# Some trigonometric equation problems

show that : $$\left(1+\cos \frac{2\pi}{13}\right)\left(1-\cos \frac{4\pi}{13}\right)\left(1+\cos \frac{6\pi}{13}\right)\left(1+\cos \frac{8\pi}{13}\right)\left(1-\cos \frac{10\pi}{13}\right)\left(1-\cos \frac{12\pi}{13}\right)=\frac{65+18\sqrt{13}}{64}$$ $$\left(1+\cos \frac{\pi}{22}\right)\left(1-\cos \frac{3\pi}{22}\right)\left(1+\cos \frac{5\pi}{22}\right)\left(1+\cos \frac{7\pi}{22}\right)\left(1+\cos \frac{9\pi}{22}\right)=\frac{10+3\sqrt{11}}{32}$$

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This is probably the hard way, but it ought to work. Multiply through by $64$, write $2+2\cos(2k\pi/13)=2+\zeta^k+\zeta^{13-k}$ where $\zeta=e^{2\pi i/13}$, multiply everything out, use $1+\zeta+\cdots+\zeta^{12}=0$ and standard formulas for the Gauss sum $\sum\zeta^{k^2}$. Similarly for the second one, but with $11$ instead of $13$. – Gerry Myerson Feb 12 '13 at 9:09