若 $0<\alpha <\dfrac{\mathrm \pi }{2}$，$ - \dfrac{\mathrm \pi }{2}<\beta <0$，$\cos \left( {\dfrac{\mathrm \pi }{4} + \alpha } \right) = \dfrac{1}{3}$，$\cos \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) = \dfrac{\sqrt 3 }{3}$，则 $\cos \left( {\alpha + \dfrac{\beta }{2}} \right) = $ \((\qquad)\)
【难度】
【出处】
2011年高考浙江卷（理）
【标注】
【答案】
C
【解析】
由 $\cos \left( {\dfrac{\mathrm \pi }{4} + \alpha } \right) = \dfrac{1}{3}$，$0 < \alpha < \dfrac{\mathrm \pi }{2}$，得 $\sin \left( {\dfrac{\mathrm \pi }{4} + \alpha } \right) = \dfrac{2\sqrt 2 }{3}$，
再由 $\cos \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) = \dfrac{\sqrt 3 }{3}$，$ - \dfrac{\mathrm \pi }{2} < \beta < 0$，得 $\sin \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) = \dfrac{\sqrt 6 }{3}$，\[ \begin{split}\therefore \cos \left( {\alpha + \dfrac{\beta }{2}} \right)& = \cos \left[ {\left( {\dfrac{\mathrm \pi }{4} + \alpha } \right) - \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right)} \right] \\& = \cos \left( {\dfrac{\mathrm \pi }{4} + \alpha } \right)\cos \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) + \sin \left( {\dfrac{\mathrm \pi }{4} + \alpha } \right)\sin \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) \\& = \dfrac{1}{3} \times \dfrac{\sqrt 3 }{3} + \dfrac{2\sqrt 2 }{3} \times \dfrac{\sqrt 6 }{3} \\&= \dfrac{5\sqrt 3 }{9}.\end{split} \]
再由 $\cos \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) = \dfrac{\sqrt 3 }{3}$，$ - \dfrac{\mathrm \pi }{2} < \beta < 0$，得 $\sin \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) = \dfrac{\sqrt 6 }{3}$，\[ \begin{split}\therefore \cos \left( {\alpha + \dfrac{\beta }{2}} \right)& = \cos \left[ {\left( {\dfrac{\mathrm \pi }{4} + \alpha } \right) - \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right)} \right] \\& = \cos \left( {\dfrac{\mathrm \pi }{4} + \alpha } \right)\cos \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) + \sin \left( {\dfrac{\mathrm \pi }{4} + \alpha } \right)\sin \left( {\dfrac{\mathrm \pi }{4} - \dfrac{\beta }{2}} \right) \\& = \dfrac{1}{3} \times \dfrac{\sqrt 3 }{3} + \dfrac{2\sqrt 2 }{3} \times \dfrac{\sqrt 6 }{3} \\&= \dfrac{5\sqrt 3 }{9}.\end{split} \]
题目
答案
解析
备注
