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PROBLEM $$ \prod_{i=0}^4 \left(1 + \cos \left(\frac{(2k+1)\pi}{10}\right)\right)$$

My Try $$ \left(1 + \cos \frac{\pi}{10}\right) \left(1 + \cos \frac{9\pi}{10}\right) \left(1 + \cos \frac{7\pi}{10}\right) \left(1 + \cos \frac{3\pi}{10}\right) = \sin^2 \left(\frac{\pi}{10}\right) \sin^2 \left(\frac{3\pi}{10}\right) $$

I am not able to proceed further. Please help me.

Michael Rozenberg
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sai saandeep
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3 Answers3

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$$\sin^2 {\frac {\pi}{10}} \sin^2 \frac {3\pi}{10}=\left(\frac {4\sin {\frac {\pi}{5}}. \cos {\frac {2\pi}{10}}\cos {\frac {4\pi}{10}}}{4\sin \frac {\pi}{5}}\right)^2=\left(\frac {\sin \frac {4\pi}{5}}{4\sin \frac {\pi}{5}}\right)^2=\frac {1}{16}$$

Rohan Shinde
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Michael Rozenberg
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    Why the hell was this downvoted. It is the best anyone can think of(+1) – Rohan Shinde Feb 23 '18 at 13:21
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    Upvoted. The use of degrees is irrelevant to the validity (and elegance) of this solution. – Deepak Feb 23 '18 at 13:23
  • @Deepak I hope this makes it better – Rohan Shinde Feb 23 '18 at 13:37
  • @Manthanein I don't know that it does. It was fine before. Some unreasonable people downvoted simply because of the use of degrees. I thought it was a lot easier to see it in degrees. – Deepak Feb 23 '18 at 13:45
  • @Manthanein Anyway, you've taken the effort, it isn't wrong by any means and if Michael's fine with it, then let it be. – Deepak Feb 23 '18 at 13:46
  • I think with degrees it seems better, but if Manthanein wishes... – Michael Rozenberg Feb 23 '18 at 13:52
  • @MichaelRozenberg No just don't think of my wishes. Keep the way you like it. I loved the answer really very much and couldn't help seeing the downvotes. On seeing comments I saw that that it was downvoted because of the degree thing and I couldn't accept it. So I edited it. But you can keep the way you wish it to be. – Rohan Shinde Feb 23 '18 at 15:57
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The roots of $\Phi_{10}(x)=1-x+x^2-x^3+x^4$ are the primitive tenth roots of unity, $\xi,\xi^3,\xi^7,\xi^9$ with $\xi=\exp\left(\frac{2\pi i}{10}\right)$. The roots of $\Phi_{20}(x)=\Phi_{10}(x^2)=1-x^2+x^4-x^6+x^8$ are the primitive $20$-th roots of unity, $\zeta,\zeta^3,\zeta^7,\zeta^9,\zeta^{11},\zeta^{13},\zeta^{17},\zeta^{19}$ with $\zeta=\exp\left(\frac{\pi i}{10}\right)$. By dividing $\Phi_{20}(x)$ by $x^4$ and by writing what we get as a polynomial in $\left(x+\frac{1}{x}\right)$ we get that

$$ x^4-\frac{5}{4}x^2+\frac{5}{16}=\prod_{k\in\{1,3,7,9\}}\left(x-\cos\frac{\pi k}{10}\right)$$ and by evaluating both sides at $x=-1$ we get: $$ \prod_{k\in\{1,3,7,9\}}\left(1+\cos\frac{\pi k}{10}\right) = 1-\frac{5}{4}+\frac{5}{16} = \frac{1}{16}.$$

Jack D'Aurizio
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use that $$\sin\left(\frac{\pi}{10}\right)=\frac{\sqrt{5}-1}{4}$$ $$\sin\left(\frac{3\pi}{10}\right)=\frac{\sqrt{5}+1}{4}$$

Dr. Sonnhard Graubner
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