Sum of Cosines of Four Times Angles of Triangle

Theorem
Let $\triangle ABC$ be a triangle.

Then:
 * $\cos 4 A + \cos 4 B + \cos 4 C = 4 \cos 2 A \cos 2 B \cos 2 C - 1$

Proof
First we note that:

That is, $2 C$ is the conjugate of $2 A + 2 B$.

Then:

Also presented as
This result can also be presented as:


 * $\cos 4 A + \cos 4 B + \cos 4 C + 1 = 4 \cos 2 A \cos 2 B \cos 2 C$