Question:
Find the principal value of each of the following :
$\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}$
Solution:
$\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}\left[\right.$ Formula: $\left.\sin ^{-1}(-x)=-\sin ^{-1} x\right]$
$=\sin \left\{\frac{\pi}{3}-\left(-\sin ^{-1} \frac{1}{2}\right)\right\}$
$=\sin \left\{\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right\}$
Putting value of $\sin ^{-1}\left(\frac{1}{2}\right)$
$=\sin \left\{\frac{\pi}{3}+\frac{\pi}{6}\right\}$
$=\sin \frac{3 \pi}{6}$
$=\sin \frac{\pi}{2}$
$=1$
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