Sine of 18 Degrees

Theorem

 * $\sin 18^\circ = \sin \dfrac {\pi} {10} = \dfrac {\sqrt{5} - 1} {4}$

where $\sin$ denotes the sine function.

Proof
From Sine of 90 Degrees, we have $\sin 5 \times 18^\circ = \sin 90^\circ = 1$.

Consider the equation $\sin 5x = 1$, where $x = 18^\circ$ is one of the solutions.

From Quintuple Angle Formula of Sine, we have:
 * $16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta = 1$

Let $s = \sin \theta$:
 * $16 s^5 - 20 s^3 + 5s - 1 = 0$

That is:
 * $(s-1) (4s^2+2s-1)^2 = 0$

Therefore, either $s=1$ or:
 * $s = \dfrac{1}{4} \left( \pm \sqrt{5} - 1 \right) $

Since $\sin 18^\circ > 0$:
 * $\sin 18^\circ = \dfrac{\sqrt{5} - 1}{4}$