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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
Question
$\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to
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Answer

$\sin ^{-1}\left(-\frac{1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right),$ as $sin^{-1} (-x) = -sin^{-1} x$ 
We all know that the principle branch of$ ~~sin^{-1}x ~~$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$  and $-\frac {\pi}{6} \in [-\frac {\pi}{2}, \frac {\pi}2]$
$\therefore$   $\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$ 
Now, $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)=\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin \left(\frac{3 \pi}{6}\right)=\sin \left(\frac{\pi}{2}\right)=1$ 
Therefore, the required value of $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)=1$ 

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