<s>[INST] 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
$$1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$$
 [/INST] Note that
$$\cos u + \cos v = 2\cos((u + v)/2)\cos((u − v)/2)$$
and
$$\cos u - \cos v = -2\sin((u + v)/2)\sin((u − v)/2).$$
Hence
\begin{align*}
1 + \cos 2C - \cos 2A - \cos 2B&=1 + \cos (2\pi-2(A+B)) - 2\cos(A+B)\cos(A-B)\\
&=2\cos^2 (A+B) - 2\cos(A+B)\cos(A-B)\\
&=2\cos(A+B)(\cos(A+B)-\cos(A-B))\\
&=2(-\cos C)(-2\sin A\sin B)=4\sin A\sin B\cos C.
\end{align*}
</s>