Trigonometry Olympiad problem: Evaluate $1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$ 
Find the value of
  $$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$

My attempt
I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$ but I have gone no where with it!
I also tried using double angle formula for the angles which are even.

Use $\sin(\theta)=\sin(180^\circ-\theta)$. Let the summation as $S$. Then
\begin{align}
2S&=\sum_{k=1}^{90}(k\sin(2k^\circ)+(90-k)\sin((180-2k)^\circ)) \\
&=90\sum_{k=1}^{90}\sin(2k^{\circ})
\end{align}
Now use $-2\sin(2k^\circ)\sin(1^\circ)=\cos((2k+1)^\circ)-\cos((2k-1)^\circ)$, then
\begin{align}
S=45\sum_{k=1}^{90}\sin(2k^{\circ})=-\frac{45}{2\sin(1^\circ)}\sum_{k=1}^{90}(\cos((2k+1)^\circ)-\cos((2k-1)^\circ)=\frac{45\cos(1^\circ)}{\sin(1^\circ)}=45\cot(1^\circ)
\end{align}
