（2011•浙江）若 0< a < π 2 ,- π 2…——2011 高考数学第 6 题答案解析

解答：解：$\because 0<\mathrm{a}<\frac{\pi}{2}, ~-\frac{\pi}{2}<\beta<0$ ，
$\therefore \frac{\pi}{4}<\frac{\pi}{4}+\alpha<\frac{3 \pi}{4}, \frac{\pi}{4}<\frac{\pi}{4}-\frac{\beta}{2}<\frac{\pi}{2}$
$\therefore \sin \left(\frac{\pi}{4}+\alpha\right)=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}, \sin \left(\frac{\pi}{4}-\frac{\beta}{2}\right)=\sqrt{1-\frac{1}{3}}=\frac{\sqrt{6}}{3}$
$\therefore \cos \left(\alpha+\frac{\beta}{2}\right)=\cos \left[\left(\frac{\pi}{4}+\alpha\right)-\left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right]=\cos \left(\frac{\pi}{4}+\alpha\right) \cos \left(\frac{\pi}{4}-\frac{\beta}{2}\right)+\sin \left(\frac{\pi}{4}+\alpha\right) \sin \left(\frac{\pi}{4}-\frac{\beta}{2}\right)=\frac{5 \sqrt{3}}{9}$
故选 C

点评：本题主要考查了三角函数的恒等变换及化简求值。关键是根据 $\cos \left(\alpha+\frac{\beta}{2}\right)=\cos \left[\left(\frac{\pi}{4}+\alpha\right)-\left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right]$ ，巧妙利用两角和公式进行求解．