Answer
$u\sqrt {1-v^2}-v\sqrt {1-u^2}$, $|u|\le1, |v|\le1$.
Work Step by Step
1. Let $cos^{-1}(u)=x$, we have $cos(x)=u, sin(x)=\sqrt {1-u^2}$
2. Let $sin^{-1}(v)=y$, we have $sin(y)=v, cos(y)=\sqrt {1-v^2}$
3. Thus $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)=u\sqrt {1-v^2}-v\sqrt {1-u^2}$ where $|u|\le1, |v|\le1$.
