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(English is not my native language, so I apologize if I fail to use the right technical terms)

I am stuck in proving the following. I'll explain how far I got and maybe someone can help me out by explaining the last part to me.


Show that:

$$\sum_{k=1}^{n-1}{\sin{\frac{2\pi k}{n}}} = 0$$


So far so good. This is how far I got:

  1. If $z$ is an $n$th root of unity, $z^n = 1$, thus $$0 = z^n - 1 = z^n - 1^n = (z - 1)\cdot \sum_{k=0}^{n-1}{z^k}$$
  2. If $z \neq 1$ this means that $\sum_{k=0}^{n-1}{z^k} = 0$
  3. Let $z$ be the root: $e^{\frac{2\pi \mathrm{i}}{n}}$, thus $$\sum_{k=0}^{n-1}{{e^{\frac{2\pi k \mathrm{i}}{n}}}} = 0$$
  4. With Euler's formula, this is $$1 + \sum_{k=1}^{n-1}{\left(\cos{\frac{2\pi k}{n}} + \mathrm{i} \sin{\frac{2\pi k}{n}}\right)} = 0$$
  5. Let's rearrange that a bit: $$1 + \sum_{k=1}^{n-1}{\cos{\frac{2\pi k}{n}}} + \mathrm{i} \sum_{k=1}^{n-1}{\sin{\frac{2\pi k}{n}}} = 0$$

Okay, so this is where I got stuck. I thought this would somehow lead to a solution, but I couldn't figure it out. After a while, and trying different approaches, I looked at the sample solution: They do exactly what I am doing until step 3 (not as detailed) and then simply conclude with:

The imaginary part provides the desired equation.

Somehow I am unable to figure out why. All I can say from the equation above is that the cosine-sum needs to be $-1$ in order for the sine-sum to be 0. Am I missing something obvious here?

Thanks a lot for any help in explaining that!

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    $\begingroup$ If $a + ib = 0$ with real $a$ and $b$, then $a = 0$ and $b = 0$. Here, $b$ is your sum of sines. $\endgroup$ Aug 23, 2014 at 20:52
  • $\begingroup$ Duh. Of course, how did I not see that? Thx. Unfortunately I cannot upvote comments. $\endgroup$
    – johnnycrab
    Aug 23, 2014 at 20:57
  • $\begingroup$ @johnnycrab, See math.stackexchange.com/questions/17966/… $\endgroup$ Aug 24, 2014 at 5:11

2 Answers 2

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Non-complex answer: let $S=\sum_{k=1}^{n-1}\sin\frac{2\pi k}{n}$. If $n=1$, the problem is trivial so assume $n>1$. Multiply $S$ by (the nonzero factor) $\sin\frac{\pi}{n}$ and use the formula $$ \sin A\sin B=\frac{1}{2}(\cos(A-B)-\cos(A+B)). $$ You will get a telescoping sum that simplifies to give you: $$ \sin(\pi/n)S=\frac{1}{2}(\cos(\pi/n)-\cos((2n-1)\pi/n))=\frac{1}{2}(\cos(\pi/n)-\cos(2\pi-\pi/n))=0 $$ because $\cos(2\pi-\pi/n)=\cos(-\pi/n)=\cos(\pi/n)$.

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Take the imaginary part of each side of the equation. On the left side you have your sum, on the right side you have zero.

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