Von zwei Winkeln \(\alpha\) und \(\beta\) ist bekannt, dass
\(\qquad\begin{array} {l c l c l } \cos(\alpha) & = & -\dfrac {\sqrt{3}}{2} & \quad & \pi \leq \alpha \leq 2 \pi \\ \sin(\beta) & = & \dfrac {1}{2} & \quad & \dfrac {\pi}{2} \leq \beta \leq \dfrac {3 \pi}{2} \end{array}\)
Bestimmen Sie \(\sin(\alpha - \beta)\).
\(\sin(\alpha - \beta) = \dfrac {\sqrt{6} - \sqrt{2}}{4} \)
\(\sin(\alpha - \beta) = 0 \)
\(\sin(\alpha - \beta) = \dfrac {\sqrt{3} }{2} \)
\(\sin(\alpha - \beta) = -\dfrac {\sqrt{3}}{2} \)
\(\sin(\alpha - \beta) = \dfrac {\sqrt{2} - \sqrt{6}}{4} \)