3}\right)-\sin ^{-1}\left(\frac{2}{3}\right) & =\left(\pi-\cos ^{-1} \frac{2}{3}\right)-\sin ^{-1}\left(\frac{2}{3}\right)\left[\text { Since } \cos ^{-1}(-x)=\pi-\cos ^{-1} x\right] \\
& =\pi-\left[\cos ^{-1} \frac{2}{3}+\sin ^{-1} \frac{2}{3}\right]=\pi-\frac{\pi}{2}=\frac{\pi}{2}
\end{aligned}
$
Hence $\cos \left(\cos ^{-1}\left(\frac{-2}{3}\right)-\sin ^{-1}\left(\frac{2}{3}\right)\right)=\cos \left(\frac{\pi}{2}\right)=0$