Inverse Trigonometric Functions Question - CBSE

Question

The value of $\sin\left(\dfrac{\pi}{3} - \sin^{-1}\left(-\dfrac{1}{2}\right)\right)$ is:
(a) $1$
(b) $\dfrac{1}{2}$
(c) $\dfrac{1}{3}$
(d) $\dfrac{1}{4}$

Accepted Answer

\sin^{-1}(-1/2) = -\pi/6 \Rightarrow \sin(\pi/3 + \pi/6) = \sin(\pi/2) = 1. [1.0 Mark]

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