Question

\(\sin^{-1}\!\big(\sin \tfrac{2\pi}{3}\big)=\ \ ?\)

• A. \(\dfrac{\pi}{3}\)
• B. \(\dfrac{2\pi}{3}\)
• C. \(\dfrac{5\pi}{6}\)
• D. \(\dfrac{\pi}{6}\)

Hint:

Always convert back to the principal angle for inverse trig.

Answer 1

The Correct Option is A

\(\sin\!\left(\tfrac{2\pi}{3}\right)=\tfrac{\sqrt3}{2}\). \(\sin^{-1}\) 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 \(2\pi/3\).