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
 A: 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.
A: Hint: Look at the Quadratic Formula:
The solution to $ax^2+bx+c=0$ is $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

The equation is based on the fact that
$$
\cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x)
$$
and that $\cos(5\cdot36^\circ)=-1$ to get
$$
16\cos^5(36^\circ)-20\cos^3(36^\circ)+5\cos(36^\circ)+1=0
$$
Factoring yields
$$
(\cos(36^\circ)+1)(4\cos^2(36^\circ)-2\cos(36^\circ)-1)^2=0
$$
We know that $\cos(36^\circ)+1\ne0$; therefore,
$$
4\cos^2(36^\circ)-2\cos(36^\circ)-1=0
$$
Deciding between the two roots of this equation is a matter of looking at the signs of the roots.
For $\cos(72^\circ)$, use the identity $\cos(2x)=2\cos^2(x)-1$.
A: Using $\cos5\theta = 16\cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$;
$\cos 180 = 16\cos^5 36 - 20 \cos^3 36 + 5 \cos 36$
Let $\cos36 = x$:
$-1 = 16x^5 -20x^3 +5x$
Its solutions are ${-1,\frac{1- \sqrt{5}}{4}},\frac{1+ \sqrt{5}}{4}$
And as $\cos36 = x$, $\cos 36$ must be equal to one of them.
$\cos 36$ must be equal to $\frac{1+\sqrt{5}}{4}$, as $-1$ and $\frac{1- \sqrt{5}}{4}$ are negative.
