Sum of Cosines of Angles in Triangle

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Theorem

Let $\triangle ABC$ be a triangle.

Then:

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


Proof

First we note that:

\(\ds A + B + C\) \(=\) \(\ds 180 \degrees\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \leadsto \ \ \) \(\ds \dfrac {A + B} 2 + \dfrac C 2\) \(=\) \(\ds 90 \degrees\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {A + B} 2\) \(=\) \(\ds 90 \degrees - \dfrac C 2\)

Then:

\(\ds \cos A + \cos B + \cos C - 1\) \(=\) \(\ds 2 \map \cos {\dfrac {A + B} 2} \map \cos {\dfrac {A - B} 2} + \cos C - 1\) Cosine plus Cosine
\(\ds \) \(=\) \(\ds 2 \map \cos {\dfrac {A + B} 2} \map \cos {\dfrac {A - B} 2} - 2 \sin^2 \dfrac C 2\) Double Angle Formula for Cosine: Corollary $5$
\(\ds \) \(=\) \(\ds 2 \map \cos {90 \degrees - \dfrac C 2} \map \cos {\dfrac {A - B} 2} - 2 \sin^2 \dfrac C 2\) from $(1)$
\(\ds \) \(=\) \(\ds 2 \sin \dfrac C 2 \map \cos {\dfrac {A - B} 2} - 2 \sin^2 \dfrac C 2\) Cosine of Complement equals Sine
\(\ds \) \(=\) \(\ds 2 \sin \dfrac C 2 \paren {\map \cos {\dfrac {A - B} 2} - \sin \dfrac C 2}\) factorising
\(\ds \) \(=\) \(\ds 2 \sin \dfrac C 2 \paren {\map \cos {\dfrac {A - B} 2} - \map \sin {90 \degrees - \dfrac {A + B} 2} }\) from $(1)$
\(\ds \) \(=\) \(\ds 2 \sin \dfrac C 2 \paren {\map \cos {\dfrac {A - B} 2} - \map \cos {\dfrac {A + B} 2} }\) Sine of Complement equals Cosine
\(\ds \) \(=\) \(\ds 2 \sin \dfrac C 2 \paren {2 \sin \dfrac A 2 \sin \dfrac B 2}\) Cosine minus Cosine
\(\ds \) \(=\) \(\ds 4 \sin \dfrac A 2 \sin \dfrac B 2 \sin \dfrac C 2\) simplification

Hence the result.

$\blacksquare$


Sources

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