Question:
\(\sin^{-1}\!\big(\sin \tfrac{2\pi}{3}\big)=\ \ ?\)
\(\sin^{-1}\!\big(\sin \tfrac{2\pi}{3}\big)=\ \ ?\)
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Always convert back to the principal angle for inverse trig.
Updated On: Oct 17, 2025
- \(\dfrac{\pi}{3}\)
- \(\dfrac{2\pi}{3}\)
- \(\dfrac{5\pi}{6}\)
- \(\dfrac{\pi}{6}\)
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The Correct Option is A
Solution and Explanation
\(\sin\!\left(\tfrac{2\pi}{3}\right)=\tfrac{\sqrt3}{2}\).
\(\sin^{-1}\) must return an angle in \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\).
In that range, the angle whose sine is \(\tfrac{\sqrt3}{2}\) is \(\tfrac{\pi}{3}\), not \(2\pi/3\).
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