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