Cosine of 36 Degrees/Proof 2

From ProofWiki
Jump to navigation Jump to search

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:

\(\ds z^4 - 3z^2 + 1\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds z^2\) \(=\) \(\ds \dfrac {3 \pm \sqrt {\paren {-3}^2 - 4 \times 1} } 2\) 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:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Cosine_of_36_Degrees/Proof_2&oldid=563111"