Cosine of 36 Degrees/Proof 1

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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

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

$\blacksquare$


Sources

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