calculate a trigonometric expression related to $\sin(\pi/5)$ I tried to solve a question which led to 
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?

Note that
$$\sin{\frac{3 \pi}{5}} = \sin{\frac{2 \pi}{5}}$$
then use a double-angle and triple-angle forumla:
$$\sin{2 x} = 2 \sin{x} \cos{x}$$
$$\sin{3 x} = 3 \sin{x} - 4 \sin^3{x}$$
In this case, $x=\pi/5$.  Setting the above two equations equal to each other results in the quadratic equation in question:
$$2 \cos{x} = 3 - 4 (1-\cos^2{x}) = 4 \cos^2{x}-1$$
which means that
$$\cos{x} = \frac{1 + \sqrt{5}}{4}$$
$$\sin{x} = \frac{\sqrt{10-2\sqrt{5}}}{4}$$
Then, using $\sin{2 x} = \sin{3 x}$ as noted above, we have
$$4 \sin{x} \sin{2 x} = 8 \sin^2{x} \cos{x} = 8 \frac{10-2 \sqrt{5}}{16} \frac{1+\sqrt{5}}{4} = \frac{10-10-2 \sqrt{5}+10\sqrt{5}}{8} = \sqrt{5}$$
