千葉大学(理系) 2021年 問題4
$以下の問いに答えよ。$
$(1)\ \ w=\cos \cfrac{2\pi}{5} + i\sin \cfrac{2\pi}{5} \ \ とする。ただし、i\ は虚数単位である。z\ を \ 0\ でない複素数とするとき、$
$\qquad 次の等式を証明せよ。$
$\hspace{4em} \big(z-\cfrac{1}{z}\big)\big(wz-\cfrac{1}{wz}\big)\big(w^2z-\cfrac{1}{w^2z}\big)\big(w^3z-\cfrac{1}{w^3z}\big)\big(w^4z-\cfrac{1}{w^4z}\big)=z^5-\cfrac{1}{z^5}$
$(2)\ \ ある定数 \ C\ に対して、等式$
$\hspace{4em} \sin \theta \sin \big(\theta+\cfrac{2\pi}{5}\big)\sin \big(\theta+\cfrac{4\pi}{5}\big)\sin (\theta+\cfrac{6\pi}{5}\big)\sin \big(\theta+\cfrac{8\pi}{5}\big)=C\sin 5\theta$
$\qquad がすべての実数 \ \theta \ で成り立つことを示せ。また、C\ の値を求めよ。$
$(3)\ \ \sin \cfrac{\pi}{10}\sin \cfrac{3\pi}{10}\ \ の値を求めよ。$
$(解説)$
$(1)\ \ w^5=1 \ \ です。与えられた等式の左辺をうまく展開します。$
$(2)\ \ z=\cos \theta +i\sin \theta \ \ とおいて(1)の等式に代入します。$
$(3)\ \ \theta =\cfrac{\pi}{10} \ \ とおくとうまく求まります。$
(1)
$\quad w^5=\cos 2\pi +i\sin 2\pi=1 \quad より \quad w^5-1=0$
$因数分解して \qquad (w-1)(w^4+w^3+w^2+w+1)=0 \qquad w \ne 0 \quad だから \quad w^4+w^3+w^2+w+1=0$
\begin{eqnarray*} & &\big(z-\cfrac{1}{z}\big)\big(wz-\cfrac{1}{wz}\big)\big(w^2z-\cfrac{1}{w^2z}\big)\big(w^3z-\cfrac{1}{w^3z}\big)\big(w^4z-\cfrac{1}{w^4z}\big)\\ \\ &=&\big(z-\cfrac{1}{z}\big)\big(wz-\cfrac{w^4}{z}\big)\big(w^2z-\cfrac{w^3}{z}\big)\big(w^3z-\cfrac{w^2}{z}\big)\big(w^4z-\cfrac{w}{z}\big)\\ \\ &=&w^6\big(z-\cfrac{1}{z}\big)\Big\{\big(z-\cfrac{w^3}{z}\big)\big(z-\cfrac{w}{z}\big)\Big\}\Big\{\big(wz-\cfrac{1}{z}\big)\big(w^3z-\cfrac{1}{z}\big)\Big\}\\ \\ &=&w\big(z-\cfrac{1}{z}\big)\big(z^2-w-w^3+\cfrac{w^4}{z^2}\big)\big(w^4z^2-w-w^3+\cfrac{1}{z^2}\big)\\ \\ &=&\big(z-\cfrac{1}{z}\big)\big\{z^2-(w^3+w)+\cfrac{w^4}{z^2}\big\} \big\{z^2-(w^4+w^2)+\cfrac{w}{z^2}\big\}\\ \\ &=&\big(z-\cfrac{1}{z}\big)\big\{z^4-(w^4+w^3+w^2+w)z^2+(w^7+2w^5+w^4+w^3+w)-\cfrac{w^8+w^6+w^4+w^2}{z^2}+\cfrac{w^5}{z^4}\big\}\\ \\ &=&\big(z-\cfrac{1}{z}\big)\big\{z^4-(w^4+w^3+w^2+w)z^2+(w^2+2+w^4+w^3+w)-\cfrac{w^3+w+w^4+w^2}{z^2}+\cfrac{1}{z^4}\big\}\\ \\ &=&\big(z-\cfrac{1}{z}\big)\big(z^4+z^2+1+\cfrac{1}{z^2}+\cfrac{1}{z^4}\big)\\ \\ &=&z^5-\cfrac{1}{z^5}\\ \end{eqnarray*}
(2)
$\quad w=\cos \cfrac{2\pi}{5} + i\sin \cfrac{2\pi}{5} \quad より \quad \cfrac{1}{w}=\cos \cfrac{2\pi}{5} - i\sin \cfrac{2\pi}{5}=\overline{w}$
$\quad z=\cos \theta +i\sin \theta \quad とおくと \quad \cfrac{1}{z}=\cos \theta -i\sin \theta =\overline{z},\qquad z^5=\cos 5\theta +i\sin 5\theta$
$このとき$
$\quad wz=(\cos \cfrac{2\pi}{5} + i\sin \cfrac{2\pi}{5})(\cos \theta +i\sin \theta)=\cos (\theta +\cfrac{2\pi}{5}) +i\sin (\theta +\cfrac{2\pi}{5})$
$同様にして$
$\quad w^2z=\cos (\theta +\cfrac{4\pi}{5}) +i\sin (\theta +\cfrac{4\pi}{5}) ,\qquad w^3z=\cos (\theta +\cfrac{6\pi}{5}) +i\sin (\theta +\cfrac{6\pi}{5}),\qquad w^4z=\cos (\theta +\cfrac{8\pi}{5}) +i\sin (\theta +\cfrac{8\pi}{5})$
$(1)の \big(z-\cfrac{1}{z}\big)\big(wz-\cfrac{1}{wz}\big)\big(w^2z-\cfrac{1}{w^2z}\big)\big(w^3z-\cfrac{1}{w^3z}\big)\big(w^4z-\cfrac{1}{w^4z}\big)=z^5-\cfrac{1}{z^5} に代入して$
$\quad (z-\overline{z})(wz-\overline{wz})(w^2z-\overline{w^2z})(w^3z-\overline{w^3z})(w^4z-\overline{w^4z})=z^5-\overline{z^5}$
$\quad (2i\sin \theta)\big(2i\sin (\theta +\cfrac{2\pi}{5})\big)\big(2i\sin (\theta +\cfrac{4\pi}{5})\big)\big((2i\sin (\theta +\cfrac{6\pi}{5})\big)\big(2i\sin (\theta +\cfrac{8\pi}{5})\big)=2i\sin 5\theta $
$\quad (2i)^5\sin \theta \sin (\theta +\cfrac{2\pi}{5})\sin (\theta +\cfrac{4\pi}{5})\sin (\theta +\cfrac{6\pi}{5})\sin (\theta +\cfrac{8\pi}{5})=2i\sin 5\theta $
$\quad \sin \theta \sin (\theta +\cfrac{2\pi}{5})\sin (\theta +\cfrac{4\pi}{5})\sin (\theta +\cfrac{6\pi}{5})\sin (\theta +\cfrac{8\pi}{5})=\cfrac{1}{(2i)^4}\sin 5\theta =\cfrac{1}{16}\sin 5\theta $
$\quad \therefore \ \ C=\cfrac{1}{16}$
(3)
$\quad \theta =\cfrac{\pi}{10} \quad とおくと \qquad \theta +\cfrac{2\pi}{5}=\cfrac{\pi}{10} + \cfrac{2\pi}{5} =\cfrac{\pi}{2}, \qquad \theta +\cfrac{4\pi}{5}=\cfrac{\pi}{10} + \cfrac{4\pi}{5} =\cfrac{9\pi}{10}$
$\quad \theta +\cfrac{6\pi}{5}=\cfrac{\pi}{10} + \cfrac{6\pi}{5} =\cfrac{13\pi}{10}, \qquad \theta +\cfrac{8\pi}{5}=\cfrac{\pi}{10} + \cfrac{8\pi}{5} =\cfrac{17\pi}{10} $
$(2)の \sin \theta \sin (\theta +\cfrac{2\pi}{5})\sin (\theta +\cfrac{4\pi}{5})\sin (\theta +\cfrac{6\pi}{5})\sin (\theta +\cfrac{8\pi}{5})=\cfrac{1}{16}\sin 5\theta \quad に代入して$
$\quad \sin \cfrac{\pi}{10}\sin \cfrac{\pi}{2}\sin \cfrac{9\pi}{10}\sin \cfrac{13\pi}{10}\sin \cfrac{17\pi}{10} =\cfrac{1}{16}\sin \cfrac{\pi}{2}$
$\quad \sin \cfrac{\pi}{10}\sin (\pi -\cfrac{\pi}{10})\sin (\pi + \cfrac{3\pi}{10})\sin (2\pi -\cfrac{3\pi}{10}) =\cfrac{1}{16}$
$\quad \sin \cfrac{\pi}{10}\sin \cfrac{\pi}{10} (-\sin \cfrac{3\pi}{10})(-\sin \cfrac{3\pi}{10}) =\cfrac{1}{16}$
$\quad \big(\sin \cfrac{\pi}{10}\sin \cfrac{3\pi}{10}\big)^2=\cfrac{1}{16}$
$\quad \sin \cfrac{\pi}{10}\sin \cfrac{3\pi}{10} > 0 \quad だから \qquad \sin \cfrac{\pi}{10}\sin \cfrac{3\pi}{10}=\cfrac{1}{4}$
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