So, how do I prove that this trigonometric sum equals to zero?
$$\sum_{k = 0}^{n - 1} \cos \left(\frac{2\pi k}{n}\right) = 0$$
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That's not true, @Brad, if $n$ is odd. – Thomas Andrews May 25 '14 at 18:16
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http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression – lab bhattacharjee May 26 '14 at 05:16
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Hint: Answer the following questions: What are the roots of $x^n-1=0$? What are the real parts? How can you find the sum of roots of a given polynomial equation by looking at the coefficients?
EPS
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Well, lol, actually I tried to solve exactly this problem, but I thought that it would be better to prove that the sum of all the roots of this equation(over Complex numbers) equals to zero using De Moivre's formula and...failed – Lebesgue May 25 '14 at 18:13
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And, yeah, I know that trick with Vieta's theorem, yet for me this method seems to be more elegant – Lebesgue May 25 '14 at 18:16
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You can factorize any complex polynomial of degree $n$ in this way: $(x-r_1)\cdots (x-r_n)=0$. So the sum of $r_i$'s has something to do with the coefficient of $x^{n-1}$. – EPS May 25 '14 at 18:17
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Hint: Write $$\sum_{k = 0}^{n - 1} \cos \left(\frac{2\pi k}{n}\right) = \sum_{k = 0}^{n - 1} \operatorname{Re}\!\left(e^{i\frac{2\pi k}{n}}\right) = \operatorname{Re}\!\left(\sum_{k = 0}^{n - 1}e^{i\frac{2\pi k}{n}}\right)$$ and recall that for any $x\in\mathbb{C}\setminus\{1\}$ $$ \sum_{k = 0}^{n - 1} x^k = \frac{x^n-1}{x-1}. $$
Clement C.
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