Prove the identity :
$\displaystyle
(\frac{1}{2}+\cos \frac{\pi }{20})(\frac{1}{2}+\cos \frac{3\pi }{20})(\frac{1}{2}+\cos \frac{9\pi }{20})(\frac{1}{2}+\cos \frac{27\pi }{20})=\frac{1}{16}
$
I expanded the left side of equation with a program, so this should be correct. From here I am sorry to say I cannot help. I thought there would be more simplification, but I do not see anything helpful.
$\displaystyle \dfrac{1}{16}+1/8cos(\dfrac{\pi}{20})+1/8cos(\dfrac{3\pi}{20})+1/4cos(\dfrac{\pi}{20})$$\displaystyle cos(\dfrac{3\pi}{20})+1/8sin(\dfrac{\pi}{20})+1/4cos(\dfrac{\pi}{20})sin(\dfrac{\pi}{20})$$\displaystyle +1/4cos(\dfrac{3\pi}{20})sin(\dfrac{\pi}{20})+1/2cos(\dfrac{\pi}{20})$$\displaystyle cos(\dfrac{3\pi}{20})sin(\dfrac{\pi}{20})-1/8sin(\dfrac{3\pi}{20})$$\displaystyle -1/4cos(\dfrac{\pi}{20})sin(\dfrac{3\pi}{20})-1/4cos(\dfrac{3\pi}{20})sin(\dfrac{3\pi}{20})-1/2cos(\dfrac{\pi}{20})$$\displaystyle cos(\dfrac{3\pi}{20})sin(\dfrac{3\pi}{20})-1/4sin(\dfrac{\pi}{20})sin(\dfrac{3\pi}{20})-1/2cos(\dfrac{\pi}{20})sin(\dfrac{\pi}{20})$$\displaystyle sin(\dfrac{3\pi}{20})-1/2cos(\dfrac{3\pi}{20})sin(\dfrac{\pi}{20})$$\displaystyle sin(\dfrac{3\pi}{20})-cos(\dfrac{\pi}{20})cos(\dfrac{3\pi}{20})sin(\dfra c{\pi}{20})sin(\dfrac{3\pi}{20})$