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HomeUP Board Class 12mathematicsinversetrigonometric › $\sin^{-1}\left(\sin\tfrac{2\pi}{3}\right)$ equals

$\sin^{-1}\left(\sin\tfrac{2\pi}{3}\right)$ equals

A$\tfrac{\pi}{3}$
B$-\tfrac{\pi}{3}$
C$\tfrac{2\pi}{3}$
D$\tfrac{\pi}{6}$
Answer & Solution
Correct answer: A. $\tfrac{\pi}{3}$
$\sin^{-1}$ must return a value in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ — and $\tfrac{2\pi}{3}$ is outside that. Use the supplement identity $\sin(\pi - \theta) = \sin\theta$: $\sin\tfrac{2\pi}{3} = \sin\left(\pi - \tfrac{2\pi}{3}\right) = \sin\tfrac{\pi}{3}$, and $\tfrac{\pi}{3}$ IS in range. So the inverse returns $\tfrac{\pi}{3}$.
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