# Cosine of 36 Degrees/Proof 1

## Theorem

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

where $\phi$ denotes the golden mean.

## Proof

Let $u = \cos 72 \degrees, v = \cos 36 \degrees$.

Recall from Cosine of Complement equals Sine that $\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$

Therefore:

 $\ds \map \cos {90 \degrees - 18 \degrees}$ $=$ $\ds \map \sin {18 \degrees}$ $\ds \leadsto \ \$ $\ds \map \cos {72 \degrees}$ $=$ $\ds \map \sin {18 \degrees}$

Therefore:

 $\ds \cos 36 \degrees$ $=$ $\ds 1 - 2 \sin^2 18 \degrees$ Double Angle Formula for Cosine: Corollary 2 $\ds \leadsto \ \$ $\ds v$ $=$ $\ds 1 - 2 u^2$ substituting for $u$ and $v$ $\ds \cos 72 \degrees$ $=$ $\ds 2 \cos^2 36 \degrees - 1$ Double Angle Formula for Cosine: Corollary 1 $\ds \leadsto \ \$ $\ds u$ $=$ $\ds 2 v^2 - 1$ substituting for $u$ and $v$ $\ds \leadsto \ \$ $\ds u + v$ $=$ $\ds 2 \paren {v^2 - u^2}$ $\ds$ $=$ $\ds 2 \paren {v + u} \paren {v - u}$ $\ds \leadsto \ \$ $\ds 1$ $=$ $\ds 2 \paren {v - u}$ dividing both sides by $\paren {v + u}$ $\ds$ $=$ $\ds 2 v - 4 v^2 + 2$ substituting $u = 2 v^2 - 1$ $\ds \leadsto \ \$ $\ds \paren {2 v}^2$ $=$ $\ds 2 v + 1$ $\ds \leadsto \ \$ $\ds 2 v$ $=$ $\ds \phi$ Square of Golden Mean equals One plus Golden Mean

$\blacksquare$