Prove that $\frac{1}{90} \sum_{n=1}^{90} 2n \cdot \sin((2n)^\circ) = \cot (1^{\circ})$ Show that $$\frac{(2\sin(2^\circ)) + (4\sin(4^\circ))+ (6\sin(6^\circ)) + \ldots +(180\sin(180^\circ))}{90} = \cot(1^\circ).$$
I used a lot of steps, and typing it all down on here would take me an hour, but here are my last few steps up to the point where I got stuck: 
$$180 (\sin(2^\circ) + \sin(4^\circ) + \sin(6^\circ) +.....+ \sin(88^\circ)) + 90$$
Using the product-to-sum formulas, I then reduced it down to 
$$180(2\sin(45^\circ)    [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90$$
Simplifying a bit more gives me: 
$$\frac{180\sqrt{2}   [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90}{90}$$ $$= 2\sqrt{2}   [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90.$$
Now I am stuck. What can I do next to achieve the final result? 
 A: $$S=\sum_{n=1}^{90}2(91-n)\sin(2n^\circ) = \sum_{n=1}^{90}2\sum_{k=1}^{n}\sin(2k^\circ).$$
Since:
$$2(\sin 1^{\circ})\sum_{k=1}^{n}\sin(2k^\circ)=\sum_{k=1}^{n}\left(\cos((2k-1)^{\circ})-\cos((2k+1)^{\circ})\right)=\cos 1^\circ-\cos((2n+1)^\circ)$$
it follows that:
$$ S = \frac{1}{\sin 1^\circ}\left(90\cos 1^\circ-\sum_{n=1}^{90}\cos((2n+1)^{\circ})\right),$$
but with the same trick shown in this other question we have that:
$$\sum_{n=1}^{90}\cos((2n+1)^{\circ})=-2\cot 1^\circ,$$
hence $\color{red}{S=92\cot 1^{\circ}}$. On the other hand, since:
$$ \sum_{n=1}^{90}2\sin(2n^\circ) = 2\cot 1^{\circ} $$
always by the same trick, it follows that:
$$\sum_{n=1}^{90}2n\sin(2n^\circ)=(91\cdot 2-92)\cot 1^\circ= \color{blue}{90\cot 1^{\circ}} $$
as wanted.

For a complex-analytic derivation, set $t=\frac{\pi}{180}, z=e^{2it}$ and consider that:
$$\sum_{n=1}^{90}2n \sin(2nt) = \Im\sum_{n=1}^{90}2n z^n=2\Im\left(\frac{z-91z^{91}+90 z^{92}}{(1-z)^2}\right).$$
Since $z^{90}=-1$, the last expression simplifies to:
$$\begin{eqnarray*}2\Im\left(\frac{92 z-90 z^{2}}{(1-z)^2}\right)&=&2\Im\left(90\frac{z}{1-z}+\frac{2z}{(1-z)^2}\right)=90\cdot\Im\left(\frac{2e^{2it}}{1-e^{2it}}\right)\\&=&90\cdot\frac{\Im(ie^{it})}{\sin t}=90\cot t.\end{eqnarray*}$$
A: Your sum is the imaginary part of a (complex) arithme-geometric series. Sum $2n\,e^{i(2n)^{\circ}}$ instead and take the imaginary part when this is done.
