> 
> Given that $y = \arccos x$, $ - 1 \le x \le 1\,and\,0 \le y \le \pi $, express $\arcsin x$ in terms of y.
> 
> 
> 

The best I know how to do this is is:

$$\eqalign{
 & \cos y = x \cr 
 & {\cos ^2}y + {\sin ^2}y = 1 \cr 
 & {\sin ^2}y = 1 - {\cos ^2}y \cr 
 & \sin y = \sqrt {1 - {x^2}} \cr 
 & \arcsin \left( {\sqrt {1 - {x^2}} } \right) = y \cr} $$

However this isn't what is asked for, How do I go about getting things in terms of y?

Thanks