Evaluate $\sin[\pi/2-\sin^{-1} (-1/2)]$

Question

My answer isn't even in options. What am I doing wrong?

Answer 1

There is an typo in the question. I google it and find "CBSE Sample Paper 2021-22". The correct question is:

Now you can solve it.

$$\sin\left[\frac{\pi}3-\sin^{-1}\left(-\frac{1}2\right)\right]{=\sin\left[\frac{\pi}3+\sin^{-1}\left(\frac{1}2\right)\right]\\ =\sin\left[\frac{\pi}3+\frac{\pi}6\right]\\ =\sin \frac{\pi}{2}=1}$$

Answer 2

The set of all angles $\theta$ for which $\sin \theta = -\frac{1}{2}$ is $$\theta \in \left\{\ldots, -\frac{17\pi}{6}, -\frac{13\pi}{6}, -\frac{5\pi}{6}, -\frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}, \ldots \right\} = \left\{-\frac{\pi}{6} + 2\pi k, -\frac{5\pi}{6} + 2\pi k\right\}_{k \in \mathbb Z}.$$ So the set $\frac{\pi}{2} - \theta$ is just $$\frac{\pi}{2} - \theta = \left\{\pm \frac{\pi}{3} + 2\pi k \right\}_{k \in \mathbb Z}.$$ Consequently $$\sin \left(\frac{\pi}{2} - \theta\right) = \pm \sin \frac{\pi}{3} = \pm \frac{\sqrt{3}}{2},$$ neither of which is among the included answer choices.