trigonometry
How to compute the sum of $\cot^2\left(\frac{\pi}{9}\right)+\cot^2\left(\frac{2\pi}{9}\right)+\cot^2\left(\frac{4\pi}{9}\right)=~?$
The answer is $9$.
I tried to use the formula $\cot (2\theta)=\dfrac{\cot^2\theta-1}{2\cot\theta}$ but it is getting more and more complicated.
Proof without words: $\displaystyle\cot(\theta/2)=\frac{\color{green}{\sin(\theta)}}{\color{red}{1-\cos(\theta)}}$
$\hspace{3.5cm}$
Recall the expansion: $$\begin{align}\tan nx &= \dfrac{\binom{n}{1}\tan x - \binom{n}{3}\tan^3 x + \cdots }{1-\binom{n}{2}\tan^2 x + \binom{n}{4}\tan^4 x + \cdots } \\&= \frac{\binom{n}{1}\cot^{n-1} x - \binom{n}{3}\cot^{n-3} x + \cdots }{\cot^{n} x - \binom{n}{2}\cot^{n-2} x + \binom{n}{4}\cot^{n-4} x + \cdots }\end{align}$$
Now, $\displaystyle \left\{\frac{k\pi}{n}\right\}_{k=1}^{n-1}$ are the roots of the equation: $\displaystyle \tan nx = 0$
Thus, $\displaystyle \binom{n}{1}\cot^{n-1} x - \binom{n}{3}\cot^{n-3} x + \cdots = 0$ whenever, $x = \dfrac{k\pi}{n}$ for $1 \le k \le n-1$.
Thus, using Vieta's formula we might write:
$$\begin{align*}\sum\limits_{k=1}^{n-1} \cot^2 \frac{k\pi}{n} &= \left(\sum\limits_{k=1}^{n-1} \cot \frac{k\pi}{n}\right)^2 - 2\sum\limits_{1 \le k_1 < k_2 \le n-1} \cot \frac{k_1\pi}{n}\cot \frac{k_2\pi}{n}\\&= 0 + 2\frac{\binom{n}{3}}{\binom{n}{1}}\\&= \frac{(n-1)(n-2)}{3}\end{align*}$$
Best Answer
Let $S= \cot^2\frac{\pi}{9}+\cot^2\frac{2\pi}{9}+\cot^2\frac{4\pi}{9}$ and note that $\cot \frac{4\pi}9=\tan\frac{\pi}{18}$
$$\begin{align} S&=\frac{1}{\sin^2\frac{\pi}9} -1+\frac{1}{\sin^2\frac{2\pi}9}-1+\frac{1}{\cos^2\frac{\pi}{18}}-1 \\ \\ &=\frac{1}{\sin^2\frac{2\pi}9}+\frac{1+4\sin^2\frac{\pi}{18}}{\sin^2\frac{\pi}{9}}-3\\ \\ &=\frac{1+4\cos^2\frac{\pi}{9}\left(1+4\sin^2\frac{\pi}{18}\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{7+6\cos\frac{2\pi}{9}-4\cos\frac{\pi}{9}-4\cos\frac{\pi}{9}\cos\frac{2\pi}{9}}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{6-6\cos\frac{\pi}{9}+6\cos\frac{2\pi}{9}}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{6\cos\frac{\pi}{9}\left(2\cos\frac{\pi}{9}-1\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{24\cos\frac{\pi}{9}\left(\sin\frac{2\pi}{9}\sin\frac{\pi}{9}\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=12-3\\ \\ &=9 \end{align}$$