Question

`sin^-1(sin  (17pi)/8)`

Answer 1

We know

`sin(sin^-1theta)=theta if - pi/2<=theta<=pi/2`

We have

`sin^-1(sin  (17pi)/8)=sin^-1{sin(2pi+pi/8)}`

`=sin^-1(sin  pi/8)`

`=pi/8`