By comparing expressions for $\displaystyle (\cos\theta+i\sin\theta)^5$ given by De Moivre's theorem and by the binomial theorem, prove that

$\displaystyle \cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$.

By considering the equation $\displaystyle \cos 5\theta=0$, prove that

$\displaystyle \cos (\frac{\pi}{10}).\cos (\frac{3\pi}{10})=\frac{1}{4}\sqrt{5}$

My problem is in the second part.

$\displaystyle \cos 5\theta=0$ means that $\displaystyle 16\cos^5\theta-20\cos^3\theta+5\cos\theta=0$, but after this I don't see how $\displaystyle \cos (\frac{\pi}{10}).\cos (\frac{3\pi}{10})=\frac{1}{2}\sqrt{5}$ relates to it.

Thanks!