Evaluation of $ \sum_{k=0}^n \cos k\theta $

I just wanted to evaluate

$$ \sum_{k=0}^n \cos k\theta $$

and I know that it should give

$$ \cos\left(\frac{n\theta}{2}\right)\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin(\theta / 2)} $$

I tried to start by writing the sum as

$$ 1 + \cos\theta + \cos 2\theta + \cdots + \cos n\theta $$

and expand each cosine by its series representation. But this soon looked not very helpful so I need some clue about how this partial sum is calculated more efficiently …

Solutions Collecting From Web of "Evaluation of $ \sum_{k=0}^n \cos k\theta $"

\sin \frac{\theta}{2} \cdot \cos(k \theta) = \underbrace{\frac{1}{2} \sin\left( \left(k+\frac{1}{2}\right)\theta\right)}_{f_{k+1}} – \underbrace{\frac{1}{2} \sin\left( \left(k-\frac{1}{2}\right)\theta\right)}_{f_{k}}
$$ \begin{eqnarray}
\sin \frac{\theta}{2} \cdot \sum_{k=1}^n \cos(k \theta) &=& \sum_{k=1}^n \left(f_{k+1} – f_k\right) = f_{n+1}-f_1 \\ &=& \frac{1}{2} \underbrace{\sin\left(\left(n+\frac{1}{2}\right)\theta\right)}_{\sin(\alpha+\beta)}-\frac{1}{2} \underbrace{\sin \frac{\theta}{2}}_{\sin(\alpha-\beta)} = \cos(\alpha) \sin(\beta) \\
&=& \cos\left(\frac{n+1}{2}\theta\right) \sin\left(\frac{n}{2} \theta\right)
where $\alpha = \frac{n+1}{2} \theta$ and $\beta = \frac{n}{2} \theta$.

Note that $\cos(n\theta) = \Re({e^{in\theta}})$. Thus, our sum can be thought of as $\Re(\sum_{n = 0}^{N}{e^{in\theta}})$. Now, $$\sum_{n = 0}^N{e^{in\theta}} = \frac{e^{i(N+1)\theta}-1}{e^{i\theta} – 1}.$$ Now $$\frac{e^{i(N+1)\theta}-1}{e^{i\theta} – 1} = \frac{e^{i(N+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(N+1)\theta/2} – e^{-i(N+1)\theta/2}}{e^{i\theta/2} – e^{-i\theta/2}}.$$ Take the real part of the right hand side of the equality and simplify and you will get the result you want. The tricky part is knowing to break up $\frac{e^{i(N+1)\theta}-1}{e^{i\theta} – 1}$ as we did in the above equation. Let me know if you get stuck or don’t understand.