Cosine of Half Angle in Triangle

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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 $C$.

Let $s$ denote the semiperimeter of $\triangle ABC$.


Then:

$\cos \dfrac C 2 = \sqrt {\dfrac {s \paren {s - c} } {a b} }$


Proof

\(\ds \cos C\) \(=\) \(\ds \dfrac {a^2 + b^2 - c^2} {2 a b}\) Law of Cosines
\(\ds \leadsto \ \ \) \(\ds 2 \cos^2 \dfrac C 2 - 1\) \(=\) \(\ds \dfrac {a^2 + b^2 - c^2} {2 a b}\) Double Angle Formula for Cosine: Corollary $1$
\(\ds \leadsto \ \ \) \(\ds 2 \cos^2 \dfrac C 2\) \(=\) \(\ds \dfrac {a^2 + b^2 - c^2} {2 a b} + 1\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {a^2 + 2 a b + b^2} - c^2} {2 a b}\) common denominator
\(\ds \) \(=\) \(\ds \dfrac {\paren {a + b}^2 - c^2} {2 a b}\) Square of Sum
\(\ds \) \(=\) \(\ds \dfrac {\paren {a + b + c} \paren {a + b - c} } {2 a b}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \dfrac {2 s \cdot 2 \paren {s - c} } {2 a b}\) Definition of Semiperimeter
\(\ds \leadsto \ \ \) \(\ds \cos^2 \dfrac C 2\) \(=\) \(\ds \dfrac {s \paren {s - c} } {a b}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \cos \dfrac C 2\) \(=\) \(\ds \sqrt {\dfrac {s \paren {s - c} } {a b} }\) taking square root of both sides

$\blacksquare$


Also see


Sources

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