# Sine of 18 Degrees

## Theorem

$\sin 18 \degrees = \sin \dfrac \pi {10} = \dfrac {\sqrt 5 - 1} 4$

where $\sin$ denotes the sine function.

## Proof

$\map \sin {5 \times 18 \degrees} = \sin 90 \degrees = 1$.

Consider the equation:

$\sin 5x = 1$

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

$16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta = 1$

Let $s = \sin \theta$:

$16 s^5 - 20 s^3 + 5s - 1 = 0$

That is:

$\paren {s - 1} \paren {4 s^2 + 2 s - 1}^2 = 0$

Therefore, either:

$s = 1$

$s = \dfrac 1 4 \paren {\pm \sqrt 5 - 1}$
Since $0 < \sin 18 \degrees < 1$:
$\sin 18 \degrees = \dfrac {\sqrt 5 - 1} 4$
$\blacksquare$