<s>[INST] 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.
 [/INST] As a complement to other answers, you should know that the following identities are valid
$$ \begin{align} 
2 \sin \frac{1}{2} x \sum_{k=1}^{n} \cos kx &= +\sin(n+\frac{1}{2})x - \sin\frac{1}{2}x \\
2\sin \frac{1}{2} x \sum_{k=1}^{n} \sin kx &= -\cos(n+\frac{1}{2})x + \cos\frac{1}{2}x
\end{align}$$
Thsese formulas can easily be proved by moving $\sin \frac{1}{2} x$ into the summation, using the identities
$$ \begin{align}
2 \sin \frac{1}{2} x \cos kx &= + \sin(k + \frac{1}{2})x - \sin(k - \frac{1}{2})x \\
2 \sin \frac{1}{2} x \sin kx &= - \cos(k + \frac{1}{2})x + \cos(k - \frac{1}{2})x
\end{align}$$
and the telescoping property of partial sums. In your example, we have
$$x=2\frac{\pi}{180}=\frac{\pi}{90}$$
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