# Trig identity

1. Jan 18, 2008

### ehrenfest

[SOLVED] trig identity

1. The problem statement, all variables and given/known data
Can someone help me prove that
$$\sum_{k=1}^{(n-1)/2}\cos(2 \pi k / n) = -1/2$$

where n is an odd number?

2. Relevant equations

3. The attempt at a solution
I don't know where to start. You can easily verify it is true for n=3. But after that things get complicated. Maybe induction... I would go to complex exponentials but I reduced a problem involving complex exponentials to this, so I don't really want to go back...

2. Jan 19, 2008

### chaoseverlasting

Actually, you would have to use complex numbers. I suppose you would start with $$x=cos\frac{\pi}{2} +i sin\frac{\pi}{2}$$.

3. Jan 19, 2008

### ehrenfest

And use deMoivre? I don't see how that will help.

4. Jan 19, 2008

### NateTG

I think there's a nice geometric argument available.

Can you show that:
$$\sum_{k=1}^{(n)}\cos(\frac{2 \pi k }{n}) = 0$$
And that:
$$\sum_{k=1}^{\frac{n-1}{2}}\cos(\frac{2 \pi k}{n})=\sum_{k=\frac{n+1}{2}}^{n}\cos(\frac{2 \pi k}{n})$$

Last edited: Jan 19, 2008
5. Jan 19, 2008

### ehrenfest

I assume the parenthesis around the n don't mean anything.

And the answer is no.That was actually the original problem. I thought I reduced the problem, but maybe I just made it worse...

I might as well state the original problem: Find the electric field at the center of regular 13-gon.

Last edited: Jan 19, 2008
6. Jan 19, 2008

### NateTG

Yeah, it's late, and I'm always sloppy.

It seems that, assuming the 13-gon has a uniform static charge, the field at the center is zero by symetry. (Assume by contradiction that it's not. Then there is at least one line from the center to a vextex that is not parralel to the field at the center of the polyon. Rotate by $\pi$ using that line as an axis. This will produce the same charge geometry, but a different electric field, which is a contradiction.)

Alternatively: The terms of the sum correspond to the x-component of the segments of a path around the perimiter of a regular n-sided polygon. Since the perimiter is closed, their sum must be zero. (Consider the vector sum of$<\cos \frac{2 \pi k}{n}, \sin \frac{2 \pi k}{n}>$.)

7. Jan 19, 2008

### ehrenfest

Hmm. Its not obvious to me why the vector sum of the vertices of a regular n-gon must be 0. Obviously, if you sum the vector connecting vertex 0 to vertex 1, the vector connecting vertex 1 to vertex 2, and so on, then you will get zero, but that is not the same thing.

It is pretty obvious when you look at the symmetry. But there has got to be a numerical way to show it!

8. Jan 19, 2008

### morphism

There's a slick way to do it with complex numbers: Consider the n nth roots of unity, generated by cos(2*pi*i/n) + i sin(2*pi*i/n). Since these n numbers are precisely the roots of the polynomial x^n - 1, their sum must be 0 (-1 * the coefficient of x^(n-1)). Now take the real part of this sum.

9. Jan 19, 2008

### Gib Z

I would have also done it complex numbers, but I would have done it another way, its probably longer though;

Write $$\cos (\frac{2\pi k}{n}) = Re\left( exp(\frac{2\pi ik}{n} )\right)$$ and then interchange the summand operator with the Re operator (extracts only the real part of the number), then it becomes an ordinary geometric series.

10. Jan 19, 2008

### ehrenfest

You mean
$$Re\left( exp(\frac{2\pi i1}{n} )\right), Re\left( exp(\frac{2\pi i2}{n} )\right),Re\left( exp(\frac{2\pi i3}{n} )\right),...$$

with common ratio

$$Re\left( exp(\frac{2\pi i1}{n} )\right)$$

?

Is that really a geometric series?

Last edited: Jan 19, 2008
11. Jan 19, 2008

### Gib Z

Yes? Why not?

12. Jan 19, 2008

### ehrenfest

Because Re(x)Re(y) is not equal to Re(xy).
Take for example x=y=i.

13. Jan 19, 2008

### Gib Z

How does that relate to this?

14. Jan 20, 2008

### ehrenfest

A geometric series has the form x,x^2,x^3,...

Letting $$x = Re\left( exp(\frac{2\pi i1}{n} )\right)$$ gives

$$Re\left( exp(\frac{2\pi i1}{n} )\right),Re\left( exp(\frac{2\pi i1}{n} )\right)Re\left( exp(\frac{2\pi i1}{n} )\right),Re\left( exp(\frac{2\pi i1}{n} )\right)Re\left( exp(\frac{2\pi i1}{n} )\right)Re\left( exp(\frac{2\pi i1}{n} )\right),...$$

which is not the same as

$$Re\left( exp(\frac{2\pi i1}{n} )\right), Re\left( exp(\frac{2\pi i2}{n} )\right),Re\left( exp(\frac{2\pi i3}{n} )\right),...$$

15. Jan 20, 2008

### Gib Z

We are interchanging the summation order with the Re operator. I'm sure you agree that we can do that?

16. Jan 21, 2008

### NateTG

Each edge vector of the n-gon can be translated into a spoke vector by means of a fixed rotation and scaling. Since linear transforms are distributive over vectors, this means that the vector sum of the spokes is the same as this rotation and scaling applied to the vector sum of the edges, but we know that that's zero.

...

Gib Z is applying the Euler Identity:
$$e^{i \theta} =\cos \theta + i \sin \theta$$
to express the sum of the x components as:
$$\rm{Real} \sum_{k=1}^{n} \left({e^{i\frac{2 \pi k}{n}}\right)^k$$

17. Jan 21, 2008

### ehrenfest

I don't know what this means.

Again, Re(x)Re(y) is not equal to Re(xy). Please write out the first few terms of the geometric series you are talking about and show the multiplication by the common factor. I am relatively sure that

$$Re\left( exp(\frac{2\pi i1}{n} )\right), Re\left( exp(\frac{2\pi i2}{n} )\right),Re\left( exp(\frac{2\pi i3}{n} )\right),...$$

is not a geometric series.

18. Jan 21, 2008

### NateTG

Gib Z means that he's distributing 'Real Part' over the sum. (The real part of a sum is the sum of the real parts provided the sum converges.)

$$\sum_{k=1}^{n}\left({e^{i\frac{2 \pi }{n}}\right)^k=e^{i\frac{2 \pi} {n}} + e^{2 i \frac{2 pi}{n}} + e ^ {3 i \frac{2 pi}{n}}...$$

$$\rm{Real Part} \left(\sum_{k=1}^{n}\left({e^{i\frac{2 \pi }{n}}\right)^k\right) = \cos \frac{1(2 \pi)}{n} + \cos \frac{2(2\pi)}{n} + \cos \frac{3 (2 \pi)}{n}...$$

19. Jan 21, 2008

### ehrenfest

I see what you two mean.

But please confirm the following statement:

$$Re\left( exp(\frac{2\pi i1}{n} )\right), Re\left( exp(\frac{2\pi i2}{n} )\right),Re\left( exp(\frac{2\pi i3}{n} )\right),...$$

IS NOT A GEOMETRIC SERIES.

20. Jan 22, 2008

### NateTG

It's not, but it's close to one.