回復 1# anson721 的帖子
\({{\tan }^{2}}{{60}^{{}^\circ }}=3\)
先求\({{\tan }^{2}}{{20}^{{}^\circ }},{{\tan }^{2}}{{40}^{{}^\circ }},{{\tan }^{2}}{{80}^{{}^\circ }}\)為三根的方程式
\(\cos {{40}^{{}^\circ }},\cos {{80}^{{}^\circ }},\cos {{160}^{{}^\circ }}\)是\(4{{x}^{3}}-3x=-\frac{1}{2}\)的三根
由於
\(\begin{align}
& {{\tan }^{2}}\theta =\frac{1-\cos 2\theta }{1+\cos 2\theta } \\
& \cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \\
\end{align}\)
令\(x=\frac{1-y}{1+y}\)
以\({{\tan }^{2}}{{20}^{{}^\circ }},{{\tan }^{2}}{{40}^{{}^\circ }},{{\tan }^{2}}{{80}^{{}^\circ }}\)為三根的方程式為
\(\begin{align}
& 4{{\left( \frac{1-y}{1+y} \right)}^{3}}-3\left( \frac{1-y}{1+y} \right)=-\frac{1}{2} \\
& 8{{\left( 1-y \right)}^{3}}-6\left( 1-y \right){{\left( 1+y \right)}^{2}}=-{{\left( 1+y \right)}^{3}} \\
& {{y}^{3}}-33{{y}^{2}}+27y-3=0 \\
\end{align}\)
所求為
\(\begin{align}
& \left( x-3 \right)\left( {{x}^{3}}-33{{x}^{2}}+27x-3 \right)=0 \\
& {{x}^{4}}-36{{x}^{3}}+126{{x}^{2}}-84x+9=0 \\
\end{align}\)
