Sum of Sides of Triangle in terms of Circumradius and Half Angle Cosines

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Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then:

$a + b + c = 8 R \cos \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2$

where $R$ denotes the circumradius of $\triangle ABC$.


Proof

From Law of Sines:

$\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$


Hence:

\(\ds a\) \(=\) \(\ds 2 R \sin A\)
\(\ds b\) \(=\) \(\ds 2 R \sin B\)
\(\ds c\) \(=\) \(\ds 2 R \sin C\)
\(\ds \leadsto \ \ \) \(\ds a + b + c\) \(=\) \(\ds 2 R \paren {\sin A + \sin B + \sin C}\)
\(\ds \) \(=\) \(\ds 2 R \paren {4 \cos \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2}\) Sum of Sines of Angles in Triangle
\(\ds \) \(=\) \(\ds 8 R \cos \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2\)

$\blacksquare$


Sources