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.

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 $\cos(2x)=2\cos^2(x)-1$.