问题及解答

求 $\sin(\frac{k}{5}\pi)$ 的值, 这里 $k\in\mathbb{Z}$.

利用三倍角公式, 可以求出

\[
\cos(36^{\circ})=\cos(\frac{1}{5}\pi)=\frac{1+\sqrt{5}}{4},
\]

\[
\sin(36^{\circ})=\sin(\frac{1}{5}\pi)=\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}.
\]

从而

\[
\sin(72^{\circ})=\sin(\frac{2}{5}\pi)=2\sin(\frac{1}{5}\pi)\cos(\frac{1}{5}\pi)=2\cdot\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}\cdot\frac{1+\sqrt{5}}{4}=\frac{1}{2}\sqrt{\frac{5+\sqrt{5}}{2}}.
\]

$\sin(108^{\circ})=\sin(\frac{3}{5}\pi)=\sin(\pi-\frac{3}{5}\pi)=\sin(\frac{2}{5}\pi)$. 当然, 也可以利用三倍角公式计算:

\[
\cos(72^{\circ})=\cos(\frac{2\pi}{5})=2\cos^2(\frac{\pi}{5})-1=2\cdot\Bigl(\frac{1+\sqrt{5}}{4}\Bigr)^2-1=\frac{\sqrt{5}-1}{4}.
\]

利用三倍角公式证明 $\sin(\frac{3}{5}\pi)=\sin(\frac{2}{5}\pi)$.

\[
\begin{split}
\sin(\frac{3}{5}\pi)&=3\sin\frac{\pi}{5}-4\sin^3(\frac{1}{5}\pi)\\
&=3\cdot\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}-4\cdot\frac{1}{4}\cdot\frac{5-\sqrt{5}}{2}\cdot\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}\\
&=\sqrt{\frac{5-\sqrt{5}}{2}}\cdot\biggl(\frac{3}{2}-\frac{5-\sqrt{5}}{4}\biggr)\\
&=\sqrt{\frac{5-\sqrt{5}}{2}}\cdot\frac{1+\sqrt{5}}{4}\\
&=\frac{1}{2}\sqrt{\frac{5+\sqrt{5}}{2}}=\sin(\frac{2}{5}\pi).
\end{split}
\]