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Help us understand the problem. What are the problem?

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@simanezumi1989

2次元の回転行列が回転行列である証明

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)
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1
Help us understand the problem. What are the problem?