# A proof

1. Nov 5, 2005

### Tony11235

Prove that for any real number $$\alpha \neq 0, \pm 2 \pi, \pm 4 \pi$$. . . one has
$$\frac{1}{2} + \cos \alpha + . . . \cos N \alpha = \frac{sin[(N+\tfrac 1 {2}) \alpha ]}{ 2 \sin (\tfrac \alpha {2})}$$
I know it involves Euler's formula that relates sin and cos to the exponential function, and the formula for the sum of a geometric series. I'm not sure how to start this simple proof. Any help with be great.

Last edited: Nov 5, 2005
2. Nov 5, 2005

### Tide

HINT: Use Euler's formula and think geometric series! :)

3. Nov 6, 2005

### Tony11235

All I've came up with is the expression

$$\frac{2\cos(N \alpha) - \cos(\alpha) -1}{2(1 - \cos(\alpha))}$$

When I plug in pi for alpha in the expression above, I get the same expression when I plug pi in for alpha in

$$\frac{sin[(N+\tfrac 1 {2}) \alpha ]}{ 2 \sin (\tfrac \alpha {2})}$$

It would be a little hard, or time consuming to simplify the first expression with trig identities to the second. Or maybe thats just what I think.

Last edited: Nov 6, 2005
4. Nov 9, 2005

### Curious3141

Sorry that this comes so late, but have you solved the problem yet ?

I tried the "direct" method with complex numbers and geometric series and it gets nowhere fast, as you observed. It's very much easier to just prove this one by induction.

Verify for N = 0 (trivial).

Do the inductive step and group the terms together under the common denominator $2\sin{\frac{\alpha}{2}}$, making the observation that you can only do this for nonzero values for the denominator (hence the constraint stated in the question). You will find that to simplify this, you need at one point to reexpress $\sin{(N+\frac{1}{2})\alpha}$ as $\sin{(N+1 - \frac{1}{2})\alpha}$ and expand that out to cancel some terms. In the end, you'll be able to reduce it very easily to the "expected" RHS.

Post again if you're able to solve it, or if you need further help.

5. Nov 9, 2005

### amcavoy

How did you get that? I am coming up with what you have times -1:

$$\frac{1}{2}+\cos{a}+\cdots +\cos{na}=\left(\sum_{k=0}^{n}\cos{kx}\right)-\frac{1}{2}$$

then I solve and come up with something that's off by a multiple of -1. Why is this? Is my series above correct?

Because once I get to your series I can get it into the other form; I'd like to know how you got there. I'll post later, I didn't mean to go off topic here, sorry.

6. Nov 9, 2005

### Tony11235

I don't know what I was doing. It had to be totally wrong. The best way is the following....

$$\frac{1}{2} + \cos \alpha + . . . \cos N \alpha$$ so

$$\frac{1}{2} (e^{-\imath N \alpha} + e^{-\imath (N - 1) \alpha} + . . . . . + e^{-\imath \alpha} + 1 + e^{\imath \alpha} + . . . . . + e^{\imath N \alpha} ) = \frac{1}{2} \cdot \frac{e^{-\imath N \alpha} (1 - e^{\imath (2N + 1) \alpha}) }{1 - e^{\imath \alpha} }$$

then

$$\frac{1}{2} \cdot \frac{e^{-\imath (N + 1) \alpha} - e^{\imath (N + 1) \alpha} }{e^{\tfrac {-\alpha} {2}} - e^{\tfrac {\alpha} {2}}}} }$$

$$\frac{1}{2} \cdot \frac{-2\imath \sin[(N + \tfrac 1 {2})\alpha] }{-2\imath \sin(\tfrac \alpha {2}) }$$

and finally

$$\frac{sin[(N+\tfrac 1 {2}) \alpha ]}{ 2 \sin (\tfrac \alpha {2})}$$

Last edited: Nov 10, 2005
7. Nov 10, 2005

### Curious3141

Yes, this is a very elegant method (I think there's a typo in the LaTex somewhere there, but the proof is evident). I tried expressing the LHS as a real part of a complex sum. I think you must've done the same sort of thing initially, that's why we got nowhere.

Much more satisfying than a proof by induction, certainly.

8. Nov 10, 2005

### quasar987

Cool, I had been wondering what the non-induction proof was too.