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:
 * $\sin \left({5 \times 18^\circ}\right) = \sin 90^\circ = 1$.

Consider the equation:
 * $\sin 5x = 1$

where $x = 18^\circ$ is one of the solutions.

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

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

That is:
 * $\left({s - 1}\right) \left({4 s^2 + 2 s - 1}\right)^2 = 0$

Therefore, either:
 * $s = 1$

or by the Quadratic Formula:
 * $s = \dfrac 1 4 \left({\pm \sqrt 5 - 1}\right)$

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