Why doesn't this work?

Question

https://i.stack.imgur.com/0DZmF.jpg

I'm trying to find the sum of $\cos(\frac{\pi}{k}) + \cos(\frac{2\pi}{k}) + \cos(\frac{3\pi}{k}) + ... + \cos(\frac{n\pi}{k})$

I multiplied it by $2\sin(\frac{\pi}{k})$ and realised that terms sort of cancelled out. I got some formula for the sum but it did not work when I tested it and I dunno why not.

Thanks.

Related : http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro — lab bhattacharjee, Dec 25 '13 at 05:49
You could also multiply by $2\sin(\tfrac{\pi}{2k})$ to get neighboring terms to cancel. — Lutz Lehmann, Dec 25 '13 at 22:30

mathlove · Answer 1 · 2013-12-25T06:14:52.383

Your calculation is correct. What you did is to lead the following :

$$2\sin\left(\frac{\pi}{k}\right)\sum_{i=1}^{n}\cos \left(\frac{i\pi}{k}\right)=\sin\left(\frac{n\pi}{k}\right)+\sin\left(\frac{(n+1)\pi}{k}\right)-\sin\left(\frac{\pi}{k}\right).$$

However, note that $$\sum_{i=1}^{n}\cos \left(\frac{i\pi}{k}\right)=\frac{\sin\left(\frac{n\pi}{k}\right)+\sin\left(\frac{(n+1)\pi}{k}\right)-\sin\left(\frac{\pi}{k}\right)}{2\sin\left(\frac{\pi}{k}\right)}.$$ holds only when $\sin\left(\frac{\pi}{k}\right)\not =0$.

Why doesn't this work?

1 Answers1