Sum of Cotangents of Half Angles in Triangle
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Theorem
Let $\triangle ABC$ be a triangle.
Then:
- $\cot \dfrac A 2 + \cot \dfrac B 2 + \cot \dfrac C 2 = \cot \dfrac A 2 \cot \dfrac B 2 \cot \dfrac C 2$
Proof
\(\ds \dfrac {\cot \frac A 2 + \cot \frac B 2 + \cot \frac C 2 - \cot \frac A 2 \cot \frac B 2 \cot \frac C 2} {1 - \cot \frac B 2 \cot \frac C 2 - \cot \frac C 2 \cot \frac A 2 - \cot \frac A 2 \cot \frac B 2}\) | \(=\) | \(\ds \map \cot {\frac A 2 + \frac B 2 + \frac C 2}\) | Cotangent of Sum of Three Angles | |||||||||||
\(\ds \) | \(=\) | \(\ds \cot 90 \degrees\) | Sum of Angles of Triangle equals Two Right Angles | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Cotangent of Right Angle | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cot \frac A 2 + \cot \frac B 2 + \cot \frac C 2 - \cot \frac A 2 \cot \frac B 2 \cot \frac C 2\) | \(=\) | \(\ds 0\) | multiplying both sides by $1 - \cot \dfrac B 2 \cot \dfrac C 2 - \cot \dfrac C 2 \cot \dfrac A 2 - \cot \dfrac A 2 \cot \dfrac B 2$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cot \frac A 2 + \cot \frac B 2 + \cot \frac C 2\) | \(=\) | \(\ds \cot \frac A 2 \cot \frac B 2 \cot \frac C 2\) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(23)$