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.
On
use that $$\sin\left(\frac{\pi}{10}\right)=\frac{\sqrt{5}-1}{4}$$ $$\sin\left(\frac{3\pi}{10}\right)=\frac{\sqrt{5}+1}{4}$$
On
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: $$ \prod_{k\in\{1,3,7,9\}}\left(1+\cos\frac{\pi k}{10}\right) = 1-\frac{5}{4}+\frac{5}{16} = \frac{1}{16}.$$
$$\sin^2 {\frac {\pi}{10}} \sin^2 \frac {3\pi}{10}=\left(\frac {4\sin {\frac {\pi}{5}}. \cos {\frac {2\pi}{10}}\cos {\frac {4\pi}{10}}}{4\sin \frac {\pi}{5}}\right)^2=\left(\frac {\sin \frac {4\pi}{5}}{4\sin \frac {\pi}{5}}\right)^2=\frac {1}{16}$$