$\sin[\frac{\pi}{2}-\sin^{-1}(\frac{-\sqrt{3}}{2})]$= | ClassNotes | Ace Your Exams

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Sep 18, 2022

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$\sin[\frac{\pi}{2}-\sin^{-1}(\frac{-\sqrt{3}}{2})]$=

Difficulty: Easy

A:

1/2

B:

$\frac{-\sqrt{3}}{2}$

C:

$\frac{1}{\sqrt{2}}$

D:

$\frac{-1}{\sqrt{2}}$

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NET NUST Entry Test Mathematics Inverse Trigonometric Functions