Cosine of 36 Degrees/Proof 3

Theorem

$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = \dfrac {1 + \sqrt 5} 4$

where $\phi$ denotes the golden mean.

$\ds \sin 108 \degrees$

$=$

$\ds 3 \sin 36 \degrees - 4 \sin^3 36 \degrees$

$\ds \sin 72 \degrees$

$=$

$\ds 3 \sin 36 \degrees - 4 \sin^3 36 \degrees$

$\ds 2 \sin 36 \degrees \cos 36 \degrees$

$=$

$\ds 3 \sin 36 \degrees - 4 \sin^3 36 \degrees$

$\ds 2 \cos 36 \degrees$

$=$

$\ds 3 - 4 \sin^2 36 \degrees$

dividing both sides by $\sin 36 \degrees$

$\ds $

$=$

$\ds 4 \cos^2 36 \degrees - 1$

$\ds \leadsto \ \ $

$\ds 4 \cos^2 36 \degrees - 2 \cos 36 \degrees - 1$

$=$

$\ds 0$

$\ds \cos 36 \degrees$

$=$

$\ds \frac {2 \pm \sqrt {2^2 + 4 \times 4} } {2 \times 4}$

$\ds $

$=$

$\ds \frac {2 \pm \sqrt {20} } 8$

$\ds $

$=$

$\ds \frac {1 + \sqrt 5} 4$

negative root is rejected as $\cos 36 \degrees > 0$

$\blacksquare$