# Thread: Help with De Moivre formula to establish identity

1. ## Help with De Moivre formula to establish identity

Hello, I am new to the forum here and am not sure if I am posting this in the correct section. I didnt see any section for complex variables and this seemed the closest to complex analysis

I need help establishing the following identity

$\displaystyle 1 + \cos \theta + \cos 2\theta +.....+ \cos n\theta$ = $\displaystyle \frac{1}{2}$ + $\displaystyle {\sin (n + \frac{1}{2}) \theta}$ / $\displaystyle {{2 sin \frac{\theta}{2}}$
That is $\displaystyle {\sin (n + \frac{1}{2}) \theta}$ divided by $\displaystyle {{2 sin \frac{\theta}{2}}$. (I didnt know how to write that out in tex)

I need help establishing the above identity using DDe Moivre's formula and
$\displaystyle 1 + z + z^2+...+z^n = \frac{z^{n+1} - 1}{z - 1}$

2. ## Re: Help with De Moivre formula to establish identity

Originally Posted by flametag2
I need help establishing the following identity
$\displaystyle 1 + \cos \theta + \cos 2\theta +.....+ \cos n\theta$ = $\displaystyle \frac{1}{2}$ + $\displaystyle {\sin (n + \frac{1}{2}) \theta}$ / $\displaystyle {{2 sin \binom{\theta}{2}}$
That is $\displaystyle {\sin (n + \frac{1}{2}) \theta}$ divided by $\displaystyle {{2 \sin \binom{\theta}{2}}$.
What in the world does $\displaystyle \sin \binom{\theta}{2}$ mean?

3. ## Re: Help with De Moivre formula to establish identity

Sorry
that should be

$\displaystyle {{2 sin \frac{\theta}{2}}$

4. ## Re: Help with De Moivre formula to establish identity

This is a bit cumbersome to do with De Moivre's formula.
Are you allowed to use Euler's formula instead?

That is:
$\displaystyle \cos x = \frac 1 2 (e^{ix} + e^{-ix})$
or rather:
$\displaystyle \cos nx = \frac 1 2 ((e^{ix})^n + (e^{ix})^{-n})$

5. ## Re: Help with De Moivre formula to establish identity

No, unfortunately not. We have to use De Moivre's formula

6. ## Re: Help with De Moivre formula to establish identity

Ah well, basically it's the same thing:

$\displaystyle (\cos x + i \sin x)^n = \cos nx + i \sin nx$
$\displaystyle (\cos(-x) + i \sin(-x))^n = \cos(-nx) + i \sin(-nx)$
To find:
$\displaystyle (\cos x + i \sin x)^n + (\cos x - i \sin x)^n = 2 \cos(nx)$
or:
$\displaystyle \cos nx = \frac 1 2 ((\cos x + i \sin x)^n + (\cos x - i \sin x)^n)$

If we define $\displaystyle z=\cos x + i \sin x$, we get:
$\displaystyle \cos nx = \frac 1 2 (z^n + {\bar z}^n)$

Perhaps you can substitute that...

7. ## Re: Help with De Moivre formula to establish identity

Originally Posted by flametag2
Hello, I am new to the forum here and am not sure if I am posting this in the correct section. I didnt see any section for complex variables and this seemed the closest to complex analysis

I need help establishing the following identity

$\displaystyle 1 + \cos \theta + \cos 2\theta +\ldots+ \cos n\theta = \tfrac{1}{2} +\sin (n + \tfrac{1}{2}) \theta / 2 sin \tfrac{\theta}2$

I need help establishing the above identity using De Moivre's formula and
$\displaystyle 1 + z + z^2+\ldots+z^n = \frac{z^{n+1} - 1}{z - 1}$
Put $\displaystyle z = \cos\theta+i\sin\theta$ in the identity $\displaystyle 1 + z + z^2+\ldots+z^n = \tfrac{z^{n+1} - 1}{z - 1}$ and then take the real part of both sides. On the left, you get $\displaystyle 1 + \cos \theta + \cos 2\theta +\ldots + \cos n\theta$ as required (by De Moivre's theorem). On the right, you want the real part of $\displaystyle \frac{\cos(n+1)\theta + i\sin(n+1)\theta - 1}{\cos\theta-1+i\sin\theta}.$ Rationalise the denominator of that fraction, by multiplying top and bottom of the fraction by $\displaystyle \cos \theta - 1 - i\sin\theta.$ Then take the real part and use some trig identities to get the answer in the required form.