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2次元の回転行列が回転行列である証明

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目標

点 $(x,y)$ を反時計回りに $\theta$ だけ回転させた点を $(X,Y)$ とする。このとき

\left(\begin{matrix}
X\\
Y
\end{matrix}\right)
=\left(\begin{matrix}
\cos{\theta} & -\sin{\theta}\\
\sin{\theta} & \cos{\theta}
\end{matrix}\right)
\left(\begin{matrix}
x\\
y
\end{matrix}\right)

となることを示す。

証明

点 $(x,y)$を極座標表示すると

\left(\begin{matrix}
x\\
y
\end{matrix}\right)
=
\left(\begin{matrix}
r\cos{\alpha}\\
r\sin{\alpha}
\end{matrix}\right)...(1)

となる。ここで反時計回りに $\theta$ だけ回転させた点を $(X,Y)$ とすると,

\left(\begin{matrix}
X\\
Y
\end{matrix}\right)
=
\left(\begin{matrix}
r\cos{(\alpha+\theta)}\\
r\sin{(\alpha+\theta)}
\end{matrix}\right)

加法定理より

\left(\begin{matrix}
X\\
Y
\end{matrix}\right)
=
\left(\begin{matrix}
r\cos{\alpha}\cos{\theta}-r\sin{\alpha}\sin{\theta}\\
r\sin{\alpha}\cos{\theta}+r\cos{\alpha}\sin{\theta}
\end{matrix}\right)

(1)より

\left(\begin{matrix}
X\\
Y
\end{matrix}\right)
=
\left(\begin{matrix}
x\cos{\theta}-y\sin{\theta}\\
y\cos{\theta}+x\sin{\theta}
\end{matrix}\right)
=
\left(\begin{matrix}
\cos{\theta} & -\sin{\theta}\\
\sin{\theta} & \cos{\theta}
\end{matrix}\right)
\left(\begin{matrix}
x\\
y
\end{matrix}\right)
simanezumi1989
大学時代に微分方程式の定性的理論に関する研究が国際誌に掲載せれたことが自慢です。
http://fromalgorithm.jimdo.com
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