Sum of Cosines of Twice Angles of Triangle

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Theorem

Let $\triangle ABC$ be a triangle.

Then:

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


Proof

First we note that:

\(\ds A + B + C\) \(=\) \(\ds 180 \degrees\) Sum of Angles of Triangle equals Two Right Angles
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds A + B\) \(=\) \(\ds 180 \degrees - C\)

That is, $C$ is the supplement of $A + B$.


Then:

\(\ds \cos 2 A + \cos 2 B + \cos 2 C\) \(=\) \(\ds 2 \map \cos {A + B} \map \cos {A - B} + \cos 2 C\) Cosine plus Cosine
\(\ds \) \(=\) \(\ds 2 \map \cos {180 \degrees - C} \map \cos {A - B} + \cos 2 C\) from $(1)$
\(\ds \) \(=\) \(\ds -2 \cos C \map \cos {A - B} + \cos 2 C\) Cosine of Supplementary Angle
\(\ds \) \(=\) \(\ds -2 \cos C \paren {\cos A \cos B + \sin A \sin B} + \cos 2 C\) Cosine of Difference
\(\ds \) \(=\) \(\ds -2 \cos C \paren {\cos A \cos B + \sin A \sin B} + \paren {2 \cos^2 C - 1}\) Double Angle Formula for Cosine: Corollary $1$
\(\ds \) \(=\) \(\ds -1 - 2 \sin A \sin B \cos C + 2 \paren {\cos C - \cos A \cos B} \cos C\) multiplying out and rearranging
\(\ds \) \(=\) \(\ds -1 - 2 \sin A \sin B \cos C + 2 \paren {\map \cos {180 \degrees - \paren {A + B} } - \cos A \cos B} \cos C\) from $(1)$
\(\ds \) \(=\) \(\ds -1 - 2 \sin A \sin B \cos C + 2 \paren {-\map \cos {A + B} - \cos A \cos B} \cos C\) Cosine of Supplementary Angle
\(\ds \) \(=\) \(\ds -1 - 2 \sin A \sin B \cos C + 2 \paren {-\paren {\cos A \cos B - \sin A \sin B} - \cos A \cos B} \cos C\) Cosine of Sum
\(\ds \) \(=\) \(\ds -1 - 2 \sin A \sin B \cos C + 2 \paren {-\cos A \cos B - \cos A \cos B} \cos C + 2 \sin A \sin B \cos C\) multiplying out
\(\ds \) \(=\) \(\ds -1 - 4 \cos A \cos B \cos C\) simplifying

$\blacksquare$


Also presented as

This result can also be presented as:

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


Sources

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