今回はこちらの問題を解いていきます.
以下の問いに答えよ.
(1) \(\tan{2\theta}\), \(\tan{3\theta}\)を\(\tan{\theta}\)を用いて表せ.
(2) \(\displaystyle\tan{\frac{\pi}{8}}\), \(\displaystyle\tan{\frac{3\pi}{8}}\)の値を求めよ.
(3) \(x^3-3(1+\sqrt{2})x^2-3x+1+\sqrt{2}=0\)の実数解をすべて求めよ.
(2022 信州大学 文系 [1])
それでは解いていきましょう.
(1) \(\tan\)の加法定理を用いて,$$
\begin{align}
\tan{2\theta}&=\tan{(\theta+\theta)}\\[1.5ex]
&=\frac{2\tan{\theta}}{1-\tan^2{\theta}}\\[2ex]
\tan{3\theta}&=\tan{(\theta+2\theta)}\\[1.5ex]
&=\frac{\tan{\theta}+\tan{2\theta}}{1-\tan{\theta}\tan{2\theta}}\\[1.5ex]
&=\frac{\tan{\theta}+\frac{2\tan{\theta}}{1-\tan^2{\theta}}}{1-\tan{\theta}\cdot\frac{2\tan{\theta}}{1-\tan^2{\theta}}}\\[1.5ex]
&=\frac{3\tan{\theta}-\tan^3{\theta}}{1-3\tan^2{\theta}}
\end{align}
$$
(2) (1)で求めた\(\tan{2\theta}\)の式で\(\displaystyle \theta=\frac{\pi}{8}\)とすると,$$
\begin{align}
&\tan{\frac{\pi}{4}}=\frac{2\tan{\frac{\pi}{8}}}{1-\tan^2{\frac{\pi}{8}}}\\[1.5ex]
\iff & 1-\tan^2{\frac{\pi}{8}}=2\tan{\frac{\pi}{8}}\\[1.5ex]
\iff & \tan^2{\frac{\pi}{8}}+2\tan{\frac{\pi}{8}}-1=0
\end{align}
$$となるから, \(\displaystyle \tan{\frac{\pi}{8}}\)は2次方程式\(x^2+2x-1=0\)の解になることがわかる. この2次方程式を解くと,$$
x=-1\pm\sqrt{2}
$$となり, \( \displaystyle 0<\frac{\pi}{8}<\frac{\pi}{2}\)より, \(\displaystyle \tan{\frac{\pi}{8}}>0\)であるから,$$
\displaystyle \tan{\frac{\pi}{8}}=\sqrt{2}-1
$$であることがわかる.
次に, (1)で求めた\(\tan{3\theta}\)の式で\(\displaystyle \theta=\frac{\pi}{8}\)とすると,$$
\begin{align}
\tan{\frac{3\pi}{8}}&=\frac{3\tan{\frac{\pi}{8}}-\tan^3{\frac{\pi}{8}}}{1-3\tan^2{\frac{\pi}{8}}}\\[1.5ex]
&=\frac{3(\sqrt{2}-1)-(\sqrt{2}-1)^3}{1-3(\sqrt{2}-1)^2}\\[1.5ex]
&=\frac{(\sqrt{2}-1)(3-3+2\sqrt{2}}{-8+6\sqrt{2}}\\[1.5ex]
&=\frac{4-2\sqrt{2}}{-8+6\sqrt{2}}\\[1.5ex]
&=\frac{2-\sqrt{2}}{-4+3\sqrt{2}}\\[1.5ex]
&=\frac{(2-\sqrt{2})(-4-3\sqrt{2})}{-2}\\[1.5ex]
&=\frac{-2-2\sqrt{2}}{-2}\\[1.5ex]
&=\sqrt{2}+1
\end{align}
$$と求まる.
(3) \(\tan{3\theta}\)の式で\(\displaystyle \theta=\frac{\pi}{8}\)とし, \(\displaystyle \tan{\frac{3\pi}{8}}=\sqrt{2}+1\)から,
$$
\begin{align}
&\sqrt{2}+1 =\frac{3\tan{\frac{\pi}{8}}-\tan^3{\frac{\pi}{8}}}{1-3\tan^2{\frac{\pi}{8}}}\\[1.5ex]
\iff & \tan^3{\frac{\pi}{8}}-3(1+\sqrt{2})\tan^2{\frac{\pi}{8}}-3\tan{\frac{\pi}{8}}+1+\sqrt{2}=0
\end{align}
$$となる. これから, \(\displaystyle \tan{\frac{\pi}{8}}=\sqrt{2}-1\)は\(3\)次方程式\(x^3-3(1+\sqrt{2})x^2-3x+1+\sqrt{2}=0\)の解になること, そして, この\(3\)次方程式の左辺は\(x-(\sqrt{2}-1)\)で割り切れること, がわかる.
\(3\)次方程式の左辺を因数分解すると,
$$
\left\{x-(\sqrt{2}-1)\right\}\left\{x^2-2(\sqrt{2}+2)x-3-2\sqrt{2}\right\}=0
$$となり. \(2\)次方程式\(x^2-2(\sqrt{2}+2)x-3-2\sqrt{2}=0\)を解くと,$$
\begin{align}
x&=\sqrt{2}+2\pm\sqrt{(\sqrt{2}+2)^2+3+2\sqrt{2}}\\[1.5ex]
&=\sqrt{2}+2\pm\sqrt{9+6\sqrt{2}}\\[1.5ex]
&=\sqrt{2}+2\pm\sqrt{9+2\sqrt{18}}\\[1.5ex]
&=\sqrt{2}+2\pm\sqrt{3+2\sqrt{3\cdot 6}+6}\\[1.5ex]
&=\sqrt{2}+2\pm\sqrt{(\sqrt{3}+\sqrt{6})^2}\\[1.5ex]
&=\sqrt{2}+2\pm(\sqrt{3}+\sqrt{6})\\[1.5ex]
\end{align}
$$となる.
よって与えられた\(3\)次方程式の実数解は,$$
x=\sqrt{2}-1, 2+\sqrt{2}+\sqrt{3}+\sqrt{6}, 2+\sqrt{2}-\sqrt{3}-\sqrt{6}
$$の\(3\)つであることがわかる.
youtubeでも解説しています.
