Sum of Sines of Angles in Triangle

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Theorem

Let $\triangle ABC$ be a triangle.

Then:

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


Proof

First we note that:

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

Then:

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

$\blacksquare$


Sources

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