设复数 $z = \cos \alpha + {\mathrm{i}}\sin \beta $，$\omega = \sin \alpha + {\mathrm{i}}\cos \beta $ 满足 $z\overline \omega = \dfrac{{\sqrt 3 }}{2}$，则 $\sin \left({\beta - \alpha } \right) =$ \((\qquad)\)
【难度】
【出处】
2010年复旦大学优秀高中生文化水平选拔测试
【标注】
【答案】
C
【解析】
因为\[\begin{split}z\overline \omega &= \left( {\cos \alpha + {\mathrm{i}}\sin \beta } \right)\left( {\sin \alpha - {\mathrm{i}}\cos \beta } \right)\\ & = \left( {\sin \alpha \cos \alpha + \sin \beta \cos \beta } \right) + \left( {\sin \alpha \sin \beta - \cos \alpha \cos \beta } \right){\mathrm{i}}\\& = \dfrac{{\sin 2\alpha + \sin 2\beta }}{2} - {\mathrm{i}}\cos \left( {\alpha + \beta } \right)\\& = \sin \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right) - {\mathrm{i}}\cos \left( {\alpha + \beta } \right),\end{split}\]所以$$\sin \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right) = \dfrac{{\sqrt 3 }}{2},\cos \left( {\alpha + \beta } \right) = 0,$$于是$$\cos \left( {\alpha - \beta } \right) = \pm \dfrac{{\sqrt 3 }}{2},$$即$$\cos \left( {\beta - \alpha } \right) = \pm \dfrac{{\sqrt 3 }}{2},$$于是$$\sin \left( {\beta - \alpha } \right) = \pm \dfrac{1}{2}.$$
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