<s>[INST] Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$ 
Question : Is the following true for any $m\in\mathbb N$?
  $$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$

Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let 
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies 
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.
 [/INST] Consider the polynomial $S_m(x)$, satisfying $S_m(\sin^2 \theta)=\sin^2(m\theta)$. 
These are known as spread polynomials, and may easily be derived from the Chebyshev polynomials $T_m(x)$, via $$1-2S_m(\sin^2(\theta)=1-2\sin^2(m\theta)=\cos(m(2\theta))=T_m(\cos(2\theta))=T_m(1-2\sin^2 \theta)$$ so $1-2S_m(x)=T_m(1-2x)$.
Note that 
\begin{align}
&S_{m+1}(\sin^2 \theta)+S_{m-1}(\sin^2 \theta) \\
& =\sin^2(m\theta+\theta)+\sin^2(m\theta-\theta) \\
&=(\sin(m\theta)\cos \theta+\cos(m\theta)\sin \theta)^2+(\sin(m\theta)\cos \theta-\cos(m\theta)\sin \theta)^2 \\
&=2\sin^2(m \theta)\cos^2 \theta+2\cos^2(m \theta) \sin^2(m\theta) \\
&=2(1-\sin^2 \theta)S_m(\sin^2 \theta)+2\sin^2 \theta(1-S_m(\sin^2 \theta))
\end{align}
Thus $S_{m+1}(x)=2(1-2x)S_m(x)-S_{m-1}(x)+2x$.
(We could also have used the more well known recurrence $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$ and derived the recurrence for $S_m$ from there.)
Observe that $\sin^2(\frac{k\pi}{m}), k=0, 1, \ldots, m-1$ are roots of the polynomial equation  $S_m(x)=0$. Put $S_m(x)=xP_m(x)$, so that $\sin^2(\frac{k\pi}{m}), k=1, 2, \ldots, m-1$ are roots of the polynomial equation  $P_m(x)=0$. The recurrence for $S_m$ gives $$P_{m+1}(x)=2(1-2x)P_m(x)-P_{m-1}(x)+2$$
Now if we write $P_m(x)=a_m+b_mx+x^2Q_m(x)$, it is clear by Vieta's formulas that $$\sum_{k=1}^{m-1}{\frac{1}{\sin^2(\frac{k\pi}{m})}}=\frac{\sum_{k=1}^{m-1}{\prod_{j \not =k}{\sin^2(\frac{j\pi}{m})}}}{\prod_{i=1}^{m-1}{\sin^2(\frac{k\pi}{m})}}=-\frac{b_m}{a_m}$$
We prove by induction on $m$ that $a_m=m^2, b_m=-\frac{(m^2-1)m^2}{3}$.
When $m=1$, we have $S_1(x)=x$ so $P_1(x)=1=(1^2)-\frac{(1^2-1)1^2}{3}x$ so the statement is true for $m=1$.
When $m=2$, we have $S_2(x)=4x(1-x)$ so $P_2(x)=4-4x=2^2-\frac{(2^2-1)2^2}{3}x$ so the statement is true for $m=2$.
Suppose that the statement holds for $m=i-1, i$, where $i \geq 2$. Then 
\begin{align}
P_{i+1}(x)&=2(1-2x)P_i(x)-P_{i-1}(x)+2 \\
&=2(1-2x)(a_i+b_ix+x^2Q_i(x))-(a_{i-1}+b_{i-1}x+x^2Q_{i-1}(x))+2 \\
&=(2a_i-a_{i-1}+2)+(2b_i-4a_i-b_{i-1})x+x^2(-4b_i+2Q_i(x)-Q_{i-1}(x))
\end{align}
Thus (after some algebra manipulation) 
$$a_{i+1}=2a_i-a_{i-1}+2=(i+1)^2$$ 
and 
\begin{align}
b_{i+1}=2b_i-4a_i-b_{i-1}&=-2\frac{(i^2-1)i^2}{3}-4i^2+\frac{((i-1)^2-1)(i-1)^2}{3} \\
&=-\frac{((i+1)^2-1)(i+1)^2}{3}
\end{align}
We are thus done by induction.
Now, 
$$\sum_{k=1}^{m-1}{\frac{1}{\sin^2(\frac{k\pi}{m})}}=-\frac{b_m}{a_m}=-\frac{-\frac{(m^2-1)m^2}{3}}{m^2}=\frac{m^2-1}{3}$$
</s>