Prove that 

$\left(1+\cos \frac{\pi }{10}\right)\left(1+\cos \frac{3\pi }{10}\right)\left(1+\cos \frac{7\pi }{10}\right)\left(1+\cos \frac{9\pi }{10}\right)=\frac{1}{16}$

LHS = $\left(1+\cos \frac{\pi }{10}\right)\left(1+\cos \frac{3\pi }{10}\right)\left(1+\cos \frac{7\pi }{10}\right)\left(1+\cos \frac{9\pi }{10}\right)$

$=\left(1+\cos \frac{\pi }{10}\right)\left(1+\cos \frac{3\pi }{10}\right)\left(1+\cos \left(\pi -\frac{3\pi }{10}\right)\right)\left(1+\cos \left(\pi -\frac{\pi }{10}\right)\right)$

$=\left(1+\cos \frac{\pi }{10}\right)\left(1+\cos \frac{3\pi }{10}\right)\left(1-\cos \frac{3\pi }{10}\right)\left(1-\cos \frac{\pi }{10}\right)$

$=\left(1-{\cos }^2\frac{\pi }{10}\right)\left(1-{\cos }^2\frac{3\pi }{10}\right)$

$={\mathrm{Sin}}^2\frac{\pi }{10}{\mathrm{Sin}}^2\frac{3\pi }{10}$

$={\mathrm{Sin}}^{2}18 {\mathrm{Sin}}^{2}54={\mathrm{Sin}}^{2}18 {\mathrm{Cos}}^{2}36$

$={\left(\frac{\sqrt{5}-1}{4}\right)}^2{\left(\frac{\sqrt{5}+1}{4}\right)}^2=\frac{(\sqrt{5}-1{)}^2(\sqrt{5}+1{)}^2}{16\times 16}$

$=\frac{(5-1{)}^2}{4^2\times 4^2}=\frac{4^2}{4^2\times 4^2}=\frac{1}{16}$