-
<<
>>
G
B
I
$ 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 $
$sin \alpha = \sqrt{1 - cos^{2} \alpha }$
1
2
3
4
5
$cos \alpha = \sqrt{1 - sin^{2} \alpha }$
1
2
3
4
5
$tan \alpha = \frac{sin \alpha }{cos \alpha }$
1
2
3
4
5
6
$sin \alpha = tan\alpha \cdot cos \alpha$
1
2
3
4
$cos \alpha = \frac{sin \alpha }{tan \alpha }$
1
2
3
4
Beispiel Nr: 02
$\begin{array}{l}
\text{Gegeben:}\\\text{Winkel} \qquad \alpha \qquad [^{\circ}] \\
\\ \text{Gesucht:} \\\text{Kosinus alpha} \qquad cos \alpha \qquad [] \\
\\ cos \alpha = \frac{sin \alpha }{tan \alpha }\\ \textbf{Gegeben:} \\ \alpha=30^{\circ} \qquad \\ \\ \textbf{Rechnung:} \\
cos \alpha = \frac{sin \alpha }{tan \alpha } \\
\alpha=30^{\circ}\\
cos \alpha = \frac{sin 30^{\circ} }{tan 30^{\circ} }\\\\cos \alpha=0,866
\\\\\\ \small \begin{array}{|l|} \hline alpha=\\ \hline 30 ° \\ \hline 1,8\cdot 10^{3} \text{'} \\ \hline 1,08\cdot 10^{5} \text{''} \\ \hline 33\frac{1}{3} gon \\ \hline 0,524 rad \\ \hline \end{array} \small \begin{array}{|l|} \hline cosalpha=\\ \hline 0,866 rad \\ \hline 866 mrad \\ \hline 49,6 ^\circ \\ \hline 2,98\cdot 10^{3} \text{'} \\ \hline 1,79\cdot 10^{5} \text{'''} \\ \hline \end{array} \end{array}$