# Cosine of 36 Degrees

## Theorem

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

where $\phi$ denotes the golden mean.

## Proof 1

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

$\blacksquare$

## Proof 2

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:

 $\ds z^4 - 3z^2 + 1$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds z^2$ $=$ $\ds \dfrac {3 \pm \sqrt {\paren {-3}^2 - 4 \times 1} } 2$ Quadratic Formula $\ds$ $=$ $\ds \dfrac 3 2 \pm \dfrac {\sqrt 5} 2$
$\paren {2 \cos 36^\circ}^2 > \paren {2 \cos 45^\circ}^2 = 2 > \dfrac 3 2 - \dfrac {\sqrt 5} 2$

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

 $\ds k^2 \paren {a + \sqrt 5}^2$ $=$ $\ds \dfrac 3 2 + \dfrac {\sqrt 5} 2$ $\ds \leadsto \ \$ $\ds 2 k^2 \paren {a^2 + 5 + 2 a \sqrt 5}$ $=$ $\ds 3 + \sqrt 5$ Square of Sum

By comparing coefficients:

 $\ds \paren {a^2 + 5} : 2 a$ $=$ $\ds 3 : 1$ $\ds \leadsto \ \$ $\ds a^2 + 5$ $=$ $\ds 6 a$ $\ds \leadsto \ \$ $\ds a^2 - 6 a + 5$ $=$ $\ds 0$ $\ds$ $=$ $\ds \paren {a - 5} \paren {a - 1}$ $\ds \leadsto \ \$ $\ds a$ $=$ $\ds 1 \text{ or } 5$

We have, from the expansion above:

$4 a k^2 = 1$

$\sqrt a = \dfrac 1 {2 k}$

so $a$ must be square.

Thus $a = 1$ and hence:

$2 \cos 36^\circ = k \paren {a + \sqrt 5} = \dfrac {1 + \sqrt 5} 2$
$\cos 36^\circ = \dfrac {1 + \sqrt 5} 4$

$\blacksquare$

## Proof 3

 $\ds \sin 108 \degrees$ $=$ $\ds 3 \sin 36 \degrees - 4 \sin^3 36 \degrees$ Triple Angle Formula for Sine $\ds \sin 72 \degrees$ $=$ $\ds 3 \sin 36 \degrees - 4 \sin^3 36 \degrees$ Sine of Supplementary Angle $\ds 2 \sin 36 \degrees \cos 36 \degrees$ $=$ $\ds 3 \sin 36 \degrees - 4 \sin^3 36 \degrees$ Double Angle Formula for Sine $\ds 2 \cos 36 \degrees$ $=$ $\ds 3 - 4 \sin^2 36 \degrees$ dividing both sides by $\sin 36 \degrees$ $\ds$ $=$ $\ds 4 \cos^2 36 \degrees - 1$ Sum of Squares of Sine and Cosine $\ds \leadsto \ \$ $\ds 4 \cos^2 36 \degrees - 2 \cos 36 \degrees - 1$ $=$ $\ds 0$ $\ds \cos 36 \degrees$ $=$ $\ds \frac {2 \pm \sqrt {2^2 + 4 \times 4} } {2 \times 4}$ Quadratic Formula $\ds$ $=$ $\ds \frac {2 \pm \sqrt {20} } 8$ $\ds$ $=$ $\ds \frac {1 + \sqrt 5} 4$ negative root is rejected as $\cos 36 \degrees > 0$

$\blacksquare$