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.

Hints: Your sum is related with:
$$ S=\sum_{n=1}^{90} n \sin\left(\frac{\pi n}{90}\right) = \sum_{n=0}^{89}(90-n)\sin\left(\frac{\pi n}{90}\right)\tag{1}$$
that fulfills:
$$ \color{red}{2\,S} = 90\sum_{n=1}^{89}\sin\left(\frac{\pi n}{90}\right) = \color{red}{90\cdot\cot\left(\frac{\pi}{180}\right)}.\tag{2}$$
