Question

Evaluate the following:

`sin(sec^-1  17/8)`

Answer 1

`=sin(sec^-1  17/8)=sin(cos^-1  8/17)`

`=sin[sin^-1sqrt(1-(8/17)^2)]`    `[thereforecos^-1x=sin^-1sqrt(1-x^2)]`

`=sin[sin^-1(sqrt(1-64/289))]`

`=sin[sin^-1(sqrt(225/289))]`

`=sin[sin^-1  15/17]`

`=15/17`