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    trigonometry proof

    Prove that cos (2π/13)+ cos (6π/13) + cos (8π/13)= (sqrt(13)-1)/4
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    Quote Originally Posted by hooke View Post
    Prove that cos (2π/13)+ cos (6π/13) + cos (8π/13)= (sqrt(13)-1)/4
    Outline proof: For 0\leqslant k\leqslant 12, let A_k denote the angle 2k\pi/13. The points (\cos A_k,\sin A_k) are equally spaced around the unit circle, so their centre of mass is at the origin, and it follows that \sum_{k=0}^{12}\cos A_k = 0.

    Since \cos A_0 = 1 and \cos A_{13-k} = \cos A_k, it follows that (\cos A_1 + \cos A_3 + \cos A_4) + (\cos A_2 + \cos A_6 + \cos A_8) = -1/2.

    Next,
    \begin{aligned}(\cos A_1 + \cos A_3 + \cos A_4)^2 = &(\cos^2 A_1 + \cos^2 A_3 + \cos^2 A_4) \\ &+  2(\cos A_3\cos A_4 + \cos A_4\cos A_1 + \cos A_1\cos A_3)\end{aligned}.

    But \cos^2 A_1 = \tfrac12(\cos A_2 + 1) (with similar results for \cos^2 A_3 and \cos^2 A_4). Also, 2\cos A_3\cos A_4 = \cos(A_3+A_4) + \cos(A_3-A_4) = \cos A_7 + \cos A_1 (with similar results for the other two products 2\cos A_4\cos A_1 and 2\cos A_1\cos A_3).

    Putting all that together, you find that
    \cos^2 A_1 + \cos^2 A_3 + \cos^2 A_4 = \tfrac12(\cos A_2 + \cos A_6 + \cos A_8) + 3/2, and
    2(\cos A_3\cos A_4 + \cos A_4\cos A_1 + \cos A_1\cos A_3) = \sum_{k=1}^6\cos A_k = -1/2.

    Now let S = \cos A_1 + \cos A_3 + \cos A_4 and T = \cos A_2 + \cos A_6 + \cos A_8. It follows from the above calculations that S+T = -\tfrac12 and S^2 = \tfrac12T+1. Eliminate T between those equations to get 4S^2+2S-3=0, a quadratic with solutions S = \frac{-1\pm\sqrt{13}}4. Now all that remains for you to do is to figure out why the square root must have the plus sign.
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    Thank you so much Opalg, as for the last piece of parcel you left for me, do i consider the quadrants cos A1, cos A2 and cos A3 belong to? I don't see a way of doing this without a calculator.
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    Quote Originally Posted by hooke View Post
    Thank you so much Opalg, as for the last piece of parcel you left for me, do i consider the quadrants cos A1, cos A2 and cos A3 belong to? I don't see a way of doing this without a calculator.
    The quantities S and T are the two roots of the quadratic equation. It should be clear from a diagram of the 13 points on the unit circle that the three cosines making up the sum T are smaller than their counterparts in S. Specifically, \cos A_2 < \cos A_1, \cos A_8 < \cos A_3 and \cos A_6 < \cos A_4. Therefore T<S and so S must correspond to the larger root of the quadratic.
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    Thanks Opalg.
    Last edited by Ackbeet; June 11th 2011 at 02:22 AM.
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