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