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Math Help - Trig question

  1. #1
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    Trig question

    This problem seems to send me down an endless path of failure. I could do with some hints.

    Many thanks, Bobak
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  2. #2
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    We will show that,
    \sum_{k=0}^{16}\cos (k\theta) = 0 for \theta = \frac{2\pi}{17}
    Note that \zeta = \cos \theta + i \sin \theta is a primitive root of unity of 17 which means all the roots are given by 1,\zeta, ... ,\zeta^{16}.

    But the sum of the roots of unity is zero, thus (by de Moivre),
    \sum_{k=0}^{16}\zeta^k = 0\implies \sum_{k=0}^{16}\cos (k\theta) = \sum_{k=0}^{16}\sin (k\theta) = 0
    .
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  3. #3
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    Yeah i actually got a bit lost on the first "show by direct substitution part" i kept going around in circles.
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