Sum of sines $\sum_{k=0}^{n} \sin(\phi +k\alpha)$ I've got the following problem. I'd like to prove that
$$\sum_{k=0}^{n} \sin(\phi +k\alpha) = \frac{\sin\left(\frac{n+1}{2}\right)\alpha + \sin\left(\phi + \frac{n\alpha}{2}\right)}{\sin\frac{\alpha}{2}}$$ Can you help me? I tried writing sines using complex numbers (that $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$), but it only made the calcuations more difficult... May somebody show me how to solve this problem? I'd be grateful.
 A: I might be wrong but I think there might be a mistake in your first post : see my answer below. The main idea here is $\Im(e^{i\theta})=\sin(\theta)$ ($\Im(z)$ denotes the imaginary part of $z$). Using this idea, you get :
$$ \sum_{k=0}^{n} \sin(\phi+k\alpha) = \Im \Big( \sum_{k=0}^{n} \exp\big( i\phi + ik \alpha \big) \Big) \; . $$
Let's work on $\displaystyle \sum_{k=0}^{n} \exp(i\phi + ik\alpha)$. Let's assume that $\alpha \notin 2\pi \mathbb{Z}$. You have :
$$ 
\begin{eqnarray*}
\sum_{k=0}^{n} \exp \big( i\phi + ik\alpha \big) & = & \exp(i\phi) \sum_{k=0}^{n} \exp(ik\alpha) \\[2mm]
 & = & \exp(i\phi) \frac{1 - \exp \big( i(n+1)\alpha \big)}{1 - \exp(i\alpha)} \\[2mm]
 & = & \exp(i\phi) \frac{\exp \big( i\frac{n+1}{2}\alpha \big) \left[ \exp \big( -i \frac{n+1}{2} \alpha \big) - \exp \big( i\frac{n+1}{2} \alpha \big) \right] }{\exp \big( i\frac{\alpha}{2} \big) \left[ \exp \big( -i\frac{\alpha}{2} \big) - \exp \big( i\frac{\alpha}{2} \big) \right] } \\[2mm]
 & = & \exp(i\phi) \exp \big( i\frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2}\alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \\[2mm]
 & = & \exp \big( i\phi + i\frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2} \alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \; . \\
\end{eqnarray*}
$$
So, 
$$ 
\begin{eqnarray*}
\Im \Big( \sum_{k=0}^{n} \exp \big( i\phi + ik \alpha \big) \Big) & = & \Im \left[ \exp \big(i\phi + i\frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2} \alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \right] \\
 & = & \sin \big( \phi + \frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2} \alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \; .\\
\end{eqnarray*}
$$

Therefore, if $\alpha \notin 2\pi \mathbb{Z}$, 
$$\sum_{k=0}^{n} \sin(\phi + k\alpha) = \sin \big( \phi + \frac{n\alpha}{2} \big) \frac{\sin \big( \frac{n+1}{2} \alpha \big)}{\sin \big( \frac{\alpha}{2} \big)} \; .$$
And if $\alpha \in 2\pi\mathbb{Z}$, 
$$ \sum_{k=0}^{n} \sin(\phi+k\alpha) = \sum_{k=0}^{n} \sin(\phi) = (n+1)\sin(\phi) \; . $$

