# Cosine of 36 Degrees/Proof 2

## Theorem

$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = \dfrac {1 + \sqrt 5} 4$

where $\phi$ denotes the golden mean.

## 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:

 $\displaystyle z^4 - 3z^2 + 1$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle z^2$ $=$ $\displaystyle \dfrac {3 \pm \sqrt {\paren {-3}^2 - 4 \times 1} } 2$ Quadratic Formula $\displaystyle$ $=$ $\displaystyle \dfrac 3 2 \pm \dfrac {\sqrt 5} 2$

Then let $z = a + \sqrt b$:

 $\displaystyle \paren {a + \sqrt b}^2$ $=$ $\displaystyle \dfrac 3 2 + \dfrac {\sqrt 5} 2$ $\displaystyle \leadsto \ \$ $\displaystyle a^2 + b + 2 a \sqrt b$ $=$ $\displaystyle \dfrac 3 2 + \dfrac {\sqrt 5} 2$ $\displaystyle \leadsto \ \$ $\displaystyle a^2 + b$ $=$ $\displaystyle \dfrac 3 2$ $\displaystyle 2 a \sqrt b$ $=$ $\displaystyle \dfrac {\sqrt 5} 2$ $\displaystyle \leadsto \ \$ $\displaystyle a$ $=$ $\displaystyle \dfrac {\sqrt 5} {4 \sqrt b}$ $\displaystyle \leadsto \ \$ $\displaystyle a^2$ $=$ $\displaystyle \dfrac 5 {16 b}$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac 5 {16 b} + b$ $=$ $\displaystyle \dfrac 3 2$ $\displaystyle \leadsto \ \$ $\displaystyle 5 + 16 b^2$ $=$ $\displaystyle 24 b$ $\displaystyle \leadsto \ \$ $\displaystyle 16 b^2 - 24 b + 5$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle b$ $=$ $\displaystyle \dfrac {24 \pm \sqrt {24^2 - 4 \times 5 \times 16} } {32}$ $\displaystyle$ $=$ $\displaystyle \dfrac {2^3 \times 3 \pm \sqrt {\paren {2^3 \times 3}^2 - 2^6 \times 5} } {2^5}$ $\displaystyle$ $=$ $\displaystyle \dfrac 3 4 \pm \dfrac {2^3 \sqrt {3^2 - 5} } {2^5}$ $\displaystyle$ $=$ $\displaystyle \dfrac 3 4 \pm \dfrac {\sqrt {9 - 5} } 4$ $\displaystyle$ $=$ $\displaystyle \dfrac 3 4 \pm \dfrac 2 4$ $\displaystyle$ $=$ $\displaystyle \dfrac {3 \pm 2} 4$ $\displaystyle$ $=$ $\displaystyle \dfrac 5 4 \text { or } \dfrac 1 4$ $\displaystyle \leadsto \ \$ $\displaystyle \sqrt b$ $=$ $\displaystyle \dfrac {\sqrt 5} 2 \text { or } \dfrac 1 2$ $\displaystyle \leadsto \ \$ $\displaystyle a$ $=$ $\displaystyle \dfrac 3 2 - \dfrac {\sqrt 5} 2 \text { or } \dfrac 3 2 - \dfrac 1 2$