Golden Ratio by definition: $$\phi = \frac{\sqrt{5} + 1}{2}$$

$$\begin{aligned} \phi & = 2\cos{\frac{\pi}{5}} \\[1em] & = 2\cos\left({\frac{\pi}{4} - \frac{\pi}{20}} \right) \\[1em] & = 2\cos\left(\frac{1}{2}\left({\frac{\pi}{2} - \frac{\pi}{10}} \right) \right) \\[1em] \end{aligned}$$

Recall that $$\cos(2x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)$$ Thus $$\begin{aligned} \phi^2 & = 4\cos^2\left(\frac{1}{2}\left({\frac{\pi}{2} - \frac{\pi}{10}} \right) \right) \\[1em] \frac{1}{2}\phi^2 - 1 & = 2\cos^2\left(\frac{1}{2}\left({\frac{\pi}{2} - \frac{\pi}{10}} \right) \right) - 1 \\[1em] \frac{1}{2}\phi^2 - 1 & = \cos\left({\frac{\pi}{2} - \frac{\pi}{10}} \right) \\[1em] \frac{1}{2}\phi^2 - 1 & = \sin\left(\frac{\pi}{10} \right) \\[1em] \frac{1}{2}\phi^2 - 1 & = \sin\left(\frac{1}{2}\left(\frac{\pi}{5} \right) \right) \\[1em] 1 - 2\left(\frac{1}{2}\phi^2 - 1 \right)^2 & = 1 - 2\sin^2\left(\frac{1}{2}\left(\frac{\pi}{5} \right) \right) \\[1em] 1 - 2\left(\frac{1}{2}\phi^2 - 1 \right)^2 & = \cos\frac{\pi}{5} \\[1em] 1 - 2\left(\frac{1}{2}\phi^2 - 1 \right)^2 & = \frac{1}{2}\phi \\[1em] \phi^4 - 4\phi^2 + \phi + 2 & = 0 \\[1em] (\phi - 2)(\phi - 1)(\phi^2 -\phi -1) & = 0 \\[1em] \end{aligned}$$

The only valid root is: $$\phi = \frac{\sqrt{5} + 1}{2}$$