Question

How do you evaluate `arcsin((sqrt3)/2)` or `sin(Arccos(-15/17))`?

Answer 1

In this way.

The range of the function `y=arcsinx` is `[-pi/2,pi/2]`, so there is only this solution:

`x=pi/3`.

`sin(arccos(-15/17))=sinalpha`

where `alpha=arccos(-15/17)`, and the angle is in the second quadrant because the range of the function `y=arccosx` is `[0,pi]` and the value is negative.

So

`cosalpha=-15/17` and than

`sinalpha=+sqrt(1-cos^2alpha)=sqrt(1-225/289)=sqrt((289-225)/289)=`

`=sqrt(64/289)=8/17`

(with the `+` because in the second quadrant the sinus is positive).