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Thread: Cosine Summation Identity

  1. #1
    Mar 2011

    Cosine Summation Identity

    prove that: cos(2pi/2n+1)+cos(4pi/2n+1)+cos(6pi/2n+1)+...+cos(2npi/2n+1)=-1/2
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  2. #2
    Super Member
    Mar 2010
    Quote Originally Posted by ralphjason007 View Post
    prove that: cos(2pi/2n+1)+cos(4pi/2n+1)+cos(6pi/2n+1)+...+cos(2npi/2n+1)=-1/2
    Consider the equation $\displaystyle f(z) = z^{2n}+z^{2n-1}+\cdots+z+1 = 0.$
    By the formula for geometric series, $\displaystyle f(z) = \frac{z^{2n+1}-1}{z-1}$ for $\displaystyle z \ne 1.$
    Now, if $\displaystyle z^{2n+1}-1 = 0$ then either $\displaystyle z = 1$ or $\displaystyle e^{\pm \frac{2k\pi}{2n+1}i}$ where $\displaystyle 1 \le k \le n.$
    Thus the roots of $\displaystyle f(z)$ are $\displaystyle z = e^{\pm \frac{2k\pi}{2n+1}i}$, where $\displaystyle 1\le k \le n.$
    The sum of these roots is $\displaystyle 2\sum_{1 \le k \le n}\cos\left(\frac{2k\pi}{2n+1}\right).$
    But by Vieta's relations the sum of the roots of $\displaystyle f(z)$ is $\displaystyle -1.$
    Thus $\displaystyle 2\sum_{1 \le k \le n}\cos\left(\frac{2k\pi}{2n+1}\right)= -1$ and hence:

    $\displaystyle \sum_{1 \le k \le n}\cos\left(\frac{2k\pi}{2n+1}\right)= -\frac{1}{2}. $
    Last edited by TheCoffeeMachine; May 9th 2011 at 08:30 PM.
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  3. #3
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    Let $\displaystyle S=\cos\alpha+\cos 2\alpha+\ldots+\cos n\alpha$

    Multiply both members with $\displaystyle 2\sin\frac{\alpha}{2}$

    $\displaystyle 2\sin\frac{\alpha}{2}S=2\sin\frac{\alpha}{2}\cos \alpha+2\sin\frac{\alpha}{2}\cos 2\alpha+\ldots+2\sin\frac{\alpha}{2}\cos n\alpha$

    Using the identity $\displaystyle 2\sin a\cos b=\sin(a+b)-\sin(b-a)$ we have

    $\displaystyle 2\sin\frac{\alpha}{2}S=\sin\frac{3\alpha}{2}-\sin\frac{\alpha}{2}+\sin\frac{5\alpha}{2}-\sin\frac{3\alpha}{2}+\ldots+\sin\frac{(2n+1) \alpha}{2}-\sin\frac{(2n-1)\alpha}{2}$

    $\displaystyle 2\sin\frac{\alpha}{2}S=\sin\frac{(2n+1)\alpha}{2}-\sin\frac{\alpha}{2}$

    $\displaystyle 2\sin\frac{\alpha}{2}S=2\sin\frac{n\alpha}{2}\cos \frac{(n+1)\alpha}{2}$

    Then $\displaystyle S=\dfrac{\sin\frac{n\alpha}{2}\cos\frac{(n+1)\alph a}{2}}{\sin\frac{\alpha}{2}}$

    Substitute $\displaystyle \alpha$ with $\displaystyle \frac{2\pi}{2n+1}$

    Then $\displaystyle S=\dfrac{\sin \frac{n\pi}{2n+1} \cos \frac{(n+1) \pi}{2n+1}}{\sin\frac{\pi}{2n+1}}=$

    $\displaystyle =\frac{1}{2}\cdot\dfrac{\sin\pi-\sin\frac{\pi}{2n+1}}{\sin\frac{\pi}{2n+1}}=\frac{ 1}{2}\cdot\left(-\dfrac{\sin\frac{\pi}{2n+1}}{\sin\frac{\pi}{2n+1}} \right)=-\frac{1}{2}$
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