Cosine of Half Angle in Triangle/Proof
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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 |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Half angle formulae