A series $10^{12} + 10^7 - 45\sum_{k=1}^{999}\csc^4\frac{k\pi}{1000}.$ There's a math clock with formulas for each of $1,\ldots,12$, most of which are easy. Number 11, however, intrigues me: $$10^{12} + 10^7 - 45\sum_{k=1}^{999}\csc^4\frac{k\pi}{1000}.$$
Wolfram Alpha agrees the answer is (around) 11. How does one prove this? How does one come up with this?
 A: Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+...+\tan^2 89^\circ = ?$,
the roots of 
$$\binom{2n}{2n-1}u^{n-1}-\binom{2n}{2n-3}u^{n-2}+\cdots+\binom{2n}3u-\binom{2n}1=0$$ are $\tan^2\dfrac{k\pi}{2n}$ where $0<k<n$
the roots of 
$$\binom{2n}1v^n-\binom{2n}3v^{n-1}+\cdots+\binom{2n}{2n-3}v^2-\binom{2n}{2n-1}v=0$$ are $\cot^2\dfrac{k\pi}{2n}$ where $0<k<n$
Now as $\csc\dfrac{k\pi}{1000}=\csc\left(\pi-\dfrac{k\pi}{1000}\right)=\csc\dfrac{(1000-k)}{1000}\pi$
$$\sum_{k=1}^{999}\csc^4\frac{k\pi}{1000}=\csc^4\frac{500\pi}{1000}+2\sum_{k=1}^{499}\csc^4\frac{k\pi}{1000}$$
Let $w=v+1\iff v=w-1$
$$0=\binom{2n}1(w-1)^n-\binom{2n}3(w-1)^{n-1}+\cdots+\binom{2n}{2n-3}(w-1)^2-\binom{2n}{2n-1}(w-1)$$
$$\iff\binom{2n}1w^n-w^{n-1}\left(\binom{2n}1\cdot\binom n1+\binom{2n}3\right)+w^{n-2}\left(\binom{2n}1\cdot\binom n2+\binom{2n}3\cdot\binom{n-1}1+\binom{2n}5\right)+\cdots=0$$
whose root are $w=1+\cot^2\dfrac{k\pi}{2n}$ where $0<k<n$
Now use $\sum_{r=1}^m a_r^2=(\sum_{r=1}^m a_r)^2-2\sum_{r,s=1;r>s}^m a_ra_s$
Here $2n=1000$
