<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] We have: $1+\cos(2C) = 2\cos^2C, \cos(2A)+\cos(2B) = 2\cos(A+B)\cos(A-B)=-2\cos C\cos(A-B)\implies LHS = 2\cos C(\cos C+ \cos(A-B))= 2\cos C(\cos(A-B) - \cos(A+B)) = 2\cos C(-2\sin A\sin (-B)) = RHS$
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