浙江 2013 · 数学 q06

题面

6．（5分）（2013•浙江）已知$\sin a + 2 \cos a = \frac{\sqrt{1 0}}{2}$，则 tan2α=（ ） A$\frac{4}{3}$ B$\frac{3}{4}$ C$- \frac{3}{4}$ D$- \frac{4}{3}$

答案

C

解析

分析

由题意结合$\sin^{2} \alpha + \cos^{2} \alpha = 1$可解得sinα，和cosα，进而可得tanα，再代入二倍角的正切公式可得答案

解答

解：$\because \sin a + 2 \cos a = \frac{\sqrt{1 0}}{2} ,$，又$\sin^{2} \alpha + \cos^{2} \alpha = 1$

联立解得$\left\{\begin{array}{l l}{\sin \mathbb{a} = - {\frac{\sqrt{1 0}}{1 0}}} \\ {\cos \mathbb{a} = {\frac{3 {\sqrt{1 0}}}{1 0}}} \end{array} \right. ,$，或$\left\{\begin{array}{l l}{\sin \alpha = {\frac{3 {\sqrt{1 0}}}{1 0}}} \\ {\cos \alpha = {\frac{\sqrt{1 0}}{1 0}}} \end{array} \right.$

故$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = - \frac{1}{3} ,$或$\tan a = 3$

代入可得$\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan^{2} \alpha} = \frac{2 \times ( - \frac{1}{3} )}{1 - ( - \frac{1}{3} ) 2} = - \frac{3}{4} \tan^{2} \alpha$

或$\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan^{2} \alpha} = \frac{2 \times 3}{1 - 3^{2}} - \frac{3}{4}$

故选 C