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