Cosine of 36 Degrees

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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 1

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


\(\displaystyle u\) \(=\) \(\displaystyle 2 v^2 - 1\) Double Angle Formula for Cosine: Corollary 1
\(\displaystyle \cos 36 \degrees\) \(=\) \(\displaystyle 1 - 2 \sin^2 18 \degrees\) Double Angle Formula for Cosine: Corollary 2
\(\displaystyle \) \(=\) \(\displaystyle 1 - 2 u^2\) Cosine of Complement equals Sine
\(\displaystyle \leadsto \ \ \) \(\displaystyle u + v\) \(=\) \(\displaystyle 2 \paren {v^2 - u^2}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {v + u} \paren {v - u}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1\) \(=\) \(\displaystyle 2 \paren {v - u}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 v - 4 v^2 + 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {2 v}^2\) \(=\) \(\displaystyle 2 v + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 v\) \(=\) \(\displaystyle \phi\) 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:

\(\displaystyle z^4 - 3z^2 + 1\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle z^2\) \(=\) \(\displaystyle \dfrac {3 \pm \sqrt {\paren {-3}^2 - 4 \times 1} } 2\) Quadratic Formula
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 3 2 \pm \dfrac {\sqrt 5} 2\)


Then let $z = a + \sqrt b$:

\(\displaystyle \paren {a + \sqrt b}^2\) \(=\) \(\displaystyle \dfrac 3 2 + \dfrac {\sqrt 5} 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^2 + b + 2 a \sqrt b\) \(=\) \(\displaystyle \dfrac 3 2 + \dfrac {\sqrt 5} 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^2 + b\) \(=\) \(\displaystyle \dfrac 3 2\)
\(\displaystyle 2 a \sqrt b\) \(=\) \(\displaystyle \dfrac {\sqrt 5} 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a\) \(=\) \(\displaystyle \dfrac {\sqrt 5} {4 \sqrt b}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^2\) \(=\) \(\displaystyle \dfrac 5 {16 b}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac 5 {16 b} + b\) \(=\) \(\displaystyle \dfrac 3 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 5 + 16 b^2\) \(=\) \(\displaystyle 24 b\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 16 b^2 - 24 b + 5\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle b\) \(=\) \(\displaystyle \dfrac {24 \pm \sqrt {24^2 - 4 \times 5 \times 16} } {32}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {2^3 \times 3 \pm \sqrt {\paren {2^3 \times 3}^2 - 2^6 \times 5} } {2^5}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 3 4 \pm \dfrac {2^3 \sqrt {3^2 - 5} } {2^5}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 3 4 \pm \dfrac {\sqrt {9 - 5} } 4\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 3 4 \pm \dfrac 2 4\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {3 \pm 2} 4\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 5 4 \text { or } \dfrac 1 4\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt b\) \(=\) \(\displaystyle \dfrac {\sqrt 5} 2 \text { or } \dfrac 1 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a\) \(=\) \(\displaystyle \dfrac 3 2 - \dfrac {\sqrt 5} 2 \text { or } \dfrac 3 2 - \dfrac 1 2\)