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  (17pi)/8)`

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

`=sin^-1(-sin  pi/8)`

`=sin^-1(sin-pi/8)`

`=-pi/8`