浙江 2011 · 数学 q06

题面

6、（2011•浙江）若0 < a < \frac{\pi}{2}$$- \frac{\pi}{2} < \beta < 0，cos$( \frac{\pi}{4} + a ) = \frac{1}{3} ,$，cos$( \frac{\pi}{4} - \frac{\beta}{2} ) = \frac{\sqrt{3}}{3}$，则$\cos ( a + \frac{\beta}{2} ) = ( )$ A、$\frac{\sqrt{3}}{3}$ B、$- \frac{\sqrt{3}}{3}$ C、$\frac{5 \sqrt{3}}{9}$ D、$- \frac{\sqrt{6}}{9}$

答案

C

解析

分析

先利用同角三角函数的基本关系分别求得$\sin ( \frac{\pi}{4} + a )$和$\sin ( \frac{\pi}{4} - \frac{\beta}{2} )$）的值，进而利用\cos ( a + \frac{\beta}{2} ) = c o s [( \frac{\pi}{4} + a )$$- ( \frac{\pi}{4} - \frac{\beta}{2} )]通过余弦的两角和公式求得答案解答：解：\because 0 < a < \frac{\pi}{2} , \quad - \frac{\pi}{2} < \beta < 0$$\therefore \frac{\pi}{4} < \frac{\pi}{4} + a < \frac{3 \pi}{4} , \frac{\pi}{4} < \frac{\pi}{4} - \frac{\beta}{2} < \frac{\pi}{2}$$\therefore \sin ( \frac{\pi}{4} + \alpha ) = \sqrt{1 - \frac{1}{9}} = \frac{2 \sqrt{2}}{3} , \sin ( \frac{\pi}{4} - \frac{\beta}{2} ) = \sqrt{1 - \frac{1}{3}} = \frac{\sqrt{6}}{3}$$\therefore \cos ( \alpha + \frac{\beta}{2} ) = \cos[( \frac{\pi}{4}{\pmb{\alpha}} ) - ( \frac{\pi}{4} - \frac{\beta}{2} )] = \cos ( \frac{\pi}{4}{\pmb{\alpha}} ) \cos ( \frac{\pi}{4} - \frac{\beta}{2} ) + \sin ( \frac{\pi}{4}{\pmb{\alpha}} ) \sin ( \frac{\pi}{4} - \frac{\beta}{2} ) = \frac{5 \sqrt{3}}{9}

故选 C