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.
 A: 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:
$$ \color{red}{2\,S} = 90\sum_{n=1}^{89}\sin\left(\frac{\pi n}{90}\right) = \color{red}{90\cdot\cot\left(\frac{\pi}{180}\right)}.\tag{2}$$
A: 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}
A: $$\sin\theta=\sin(\pi-\theta)$$
$$\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}\cdots 88\sin4^{\circ}+89\sin 2^{\circ}+ 90\sin 0^{\circ}$$
$$90(\sin 2^{\circ} +\sin 4^{\circ} + \sin6 ^{\circ}\cdots \sin 90^{\circ}) -45\sin90^{\circ} $$
Now it is easy to evaluate this sum .
$\sin \alpha +\sin (\alpha+\beta)+\sin (\alpha+2\beta) +\cdots n$ terms
$$=\frac{\sin{\frac{n\beta}{2}}}{\sin{\frac{\beta}{2}}}{\sin\left[ {\alpha + \frac{\beta}{2}{(n-1)}}\right]}$$
$$90\sin 45^{\circ}\left[\frac{\sin 46^{\circ}}{\sin 1^{\circ}}\right] -90\sin 45^{\circ}.\cos45^{\circ}$$
$$90\sin 45^{\circ}\left[\frac{\sin 46^{\circ}}{\sin 1^{\circ}} -\cos45^{\circ}\right]$$
Now it is almost done simplify it and get
$$45\cot 1^{\circ} $$
A: An approach. One may write
$$
\begin{align}
\sum_{k=1}^nk\sin(ka)&=\sum_{k=1}^n\frac{d}{da}(-\cos(ka))
\\\\&=-\frac{d}{da}\sum_{k=0}^n\cos(ka)
\\\\&=-\frac{d}{da}\left(\frac{1}{2}+\frac{\sin\left[(n+\frac12)a\right]}{2\sin \frac a2} \right)
\\\\&=\frac{(n+1) \sin(na)-n \sin((n+1)a)}{4\sin^2 \frac a2}
\end{align}
$$ then one may take $a:=2^{\circ}, \, n:=90$.
A: 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
$$x=2\frac{\pi}{180}=\frac{\pi}{90}$$
