The principal value of `sin^(-1)[cos(sin^(-1)'1/2)]` is `pi/3`.

Answer

1

Explanation

Given that, ` sin^(-1)[cos(sin^(-1)'1/2)] = sin^(-1)[cossin^(-1)(sin'(pi)/(6))]`
`= sin^(-1)[cos'(pi)/(6)] , [:' sin^(-1)(sinx) = x]`
`= sin^(-1)'(sqrt(3))/(2)`
`= sin^(-1)sin'(pi)/(3)= (pi)/(3)`