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

\(\ds \)

\(=\)

\(\ds \dfrac 3 2 \pm \dfrac {\sqrt 5} 2\)

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

\(\ds k^2 \paren {a + \sqrt 5}^2\)

\(=\)

\(\ds \dfrac 3 2 + \dfrac {\sqrt 5} 2\)

\(\ds \leadsto \ \ \)

\(\ds 2 k^2 \paren {a^2 + 5 + 2 a \sqrt 5}\)

\(=\)

\(\ds 3 + \sqrt 5\)

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:

$4 a k^2 = 1$

which leads to:

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

so $a$ must be square.

Thus $a = 1$ and hence:

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

$\cos 36^\circ = \dfrac {1 + \sqrt 5} 4$

$\blacksquare$

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 $\cos 36 \degrees > 0$

$\blacksquare$