1. ## trigonometry proof

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

2. 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.

3. 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.

4. 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.

5. Thanks Opalg.