Prove $1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$ for $A$, $B$, $C$ the angles of a triangle Given that $A$,$B$ and $C$ are angles of a triangle, show that
$$1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$$

Note:
$$RHS=4\sin A\sin B\cos C=$$
$$-4\sin A\sin B\cos (A+B)=$$
$$-4\sin A\sin B(\cos A \cos B-\sin A \sin B)=$$
$$-\sin 2A\sin2B+(1-\cos2A)(1-\cos2B)=$$
$$1-\cos2A-\cos2B+\cos(2A+2B)=$$
$$1-\cos2A-\cos2B+\cos2C=LHS.$$
