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?
As in this answer, we will use the fact that $$ \frac{n/z}{z^n-1} $$ has a residue of $1$ when $z^n=1$ and residue $-n$ when $z=0$.
The size of the integrand is $\sim\frac{32}{z^{10}}$ when $|z|$ is large. Therefore, $$ \begin{align} 0=\lim_{r\to\infty}\oint_{|z|=r}\color{#C00000}{\left(\frac{2000/z}{z^{2000}-1}-\frac{2/z}{z^2-1}\right)}\color{#00A000}{\left(\frac{2z}{z^2-1}\right)^4}\,\mathrm{d}z \end{align} $$ The red factor of the integrand has singularities at $e^{\pi ik/1000}$ for $1\le k\le 999$ and $1001\le k\le1999$ with a residue of $1$ and a singularity at $z=0$ with a residue of $-1998$. The green factor of the integrand removes the singularities at $z=0$, but introduces a couple at $z^2=1$. The green factor also modifies the residues at the singularities of the red factor so that the singularity at $e^{\pi ik/1000}$ has a residue of $\csc^4\left(\frac{\pi k}{1000}\right)$.
Since the sum of the residues of the product at $z^2=1$ added to the sum of the residues at the other singularities is $0$, we get that the sum of the residues at $z^2=1$ is $$ -2\sum_{k=1}^{999}\csc^4\left(\frac{\pi k}{1000}\right) $$ Expanding the integrand at $z=1$ we get $$ -\frac{999}{(z-1)^4}+\frac{332334}{(z-1)^3}+\frac{167166}{(z-1)^2}-\frac{111112222221}{5(z-1)}+O(1) $$ Expanding the integrand at $z=-1$ we get $$ \frac{999}{(z+1)^4}+\frac{332334}{(z+1)^3}-\frac{167166}{(z+1)^2}-\frac{111112222221}{5(z+1)}+O(1) $$ Thus, $$ \sum_{k=1}^{999}\csc^4\left(\frac{\pi k}{1000}\right)=\frac{111112222221}{5} $$ Finally, $$ 10^{12}+10^7-45\cdot\frac{111112222221}{5}=11 $$
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 $2n=1000$
