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


Updated By SaraNextGen
On April 24, 2024, 11:35 AM

Additional Problems
Question 1.

Find the principal value of: $\sin ^{-1}\left(\sin \frac{4 \pi}{3}\right)$
Solution:
$
\sin ^{-1}\left(\sin \frac{4 \pi}{3}\right)
$
The range of the principal value of $\sin ^{-1} x$ is $[0, \pi]$
$
=\sin ^{-1}\left(\sin \left(\pi+\frac{\pi}{3}\right)\right)=\sin ^{-1}\left(-\sin \frac{\pi}{3}\right)=\sin ^{-1}\left(\sin \left(-\frac{\pi}{3}\right)\right)=-\frac{\pi}{3}
$
Since $-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Question 2.
Find the value of: $\sin ^{-1}\left(\frac{\sqrt{5}-1}{4}\right)$
Solution:
$
\begin{aligned}
\sin ^{-1}\left(\frac{\sqrt{5}-1}{4}\right) & =\sin ^{-1}\left(\sin \frac{\pi}{10}\right) \\
& =\frac{\pi}{10}
\end{aligned}
$
(We know that $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$ )
$
\left(18^{\circ}=\frac{\pi}{10}\right)
$

Also Read : Exercise-4.2-Chapter-4-Inverse-Trigonometric-Functions-12th-Maths-Guide-Samacheer-Kalvi-Solutions

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