Question

sin^-1(-\(\frac { 1 }{ 2 }\))

Answer 1

Let sin^-1(-\(\frac { 1 }{ 2 }\)) = y
∴ sin y = \(\frac { 1 }{ 2 }\) = – sin\(\frac { π }{ 6 }\) = sin(-\(\frac { π }{ 6 }\))
The range of principal value branch of sin^-1 is [-\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\)].
Hence, principal value of sin^-1(-\(\frac { 1 }{ 2 }\)) is – \(\frac { π }{ 6 }\).