Lösung Winkel am Einheitskreis

Version 3.2 von Thomas Köhler am 2024/07/18 14:33

Winkel \(\alpha\)30°60°90°120°150°180°210°240°270°300°330°360°
\(\sin(\alpha)\)
\(\frac{1}{2}\)

\(\frac{\sqrt{3}}{2}\)

\(1\)

\(\frac{\sqrt{3}}{2}}\)

\(\frac{1}{2}\)

\(0\)

\(-\frac{1}{2}\)

\(-\frac{\sqrt{3}}{2}}\)

\(-1\)

\(-\frac{\sqrt{3}}{2}}\)

\(-\frac{1}{2}}\)

\(0\)
\(\cos(\alpha)\)
\(\frac{\sqrt{3}}{2}\)

\(\frac{1}{2}\)

\(0\)

\(-\frac{1}{2}}\)

\(-\frac{\sqrt{3}}{2}}\)

\(-1\)

\(-\frac{\sqrt{3}}{2}}\)

\(-\frac{1}{2}}\)

\(0\)

\(\frac{1}{2}\)

\(\frac{\sqrt{3}}{2}}\)

\(1\)

Einheitskreis_winkel.png

zu 2.

\(\sin(360 + \beta) = \sin(\beta)\) bzw. \(\cos(360 + \beta)=\cos(\beta)\)