Lösung Winkel am Einheitskreis

Version 1.16 von Thomas Köhler am 2024/07/18 14:07

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

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

\(1\)

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

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