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$ sin \alpha = \sqrt{1 - cos^{2} \alpha } $
$ cos \alpha = \sqrt{1 - sin^{2} \alpha } $
$ tan \alpha = \frac{sin \alpha }{cos \alpha } $
$ sin \alpha = tan\alpha \cdot cos \alpha $
$ cos \alpha = \frac{sin \alpha }{tan \alpha } $
Geometrie-Trigonometrie-Umrechnungen
$ sin^{2} \alpha + cos^{2} \alpha = 1 $
$sin \alpha = \sqrt{1 - cos^{2} \alpha }$
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$cos \alpha = \sqrt{1 - sin^{2} \alpha }$
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$tan \alpha = \frac{sin \alpha }{cos \alpha }$
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$sin \alpha = tan\alpha \cdot cos \alpha$
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$cos \alpha = \frac{sin \alpha }{tan \alpha }$
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Beispiel Nr: 05
$\begin{array}{l}
\text{Gegeben:}\\\text{Winkel} \qquad \alpha \qquad [^{\circ}] \\
\\ \text{Gesucht:} \\\text{Tangens alpha} \qquad tan \alpha \qquad [] \\
\\ tan \alpha = \frac{sin \alpha }{cos \alpha }\\ \textbf{Gegeben:} \\ \alpha=60^{\circ} \qquad \\ \\ \textbf{Rechnung:} \\
tan \alpha = \frac{sin \alpha }{cos \alpha } \\
\alpha=60^{\circ}\\
tan 60^{\circ} = \frac{sin 60^{\circ} }{cos 60^{\circ} }\\
\\
tan 60^{\circ}=1,73
\\\\\\ \small \begin{array}{|l|} \hline alpha=\\ \hline 60 ° \\ \hline 3,6\cdot 10^{3} \text{'} \\ \hline 2,16\cdot 10^{5} \text{''} \\ \hline 66\frac{2}{3} gon \\ \hline 1,05 rad \\ \hline \end{array} \small \begin{array}{|l|} \hline Tanalpha=\\ \hline 1,73 rad \\ \hline 1,73\cdot 10^{3} mrad \\ \hline 99,2 ^\circ \\ \hline 5,95\cdot 10^{3} \text{'} \\ \hline 3,57\cdot 10^{5} \text{'''} \\ \hline \end{array} \end{array}$