Prove that $\cos\alpha+\cos2\alpha+\cdots+\cos n\alpha=\frac{1}{2}\left(\frac{\sin\left(n+\frac{1}{2}\right)\alpha}{\sin\frac{1}{2}\alpha}-1\right)$ I have to prove using mathematical induction that:
$$\cos\alpha+\cos2\alpha+\cdots+\cos n\alpha=\frac{1}{2}\left(\frac{\sin\left(n+\frac{1}{2}\right)\alpha}{\sin\frac{1}{2}\alpha}-1\right)$$
If I substitute n equals one then I'm giving a such thing as:
$$\cos\alpha=\frac{1}{2}\left(\frac{\sin\frac{3}{2}\alpha}{\sin\frac{1}{2}\alpha}-1\right)$$
But I don't what I should do to prove nextly and that this equation is completed for n+1.

Hint: Write
\begin{align}
\cos\alpha = \operatorname{Re} e^{i\alpha}
\end{align}
then the sum becomes
\begin{align}
\operatorname{Re}\left(e^{i\alpha}+e^{i2\alpha}+\ldots+e^{in\alpha} \right)
\end{align}
which is a geometric series.
Edit: It's not hard to see
\begin{align}
e^{i\alpha}+e^{i2\alpha} + \ldots + e^{in\alpha} = \frac{e^{i\alpha}-e^{i(n+1)\alpha}}{1-e^{i\alpha}}.
\end{align} 
I will leave it to the reader to put it in real form. 
