# Evaluation of a trigonometric partial sum

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 ...

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Whenever you already know the answer in terms of a natural number n, the first thought to prove the assertion should always be induction. In this case, it solves it rather simply. Try it. – Rijul Saini Sep 6 '12 at 22:44
Wait, you wrote the expanded sum with a 1 at the beginning, but there's no $k=0$ term in the sigma form... – Ben Millwood Sep 7 '12 at 0:48
Thanks @BenMillwood, the index k should start from zero of course ... – Dilaton Sep 7 '12 at 8:16

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

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Ah thanks Sasha, I almost suspected that this involves some nifty trigonometric tricks ... – Dilaton Sep 6 '12 at 21:04

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.

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That's one way of doing it... why not just multiply by the complex conjugate of the denominator instead? – Ben Millwood Sep 7 '12 at 0:46
Also, I'm not sure your geometric formula is quite right. I think your result is a factor of $e^{i\theta}$ short. – Ben Millwood Sep 7 '12 at 0:49
And the next line is wrong as well – that numerator is zero! I started off thinking this was good, but now I'm downvoting it for containing an arithmetic error in pretty much every equation. – Ben Millwood Sep 7 '12 at 0:51
Hi Ben. I'm sorry for the mistakes. I was really sloppy with the derivation because i wrote this fairly quickly. I hope this is right now. – Shankara Pailoor Sep 7 '12 at 1:28
It still has mistakes and typos. You should have $\mathrm{e}^{i k \theta}$ instead of $\mathrm{e}^{i n \theta}$ in the sum. Moreover the correct evaluation of the sum is: $$\sum_{k=1}^N \exp(i k \theta) = \frac{\exp(i (N+1)\theta) - \exp(i \theta)}{\exp(i \theta)-1}$$ – Sasha Sep 7 '12 at 2:34