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Math Help - Help with De Moivre formula to establish identity

  1. #1
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    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

    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}
    Last edited by flametag2; December 24th 2011 at 08:41 AM.
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  2. #2
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    Re: Help with De Moivre formula to establish identity

    Quote Originally Posted by flametag2 View Post
    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?
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  3. #3
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    Re: Help with De Moivre formula to establish identity

    Sorry
    that should be

    {{2 sin \frac{\theta}{2}}
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  4. #4
    Super Member ILikeSerena's Avatar
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    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:
    \cos x = \frac 1 2 (e^{ix} + e^{-ix})
    or rather:
    \cos nx = \frac 1 2 ((e^{ix})^n + (e^{ix})^{-n})
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  5. #5
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    Re: Help with De Moivre formula to establish identity

    No, unfortunately not. We have to use De Moivre's formula
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  6. #6
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    Re: Help with De Moivre formula to establish identity

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

    Quote Originally Posted by flametag2 View Post
    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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