Question

If sin^-1(sin\(\frac { 2π }{ 3}\))

Answer 1

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