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Answer: \(-\dfrac{\pi}{3}\)

\(\dfrac{5\pi}{3}\) is not in the principal range \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). Now \(\sin\dfrac{5\pi}{3}=\sin\left(2\pi-\dfrac{\pi}{3}\right)=-\sin\dfrac{\pi}{3}=\sin\left(-\dfrac{\pi}{3}\right)\), and \(-\dfrac{\pi}{3}\) lies in the principal range. So \(\sin^{-1}\left(\sin\dfrac{5\pi}{3}\right)=-\dfrac{\pi}{3}\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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