Sum of Cosines of Twice 2 Angles minus Cosine of Twice Third Angle of Triangle

Theorem

Let $\triangle ABC$ be a triangle.

Then:

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

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 $1$

\(\ds \)

\(=\)

\(\ds 1 - 2 \sin A \sin B \cos C - 2 \paren {\cos A \cos B + \cos C} \cos C\)

multiplying out and rearranging

\(\ds \)

\(=\)

\(\ds 1 - 2 \sin A \sin B \cos C - 2 \paren {\cos A \cos B + \map \cos {180 \degrees - \paren {A + B} } } \cos C\)

from $(1)$

\(\ds \)

\(=\)

\(\ds 1 - 2 \sin A \sin B \cos C - 2 \paren {\cos A \cos B - \map \cos {A + B} } \cos C\)

Cosine of Supplementary Angle

\(\ds \)

\(=\)

\(\ds 1 - 2 \sin A \sin B \cos C - 2 \paren {\cos A \cos B - \paren {\cos A \cos B - \sin A \sin B} } \cos C\)

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

\(\ds \)

\(=\)

\(\ds 1 - 4 \sin A \sin B \cos C\)

simplifying

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Exercises $\text {XXXII}$: $15$.