Lösung Skizzieren

Zuletzt geändert von Miriam Erdmann am 2024/07/19 14:01

Winkel \alpha-90°-60°-30°30°60°90°120°150°180°210°240°270°300°330°360°390°420°
Bogenlänge x-\frac{\pi}{2}-\frac{\pi}{3}-\frac{\pi}{6}0\frac{\pi}{6}\frac{\pi}{3}\frac{\pi}{2}\frac{2\pi}{3}\frac{5\pi}{6}{\pi}\frac{7\pi}{6}\frac{4\pi}{3}\frac{3\pi}{2}\frac{5\pi}{3}\frac{11\pi}{6}{2\pi}\frac{13\pi}{6}\frac{7\pi}{3}
f(x)=\sin(x)-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-\frac{1}{2}\frac{-\sqrt{3}}{2}-1\frac{-\sqrt{3}}{2}-\frac{1}{2}0\frac{1}{2}\frac{\sqrt{3}}{2}

Winkel \alpha-90°-60°-30°30°60°90°120°150°180°210°240°270°300°330°360°390°420°
Bogenlänge x-\frac{\pi}{2}-\frac{\pi}{3}-\frac{\pi}{6}0\frac{\pi}{6}\frac{\pi}{3}\frac{\pi}{2}\frac{2\pi}{3}\frac{5\pi}{6}{\pi}\frac{7\pi}{6}\frac{4\pi}{3}\frac{3\pi}{2}\frac{5\pi}{3}\frac{11\pi}{6}{2\pi}\frac{13\pi}{6}\frac{7\pi}{3}
f(x)=\cos(x)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\frac{-\sqrt{3}}{2}-\frac{1}{2}0\frac{1}{2}\frac{\sqrt{3}}{2}1\frac{\sqrt{3}}{2}\frac{1}{2}

LösungAufgabe7.png