Cosine of 36 Degrees/Proof 2

Proof
From Complex Algebra: $z^4 - 3z^2 + 1 = 0$, the roots of $z^4 - 3z^2 + 1 = 0$ are:
 * $2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$

Then:

By Shape of Cosine Function:
 * $\paren {2 \cos 36^\circ}^2 > \paren {2 \cos 45^\circ}^2 = 2 > \dfrac 3 2 - \dfrac {\sqrt 5} 2$

so $\paren {2 \cos 36^\circ}^2 = \dfrac 3 2 + \dfrac {\sqrt 5} 2$.

Let $z = k \paren {a + \sqrt 5}$, where $k \in \Q$ and $a \in \Z$.

Then for $z = 2 \cos 36^\circ$:

By comparing coefficients:

We have, from the expansion above:
 * $4 a k^2 = 1$

which leads to:
 * $\sqrt a = \dfrac 1 {2 k}$

so $a$ must be square.

Thus $a = 1$ and hence:
 * $2 \cos 36^\circ = k \paren {a + \sqrt 5} = \dfrac {1 + \sqrt 5} 2$
 * $\cos 36^\circ = \dfrac {1 + \sqrt 5} 4$