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