Cosine of 36 Degrees/Proof 2
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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 | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 3 2 \pm \dfrac {\sqrt 5} 2\) |
- $\paren {2 \cos 36^\circ}^2 > \paren {2 \cos 45^\circ}^2 = 2 > \dfrac 3 2 - \dfrac {\sqrt 5} 2$
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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $92 \ \text {(a)}$