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