How to show that $\cos\frac{2\pi}{n} + \cos\frac{4\pi}{n} + \ldots+ \cos\frac{2\pi(n-1)}{n} = -1$ for all positive integers $n$?

Question

It doesn't make sense to me, how we will even show it works for $n =1$ and we can work out induction from there but how can I show even the base case.

Thanks in Advance.

Hint: write $\cos\frac{2\pi k}{n}$ as $\frac{1}{2} \left(e^{\frac{2\pi i k}{n}} + e^{-\frac{2\pi i k}{n}}\right)$ and use summation formula for finite geometric series. — Start wearing purple, Sep 23 '13 at 19:23
Related: http://math.stackexchange.com/questions/292395/trigonometric-inequality-sin1-sin2-ldots-sinn-2/292427 (and almost certainly some others I'm forgetting about), http://en.wikipedia.org/wiki/Dirichlet_kernel — Micah, Sep 23 '13 at 20:05

score 8 · Answer 1 · answered Sep 23 '13 at 19:31

8

The result is not true for $n=1$ but is true for $n>1$.

Let $\zeta = \text{cos}(\frac{2\pi}{n}) + i\text{sin}(\frac{2\pi}{n})$.

Then $\zeta\neq 1$ and $\zeta^n = 1$ by De Moivre. These two facts tell us that $1+\zeta+\zeta^2...+\zeta^{n-1} = 0$.

Taking real parts and using De Moivre gives the result.

answered Sep 23 '13 at 19:31

fretty

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Hi, could you elaborate on why $\zeta\ne1$ and $\zeta^n=1$ implies $1+\zeta+\zeta^2+\cdots+\zeta^{n-1}=0$? – yroc Dec 01 '14 at 00:32
Can you factorise $x^n - 1$? – fretty Dec 02 '14 at 09:30
That does it. Thanks. – yroc Dec 03 '14 at 22:13

score 4 · Answer 2 · answered Sep 23 '13 at 20:14

If you want to prove the formula without complex analysis, here is the trick: Let

$$S:= \cos\frac{2\pi}{n} + \cos\frac{4\pi}{n} + \ldots+ \cos\frac{2\pi(n-1)}{n}$$

Then

$$\sin(\frac{\pi}{n}) S= \sin(\frac{\pi}{n}) \cos\frac{2\pi}{n} +\sin(\frac{\pi}{n}) \cos\frac{4\pi}{n} + \ldots+ \sin(\frac{\pi}{n}) \cos\frac{2\pi(n-1)}{n}$$

Now, as

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

you get a telescopic sum.

score 2 · Answer 3 · answered Sep 23 '13 at 20:08

2

Move the one to the left hand side, and you get the sum of the x-components of vectors arranged symmetrically around the unit circle. Of course, this sum is 0.

answered Sep 23 '13 at 20:08

Per Alexandersson

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dotslash · Answer 4 · 2013-09-24T02:41:30.043

Complex numbers with unit length are written as $e^{i\theta} = i*sin(\theta) + cos(\theta)$

The roots of the equation $x^n = 1$ are

$e^{(2k\pi/n)i}$ where $0\le k\lt n$

Each of these n values solve the equation $x^n=1$. So the summation of the roots is 0 (co-efficient of the $x^{n-1}$ multiplied by -1) .

$\sum\limits_{i=0}^{n-1} e^{(2k\pi/n)i}$ = 0
=> $ 1 + \sum\limits_{i=1}^{n-1} e^{(2k\pi/n)i}$ = 0

By equating the real part of both sides the required result is obtained.

For the given problem when $n=1$ the equation becomes $x=1$ and that is the only root. For an nth degree polynomial the summation of roots of the equation is the co-efficient of the $x^{n-1}$ multiplied with -1. In this case the summation is -1 and the equation does not hold.

From $n=2$ this co-efficient becomes zero and the equation holds.

How to show that $\cos\frac{2\pi}{n} + \cos\frac{4\pi}{n} + \ldots+ \cos\frac{2\pi(n-1)}{n} = -1$ for all positive integers $n$?

4 Answers4