Sine of 18 Degrees
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Theorem
- $\sin 18 \degrees = \sin \dfrac \pi {10} = \dfrac {\sqrt 5 - 1} 4$
where $\sin$ denotes the sine function.
Proof
From Sine of $90 \degrees$:
- $\map \sin {5 \times 18 \degrees} = \sin 90 \degrees = 1$.
Consider the equation:
- $\sin 5x = 1$
where $x = 18 \degrees$ is one of the solutions.
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:
- $\paren {s - 1} \paren {4 s^2 + 2 s - 1}^2 = 0$
Therefore, either:
- $s = 1$
or by the Quadratic Formula:
- $s = \dfrac 1 4 \paren {\pm \sqrt 5 - 1}$
Since $0 < \sin 18 \degrees < 1$:
- $\sin 18 \degrees = \dfrac {\sqrt 5 - 1} 4$
$\blacksquare$