Find formula of sum $\sin (nx)$ I wonder if there is a way to calculate the
$$S_n=\sin x + \sin 2x + … + \sin nx$$ 
but using only derivatives ? 
 A: Using telescopic sums:
$$ \sin(mx)\sin(x/2) = \frac{1}{2}\left(\cos\left((m-1/2)x\right)-\cos\left((m+1/2)x\right)\right)$$
Hence:
$$ S_n \sin\frac{x}{2} = \frac{1}{2}\left(\cos\frac{x}{2}-\cos\left(\left(n+\frac{1}{2}\right)x\right)\right)=\sin\frac{nx}{2}\cdot\sin\frac{(n+1)x}{2}.$$
A: You may write
$$
\begin{align}
\sum_{k=1}^{n} \sin (k\theta)&=\Im \sum_{k=1}^{n} e^{ik\theta}\\\\
&=\Im\left( e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}\right)\\\\
&=\Im\left( e^{i\theta}\frac{e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\
&=\Im\left( e^{i\theta}\frac{e^{in\theta/2}\left(2i\sin(n\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\
&=\Im\left( e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\
&=\Im\left( \left(\cos ((n+1)\theta/2)+i\sin ((n+1)\theta/2)\right)\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\
&=\frac{\sin ((n+1)\theta/2)}{\sin(\theta/2)}\sin(n\theta/2).
\end{align}
$$
