Prove that $\cos\alpha+\cos2\alpha+\cdots+\cos n\alpha=\frac{1}{2}\left(\frac{\sin\left(n+\frac{1}{2}\right)\alpha}{\sin\frac{1}{2}\alpha}-1\right)$

Question

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.

Answer 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.