LineareAlgebra.Abbildungen03 History

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July 22, 2016, at 07:32 AM by 2003:7a:341:b900:9ca0:d321:8821:2199 -
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November 14, 2013, at 11:40 PM by 217.250.79.140 -
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October 26, 2012, at 08:21 AM by 79.219.84.144 -
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\left(\begin{array}{cc} a&b\\c&d\end{array}\right) =
to:
D_\alpha=\left(\begin{array}{cc} a&b\\c&d\end{array}\right) =
Changed line 49 from:
\cos(\alpha+\beta)=cos(\alpha)\cdot\cos(\beta)+\sin(\alpha)\cdot\sin(\beta)
to:
\cos(\alpha+\beta)=cos(\alpha)\cdot\cos(\beta)-\sin(\alpha)\cdot\sin(\beta)
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D_{$\alpha+\beta}=D_{$\beta}\cdot D_{$\alpha}
to:
D_{\alpha+\beta}=D_{\beta}\cdot D_{\alpha}
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Hier liest man unmittelbar zwei Additionstheoreme ab:
$$
\sin(\alpha+\beta)=\sin(\alpha)\cdot\cos(\beta)+\sin(\beta)\cdot\cos(\alpha)
$$
$$
\cos(\alpha+\beta)=cos(\alpha)\cdot\cos(\beta)+\sin(\alpha)\cdot\sin(\beta)
$$


Changed lines 42-46 from:
=\left(\begin{array}{cc} \cos(\beta)\cdot\cos(\alpha)-\sin(\beta)\cdot\sin(\alpha)&
-\cos(\beta)\cdot\sin(\alpha)-\sin(\beta)\cdot\cos(\alpha)\\
\sin(\beta)\cdot\cos(\alpha)+\cos(\beta)\cdot\sin(\alpha)
&-\sin(\beta)\cdot\sin(\alpha)+\cos(\beta)\cdot\cos(\alpha)
\end{array}\right)
to:
=\left(\begin{array}{cc} \cos(\beta)\cdot\cos(\alpha)-\sin(\beta)\cdot\sin(\alpha)&-\cos(\beta)\cdot\sin(\alpha)-\sin(\beta)\cdot\cos(\alpha)\\\sin(\beta)\cdot\cos(\alpha)+\cos(\beta)\cdot\sin(\alpha)&-\sin(\beta)\cdot\sin(\alpha)+\cos(\beta)\cdot\cos(\alpha)\end{array}\right)
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\end{array}\right)
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\left(\begin{array}{cc} \cos(\alpha+\beta)&-\sin(\alpha+\beta)\\\sin(\alpha+\beta)&cos(\alpha+\beta)\end{array}\right)=D_{$\alpha+\beta}=D_{$\beta}\cdot D_{$\alpha}
to:
\left(\begin{array}{cc} \cos(\alpha+\beta)&-\sin(\alpha+\beta)\\\sin(\alpha+\beta)&cos(\alpha+\beta)\end{array}\right)=D_{\alpha+\beta}=D_{\beta}\cdot D_{\alpha}
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to:
$$
=\left(\begin{array}{cc} \cos(\beta)\cdot\cos(\alpha)-\sin(\beta)\cdot\sin(\alpha)&
-\cos(\beta)\cdot\sin(\alpha)-\sin(\beta)\cdot\cos(\alpha)\\
\sin(\beta)\cdot\cos(\alpha)+\cos(\beta)\cdot\sin(\alpha)
&-\sin(\beta)\cdot\sin(\alpha)+\cos(\beta)\cdot\cos(\alpha)
$$
Changed line 31 from:
D_{$\alpha+\beta}=D_{$\beta}\cdotD_{$\alpha}
to:
D_{$\alpha+\beta}=D_{$\beta}\cdot D_{$\alpha}
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\left(\begin{array}{cc} \cos(\alpha+\beta)&-\sin(\alpha+\beta)\\\sin(\alpha+\beta)&cos(\alpha+\beta)\end{array}\right)=D_{$\alpha+\beta}=D_{$\beta}\cdotD_{$\alpha}
=\left(\begin{array}{cc} \cos(\beta)&-\sin(\beta)\\\sin(\beta)&cos(\beta)\end{array}\right)\cdot
to:
\left(\begin{array}{cc} \cos(\alpha+\beta)&-\sin(\alpha+\beta)\\\sin(\alpha+\beta)&cos(\alpha+\beta)\end{array}\right)=D_{$\alpha+\beta}=D_{$\beta}\cdot D_{$\alpha}
=\left(\begin{array}{cc} \cos(\beta)&-\sin(\beta)\\\sin(\beta)&cos(\beta)\end{array}\right) \cdot
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!!Drehmatrizen und Additionstheoreme

Man kann mit Hilfe von Drehmatrizen leicht die wichtigsten Additionstheoreme für Sinus und Cosinus herleiten. Eine Drehung um den Winkel {$\alpha+\beta$} kann man ersetzen durch eine Drehung um den Winkel {$\alpha$} und eine anschließende Drehung um den Winkel {$\beta$}, also
$$
D_{$\alpha+\beta}=D_{$\beta}\cdotD_{$\alpha}
$$
Schreibt man die Drehmatrizen in der oben hergeleiteten Form, so erhält man:
$$
\left(\begin{array}{cc} \cos(\alpha+\beta)&-\sin(\alpha+\beta)\\\sin(\alpha+\beta)&cos(\alpha+\beta)\end{array}\right)=D_{$\alpha+\beta}=D_{$\beta}\cdotD_{$\alpha}
=\left(\begin{array}{cc} \cos(\beta)&-\sin(\beta)\\\sin(\beta)&cos(\beta)\end{array}\right)\cdot
\left(\begin{array}{cc} \cos(\alpha)&-\sin(\alpha)\\\sin(\alpha)&cos(\alpha)\end{array}\right)
$$

Changed lines 25-26 from:
#{$\frac{1}{2}\cdot\left(\begin{array}{cc} \sqrt{3}&-1\\1&\sqrt{3}\end{array}\right)$} Drehung um 30°
to:
#{$\frac{1}{\sqrt{2}}\cdot\left(\begin{array}{cc} 1&-1\\1&1\end{array}\right)$} Drehung um 45°
Changed lines 5-6 from:
Bei der Suche nach der Drehmatrix ist es geschickt, die Bildpunkte von $P(1|0)$} und {$Q(0|1)$} zu bestimmen. Hier erhält man in Abhängigkeit vom Drehwinkel {$\alpha$}: {$P'(\cos(\alpha)|\sin(\alpha))$} und {$Q'(-\sin(\alpha)|\cos(\alpha))$}.
to:
Bei der Suche nach der Drehmatrix ist es geschickt, die Bildpunkte von {$P(1|0)$} und {$Q(0|1)$} zu bestimmen. Hier erhält man in Abhängigkeit vom Drehwinkel {$\alpha$}: {$P'(\cos(\alpha)|\sin(\alpha))$} und {$Q'(-\sin(\alpha)|\cos(\alpha))$}.
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{$\left(\begin{array}{cc} -1&0\\0&-1\end{array}\right)$} Drehung um 180° (Punktspiegelung)
to:
#{$\left(\begin{array}{cc} -1&0\\0&-1\end{array}\right)$} Drehung um 180° (Punktspiegelung)
#{$\frac{1}{2}\cdot\left(\begin{array}{cc} \sqrt{3}&-1\\1&\sqrt{3}\end{array}\right)$} Drehung um 30°
#{$\frac{1}{2}\cdot\left(\begin{array}{cc} \sqrt{3}&-1\\1&\sqrt{3}\end{array}\right)$} Drehung um 30°

January 25, 2010, at 09:01 PM by 84.173.72.200 -
Changed lines 22-24 from:
to:
'''Beispiele:'''
{$\left(\begin{array}{cc} -1&0\\0&-1\end{array}\right)$} Drehung um 180° (Punktspiegelung)

January 25, 2010, at 08:58 PM by 84.173.72.200 -
Added lines 1-23:
%right% [[Abbildungen02|<<]] Abbildungen03 [[Abbildungen04|>>]]
!!Drehungen um den Ursprung


Bei der Suche nach der Drehmatrix ist es geschickt, die Bildpunkte von $P(1|0)$} und {$Q(0|1)$} zu bestimmen. Hier erhält man in Abhängigkeit vom Drehwinkel {$\alpha$}: {$P'(\cos(\alpha)|\sin(\alpha))$} und {$Q'(-\sin(\alpha)|\cos(\alpha))$}.

%center% %width=300px%Attach:abbildungen03_1.png

Dies führt wieder zu zwei Bedingungen an die gesuchte Matrix, nämlich:
$$
\left(\begin{array}{cc} a&b\\c&d\end{array}\right)\cdot \left(\begin{array}{c} 1\\0\end{array}\right) =
\left(\begin{array}{c} \cos(\alpha)\\\sin(\alpha)\end{array}\right)\quad\mbox{und}\quad
\left(\begin{array}{cc} a&b\\c&d\end{array}\right)\cdot \left(\begin{array}{c} 0\\1\end{array}\right) =
\left(\begin{array}{c} -\sin(\alpha)\\\cos(\alpha)\end{array}\right)
$$
Damit erhält man für die Drehmatrix unmittelbar:
$$
\left(\begin{array}{cc} a&b\\c&d\end{array}\right) =
\left(\begin{array}{cc} \cos(\alpha)&-\sin(\alpha)\\\sin(\alpha)&cos(\alpha)\end{array}\right)
$$


%right% [[Abbildungen02|<<]] Abbildungen03 [[Abbildungen04|>>]]