Answer
$\cos 100^\circ-\cos 200^\circ=\sin 50^\circ$
Work Step by Step
Start with the left side:
$\cos 100^\circ-\cos 200^\circ$
Use the Sum-to-Product Formulas on page 560:
$=-2\sin\frac{100^\circ+200^\circ}{2}\sin\frac{100^\circ-200^\circ}{2}$
Simplify:
$=-2\sin\frac{300^\circ}{2}\sin\frac{-100^\circ}{2}$
$=-2\sin 150^\circ\sin (-50^\circ)$
$=-2*\frac{1}{2}\sin (-50^\circ)$
$=-\sin (-50^\circ)$
Use the fact that sine is an odd function (i.e. $\sin (-x)=-\sin x$):
$=-(-\sin 50^\circ)$
$=\sin 50^\circ$
Since this equals the right side, the identity has been proven.
