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How can I prove that: $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$

Without using complex numbers?

I tried to raise by 2 and to multipy by 2, and got:

$2y^2=3+3\cos\frac{4\pi}{13}+2\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}+ 2\cos\frac{14\pi}{13}+2y$

But I'm stuck from here.

Thanks.

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Let $t = \frac{\pi}{13}$

then you know you want $$\cos(2t)+\cos(6t)+\cos(8t)$$

and if you call that $x$, square it, then you should have $$x^2 = \cos^2(2t)+\cos^2(6t)+\cos^2(8t)[\cos(2t)+\cos(4t)+\cos(6t)+\cos(8t)+\cos(10t)+\cos(12t)]$$

Note that,

$$[\cos(2t)+\cos(4t)+\cos(6t)+\cos(8t)+\cos(10t)+\cos(12t)] = -0.5$$

Hence,

$$x^2 = \cos^2(2t)+\cos^2(6t)+\cos^2(8t) - 0.5 $$

By the law of cosines,

$$x^2 = \frac{1}{2}(\cos(4t)+\cos(12t)+\cos(16t) + 3) - 0.5 $$

Hence, $$2x^2 = \cos(4t)+\cos(12t)+\cos(16t) + 2 $$

and as we know $$x = \cos(2t) + \cos(6t) + \cos(8t)$$

Add $$2x^2 + x = \frac{-1}{2} +2 = \frac{3}{2}$$

Then we have,

$$4x^2 + 2x -3 = 0$$

Use Quadratic formula, then you will get two solutions, but you know the answer is positive.

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  • $\begingroup$ How does it help? $\endgroup$ – Karen Mar 28 '16 at 19:07
  • $\begingroup$ @Karen i edited it :) $\endgroup$ – Allie Mar 28 '16 at 19:10
  • $\begingroup$ How do you know that it's $-0.5$? $\endgroup$ – Karen Mar 28 '16 at 19:16
  • $\begingroup$ @Karen its because $cos(2t) + cos(4t) + .....+ cos(24t) = -1$ $\endgroup$ – Allie Mar 28 '16 at 19:25
  • $\begingroup$ Is that a well know? I never heard of it. Also I didn't get your first simplification. I got different things $\endgroup$ – Karen Mar 28 '16 at 19:29
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Let $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=x$ and $\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}=y$.

Hence, $x=2\cos\frac{5\pi}{13}\cos\frac{3\pi}{13}+\cos\frac{6\pi}{13}>0$.

Now, $$x+y=\cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}=$$ $$=\tfrac{2\sin\frac{\pi}{13}\cos\frac{2\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{4\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{6\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{8\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{10\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{12\pi}{13}}{2\sin\frac{\pi}{13}}=$$ $$=\tfrac{\sin\frac{3\pi}{13}-\sin\frac{\pi}{13}+\sin\frac{5\pi}{13}-\sin\frac{3\pi}{13}+\sin\frac{7\pi}{13}-\sin\frac{5\pi}{13}+\sin\frac{9\pi}{13}-\sin\frac{7\pi}{13}+\sin\frac{11\pi}{13}-\sin\frac{9\pi}{13}+\sin\frac{13\pi}{13}-\sin\frac{11\pi}{13}}{\sin\frac{\pi}{13}}=-\frac{1}{2}$$ and $$xy=\left(\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}\right)\left(\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right)=$$ $$=\frac{3}{2}\left(\cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right)=-\frac{3}{4}.$$ Id est, $x$ and $y$ are roots of the following equation $$z^2+\frac{1}{2}z-\frac{3}{4}=0$$ and since $x>0$, we obtain $x=\frac{\sqrt{13}-1}{4}$.

Done!

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