trigonometry proof

**hooke** · June 8th 2011, 07:45 AM

Prove that cos (2π/13)+ cos (6π/13) + cos (8π/13)= (sqrt(13)-1)/4

**Opalg** · June 8th 2011, 01:20 PM

Originally Posted by hooke

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.

**hooke** · June 9th 2011, 08:00 AM

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.

**Opalg** · June 9th 2011, 08:53 AM

Originally Posted by hooke

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.

**hooke** · June 10th 2011, 08:05 PM

Thanks Opalg.

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