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Math Help - Cosine Summation Identity

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    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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    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 f(z) = z^{2n}+z^{2n-1}+\cdots+z+1 = 0.
    By the formula for geometric series, f(z) = \frac{z^{2n+1}-1}{z-1} for z \ne 1.
    Now, if z^{2n+1}-1 = 0 then either z = 1 or e^{\pm \frac{2k\pi}{2n+1}i} where 1 \le k \le n.
    Thus the roots of f(z) are z = e^{\pm \frac{2k\pi}{2n+1}i}, where 1\le k \le n.
    The sum of these roots is 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 f(z) is -1.
    Thus 2\sum_{1 \le k \le n}\cos\left(\frac{2k\pi}{2n+1}\right)= -1 and hence:

    \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
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    Let S=\cos\alpha+\cos 2\alpha+\ldots+\cos n\alpha

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

    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 2\sin a\cos b=\sin(a+b)-\sin(b-a) we have

    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}

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

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

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

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

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

    =\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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