CUET Preparation Today
The value of $sin[\frac{π}{2}-sin^{-1}(-\frac{\sqrt{3}}{2})]$ is: |
$\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ $-\frac{\sqrt{3}}{2}$ |
$\frac{1}{2}$ |
The correct answer is option 1: $\frac{1}{2}$ The principal value of $\sin^{-1}\left( -\frac{\sqrt{3}}{2} \right)$ is $-\frac{\pi}{3}$. The principal value range of $\sin^{-1} x$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Substituting $-\frac{\pi}{3}$ : $\sin\left[ \frac{\pi}{2} - \left( -\frac{\pi}{3} \right) \right] = \sin\left( \frac{\pi}{2} + \frac{\pi}{3} \right)$
$\sin\left( \frac{3\pi + 2\pi}{6} \right) = \sin\left( \frac{5\pi}{6} \right)$
$\sin\left( \frac{5\pi}{6} \right) = \frac{1}{2}$
|
