Achsenspiegelung/Winkelhalbierung/Normiert/Aufgabe/Lösung

Mit ${\displaystyle {}\beta ={\frac {\alpha }{2}}}$ gilt

${\displaystyle {}{\begin{aligned}{\begin{pmatrix}\cos \alpha &\sin \alpha \\\sin \alpha &-\cos \alpha \end{pmatrix}}{\begin{pmatrix}\cos \left({\frac {\alpha }{2}}\right)\\\sin \left({\frac {\alpha }{2}}\right)\end{pmatrix}}&={\begin{pmatrix}\cos 2\beta &\sin 2\beta \\\sin 2\beta &-\cos 2\beta \end{pmatrix}}{\begin{pmatrix}\cos \beta \\\sin \beta \end{pmatrix}}\\&={\begin{pmatrix}2\cos ^{2}\beta -1&2\sin \beta \cos \beta \\2\sin \beta \cos \beta &-2\cos ^{2}\beta +1\end{pmatrix}}{\begin{pmatrix}\cos \beta \\\sin \beta \end{pmatrix}}\\&={\begin{pmatrix}{\left(2\cos ^{2}\beta -1\right)}\cos \beta +2\sin \beta \cos \beta \sin \beta \\2\sin \beta \cos \beta \cos \beta +{\left(-2\cos ^{2}\beta +1\right)}\sin \beta \end{pmatrix}}\\&={\begin{pmatrix}{\left(2\cos ^{2}\beta +2\sin ^{2}\beta -1\right)}\cos \beta \\{\left(2\cos ^{2}\beta -2\cos ^{2}\beta +1\right)}\sin \beta \end{pmatrix}}\\&={\begin{pmatrix}\cos \beta \\\sin \beta \end{pmatrix}}.\end{aligned}}}$