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

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 $a:=2^{\circ}, \, n:=90$.

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: $$ \color{red}{2\,S} = 90\sum_{n=1}^{89}\sin\left(\frac{\pi n}{90}\right) = \color{red}{90\cdot\cot\left(\frac{\pi}{180}\right)}.\tag{2}$$

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}