Question

If sin^-1(sin$\frac{2\pi }{3}$)

Answer 1

sin^-1(sin$\frac{2\pi }{3}$) = sin^-1[ sin(π – $\frac{\pi }{3}$) ]
= sin^-1[sin$\frac{\pi }{3}$] = $\frac{\pi }{3}$.
Please note : sin^-1(sin$\frac{2\pi }{3}$) ≠ sin$\frac{2\pi }{3}$, since the range of the principal values branch of sin^-1 is (-$\frac{\pi }{2}$, $\frac{\pi }{2}$)