Question

`sin^-1(sin  (13pi)/7)`

Answer 1

We know

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

We have

`sin^-1(sin  (13pi)/7)=sin^-1{sin(2pi+pi/7)}`

`=sin^-1(sin-pi/7)`

`=-pi/7`