20. \sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) is equal to

        (A)    \frac{1}{2}

        (B)    \frac{1}{3}

        (C)    \frac{1}{4}

        (D)    1

Answers (1)

Solving the inner bracket of \sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right );

\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )  or

Take \sin^{-1}\left(-\frac{1}{2} \right ) = x  then,

\sin x =-\frac{1}{2}  and we know the range of principal value of \sin^{-1}x\ is\ \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ].

Therefore we have \sin^{-1}\left ( -\frac{1}{2} \right ) = -\frac{\pi}{6}.

Hence, \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

Hence the correct answer is D.

 

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