How to prove: $11=10^{12}+10^{7}-45\sum_{n=1}^{999}\csc^4\frac{n\pi}{1000}$

Question

How to prove: \begin{equation} 11=10^{12}+10^{7}-45\sum_{n=1}^{999}\csc^4\frac{n\pi}{1000}\;. \end{equation}

This identity appears on my clock:

Answer 1

Write $\cot\dfrac{n\pi}{1000}=c_n$

Now from Sum of tangent functions where arguments are in specific arithmetic series and Question regarding $f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$

$c_n,1\le n\le1000-1$ are the roots of $$\binom{1000}1c^{999}-\binom{1000}3c^{997}+\binom{1000}5c^{995}+\cdots-\binom{1000}{999}c=0$$

$\cot\dfrac{n\pi}{1000}=0,\dfrac{n\pi}{1000}=\dfrac{(2m+1)\pi}2\iff n=(2m+1)500$

$\implies c_n,1\le n\le1000-1,n\ne500$ are the roots of $$\binom{1000}1c^{998}-\binom{1000}3c^{998}+\binom{1000}5c^{994}+\cdots-\binom{1000}{999}=0$$

So, if $d_n=c^2_n$

$d_n,1\le n\le499$ are the roots of $$\binom{1000}1d^{499}-\binom{1000}3d^{498}+\binom{1000}5c^{497}+\cdots-\binom{1000}{999}=0$$

Again $\csc^4\dfrac{n\pi}{1000}=(d_n+1)^2=d_n^2+2d_n+1$

$$\sum_{n=1}^{999}\csc^4\dfrac{n\pi}{1000}=\csc^4\dfrac{500\cdot\pi}{1000}+2\sum_{n=1}^{499}\csc^4\dfrac{n\pi}{1000}=1+2\sum_{n=1}^{499}(d_n^2+2d_n+1)$$

Now $$\sum_{n=1}^{499}d_n=\dfrac{\binom{1000}3}{\binom{1000}1}$$

$$\sum_{n=1}^{499}d^2_n=\left(\sum_{n=1}^{499}d_n\right)^2-2\cdot\dfrac{{\binom{1000}5}}{\binom{1000}1}$$