Help with De Moivre formula to establish identity

**flametag2** · December 24th 2011, 07:56 AM

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

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

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

**Plato** · December 24th 2011, 08:07 AM

Originally Posted by flametag2

I need help establishing the following identity
$1 + \cos \theta + \cos 2\theta +.....+ \cos n\theta$ = $\frac{1}{2}$ + ${\sin (n + \frac{1}{2}) \theta}$ / ${{2 sin \binom{\theta}{2}}$
That is ${\sin (n + \frac{1}{2}) \theta}$ divided by ${{2 \sin \binom{\theta}{2}}$ .

What in the world does $\sin \binom{\theta}{2}$ mean?

**flametag2** · December 24th 2011, 08:41 AM

Sorry
that should be

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

**ILikeSerena** · December 24th 2011, 09:05 AM

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

That is:

$\cos x = \frac 1 2 (e^{ix} + e^{-ix})$

or rather:

$\cos nx = \frac 1 2 ((e^{ix})^n + (e^{ix})^{-n})$

**flametag2** · December 24th 2011, 10:31 AM

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

**ILikeSerena** · December 24th 2011, 10:41 AM

Ah well, basically it's the same thing:

Add:

$(\cos x + i \sin x)^n = \cos nx + i \sin nx$
$(\cos(-x) + i \sin(-x))^n = \cos(-nx) + i \sin(-nx)$

To find:

$(\cos x + i \sin x)^n + (\cos x - i \sin x)^n = 2 \cos(nx)$

or:

$\cos nx = \frac 1 2 ((\cos x + i \sin x)^n + (\cos x - i \sin x)^n)$

If we define $z=\cos x + i \sin x$ , we get:

$\cos nx = \frac 1 2 (z^n + {\bar z}^n)$

Perhaps you can substitute that...

**Opalg** · December 24th 2011, 10:48 AM

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

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

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

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

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