How to prove $\cos 36^{\circ} = (1+ \sqrt 5)/4$? Given $4 \cos^2 x -2\cos x -1 = 0$. 
Use this to show that $\cos 36^{\circ} = (1+ \sqrt 5)/4$, $\cos 72^{\circ} = (-1+\sqrt 5)/4$
Your help is greatly appreciated! Thanks

To derive this from fundamentals, note that
$$\sin{108^{\circ}} = \sin{72^{\circ}}$$
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, $x=36^{\circ}$.  Setting the above two equations equal to each other results in the quadratic equation in question:
$$2 \cos{x} = 3 - 4 (1-\cos^2{x}) = 4 \cos^2{x}-1$$
The rest follows from the above discussion.
