How prove $\cos(\frac{2\pi}{17}) + \cos(\frac{18\pi}{17})+\cos(\frac{26\pi}{17})+\cos(\frac{30\pi}{17}) = \frac{\sqrt{17}-1}{4}$

Question

Prove that $\cos(\frac{2\pi}{17}) + \cos(\frac{18\pi}{17})+\cos(\frac{26\pi}{17})+\cos(\frac{30\pi}{17}) = \frac{\sqrt{17}-1}{4}$

Regards that value of $\cos(2\pi/17)$, I can't find the easy way to solve that expression.

Even if I had time, I wouldn't try that method to find the all roots others cosines expressions. IMHO

https://math.stackexchange.com/questions/1717536/proving-that-cos-frac2-pi13-cos-frac6-pi13-cos-frac8-pi13-frac — lab bhattacharjee, Apr 27 '18 at 11:17

score 4 · Accepted Answer · answered Apr 27 '18 at 11:19

4

Let $p$ be an odd prime number. Then $$g_p=\sum_{k=0}^{p-1}\exp(2\pi i k^2/p)$$ is a quadratic Gauss sum. Gauss proved that $g_p=\sqrt p$ or $i\sqrt p$ according to whether $p\equiv1$ or $p\equiv3\pmod 4$. It is quite easy to prove this up to sign, but hard to prove the sign.

So $g_{17}=\sqrt{17}$. Therefore \begin{align} \sqrt{17}&=1+2\exp(2\pi i/17)+2\exp(8\pi i/17)+2\exp(18\pi i/17) +2\exp(32\pi i/17)\\ &+2\exp(16\pi i/17)+2\exp(4\pi i/17)+2\exp(30\pi i/17) +2\exp(26\pi i/17)\\ &=1+4\cos(2\pi/17)+4\cos(18\pi/17)+4\cos(26\pi/17)+4\cos(30\pi/17). \end{align}

answered Apr 27 '18 at 11:19

Angina Seng

158,341

Where does the constant 2 (in all those terms) come from that sum? – miguel747 Apr 27 '18 at 13:41
1

He folded the terms together using the periodicity of the cosine function. Example: $15^2\equiv 2^2 \bmod 17$ because $(15+2)|(15^2-2^2)$, thus $\cos(2^2×(2\pi/17))+\cos(15^2×(2\pi/17))=2\cos(2^2×(2\pi/17))$. – Oscar Lanzi Apr 27 '18 at 13:55
Last question. How can you cancel the $\sin$ component into the exponential complex on the last step? – miguel747 Apr 27 '18 at 16:28
@miguel747 $2\cos x=\exp(ix)+\exp(-ix)$. – Angina Seng Apr 27 '18 at 16:29

How prove $\cos(\frac{2\pi}{17}) + \cos(\frac{18\pi}{17})+\cos(\frac{26\pi}{17})+\cos(\frac{30\pi}{17}) = \frac{\sqrt{17}-1}{4}$

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