==定理==
$\displaystyle \sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$。
==推導==
首先注意$18^{\circ} \times 5 = 90^{\circ}$。命$\theta = 18^{\circ}$,即有$5\theta = 90^{\circ}$。
我們已知的基礎公式有:
- $\sin \varphi = \cos (90^{\circ} - \varphi)$;
- $\sin 3 \varphi = 3 \sin \varphi - 4 \sin^3 \varphi$;
- $\cos 2 \varphi = \cos^2 \varphi - \sin^2 \varphi = 2 \cos^2 \varphi -1 = 1 - 2 \sin^2 \varphi$。
\begin{align*}
\sin 3 \theta &= \sin (90^{\circ} - 2 \theta), \\
\sin 3 \theta &= \cos 2 \theta, \\
3 \sin \theta - 4 \sin^3 \theta &= \cos 2 \theta, \\
3 \sin \theta - 4 \sin^3 \theta &= 1 - 2 \sin^2 \theta.
\end{align*}
\sin 3 \theta &= \sin (90^{\circ} - 2 \theta), \\
\sin 3 \theta &= \cos 2 \theta, \\
3 \sin \theta - 4 \sin^3 \theta &= \cos 2 \theta, \\
3 \sin \theta - 4 \sin^3 \theta &= 1 - 2 \sin^2 \theta.
\end{align*}
得方程式
$$4 \sin^3 \theta - 2 \sin^2 \theta - 3 \sin \theta + 1 = 0.$$
此地我們要解出$\sin \theta$。注意到方程式係數和為$0$,所以容易因式分解得
$$(\sin \theta - 1)(4 \sin^2 \theta + 2 \sin \theta - 1) = 0,$$
從而得
$$\displaystyle \sin \theta = 1, \frac{-1 \pm \sqrt{5}}{4}.$$
但$\sin 18^{\circ} > 0$且$< 1$,所以
$$\displaystyle \sin 18^{\circ} = \frac{-1 + \sqrt{5}}{4}.$$
此地我們要解出$\sin \theta$。注意到方程式係數和為$0$,所以容易因式分解得
$$(\sin \theta - 1)(4 \sin^2 \theta + 2 \sin \theta - 1) = 0,$$
從而得
$$\displaystyle \sin \theta = 1, \frac{-1 \pm \sqrt{5}}{4}.$$
但$\sin 18^{\circ} > 0$且$< 1$,所以
$$\displaystyle \sin 18^{\circ} = \frac{-1 + \sqrt{5}}{4}.$$
(證明終了)
==其他相關數值==
- $\displaystyle \cos 18^{\circ} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$
[証]. 利用$\sin^2 \theta + \cos^2 \theta = 1$,同時注意$\cos 18^{\circ} > 0 $,所以$\cos 18^{\circ} = + \sqrt{1 - \sin^2 18^{\circ}} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$。
- $\displaystyle \sin 72^{\circ} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$
[証]. 利用$\sin 72^{\circ} = \sin (90^{\circ} - 18^{\circ}) = \cos 18^{\circ} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$。
- $\displaystyle \cos 72^{\circ} = \frac{\sqrt{5} - 1}{4}$
[証]. 利用$\cos 72^{\circ} = \cos (90^{\circ} - 18^{\circ}) = + \sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$。
- $\displaystyle \sin 108^{\circ} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$
[証]. 利用$\sin 108^{\circ} = \sin (90^{\circ} + 18^{\circ}) = + \cos 18^{\circ} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$。
- $\displaystyle \cos 108^{\circ} = \frac{1 - \sqrt{5}}{4}$
[証]. 利用$\cos 108^{\circ} = \cos (90^{\circ} + 18^{\circ}) = - \sin 18^{\circ} = - \frac{\sqrt{5} - 1}{4} = \frac{1 - \sqrt{5}}{4}$。
- $\displaystyle \sin 36^{\circ} = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$
- $\displaystyle \cos 36^{\circ} = \frac{1 + \sqrt{5}}{4}$
[証]. $\cos 36^{\circ} = \cos 2 \cdot 18^{\circ} = 2(\cos 18^{\circ})^2 - 1 = 2 \left( \frac{\sqrt{10 + 2\sqrt{5}}}{4} \right)^2 -1 = \frac{1 + \sqrt{5}}{4}$。
- $\displaystyle \sin 42^{\circ} = \frac{\sqrt{30 + 6 \sqrt{5}} - \sqrt{5} + 1}{8}$
[証]. 利用$\sin 42^{\circ} = \sin (60^{\circ} - 18^{\circ}) = \sin 60^{\circ} \cos 18^{\circ} - \cos 60^{\circ} \sin 18^{\circ}$。
- $\displaystyle \cos 42^{\circ} = \frac{\sqrt{10 + 2\sqrt{5}} + \sqrt{15} - \sqrt{3}}{8}$
[証]. 利用$\cos 42^{\circ} = \cos (60^{\circ} - 18^{\circ}) = \cos 60^{\circ} \cos 18^{\circ} + \sin 60^{\circ} \sin 18^{\circ}$。
\text{角度} \theta & \sin \theta & \cos \theta \\
\hline
18^{\circ} & \frac{\sqrt{5} - 1}{4} & \frac{\sqrt{10 + 2 \sqrt{5}}}{4} \\
36^{\circ} & \frac{\sqrt{10 - 2\sqrt{5}}}{4} & \frac{1 + \sqrt{5}}{4} \\
42^{\circ} & \frac{\sqrt{30 + 6 \sqrt{5}} - \sqrt{5} + 1}{8} & \frac{\sqrt{10 + 2\sqrt{5}} + \sqrt{15} - \sqrt{3}}{8} \\
72^{\circ} & \frac{\sqrt{10 + 2 \sqrt{5}}}{4} &\frac{\sqrt{5} - 1}{4} \\
108^{\circ} & \frac{\sqrt{10 + 2 \sqrt{5}}}{4} & \frac{1 - \sqrt{5}}{4}
\end{array}
==整理==
\begin{array}{ccc}\text{角度} \theta & \sin \theta & \cos \theta \\
\hline
18^{\circ} & \frac{\sqrt{5} - 1}{4} & \frac{\sqrt{10 + 2 \sqrt{5}}}{4} \\
36^{\circ} & \frac{\sqrt{10 - 2\sqrt{5}}}{4} & \frac{1 + \sqrt{5}}{4} \\
42^{\circ} & \frac{\sqrt{30 + 6 \sqrt{5}} - \sqrt{5} + 1}{8} & \frac{\sqrt{10 + 2\sqrt{5}} + \sqrt{15} - \sqrt{3}}{8} \\
72^{\circ} & \frac{\sqrt{10 + 2 \sqrt{5}}}{4} &\frac{\sqrt{5} - 1}{4} \\
108^{\circ} & \frac{\sqrt{10 + 2 \sqrt{5}}}{4} & \frac{1 - \sqrt{5}}{4}
\end{array}
沒有留言:
張貼留言