《2025年成才之路高中新课程学习指导高中数学必修第二册北师大版》

知识点 二倍角公式

$= \underline{\underline{2\cos^{2}\alpha - 1}}$

答案： $2\sin \alpha\cos \alpha$ $\cos^2 \alpha - \sin^2 \alpha$ $2\cos^2 \alpha - 1$ $1 - 2\sin^2 \alpha$

例 1. 求下列各式的值：
(1)$\sin \frac{\pi}{12}\cos \frac{\pi}{12}$; (2)$1 - 2\sin^{2}750^{\circ}$; (3)$\frac{2\tan 150^{\circ}}{1 - \tan^{2}150^{\circ}}$; (4)$\cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}$.
【分析】 观察角的特点$\to$寻求角的联系$\to$选择公式$\to$化简求值

答案： 例1：
(1)原式$=\frac{2\sin \frac{\pi}{12}\cos \frac{\pi}{12}}{2} = \frac{\sin \frac{\pi}{6}}{2} = \frac{1}{4}$.
(2)原式$=\cos(2 × 750°) = \cos 1500°$
$=\cos(4 × 360° + 60°) = \cos 60° = \frac{1}{2}$.
(3)原式$=\tan(2 × 150°) = \tan 300° =$
$\tan(360° - 60°)$
$= -\tan 60° = -\sqrt{3}$.
(4)原式$=\frac{2\sin 20° · \cos 20° · \cos 40° · \cos 80°}{2\sin 20°}$
$=\frac{2\sin 40° · \cos 40° · \cos 80°}{4\sin 20°} = \frac{2\sin 80° · \cos 80°}{8\sin 20°}$
$=\frac{\sin 160°}{8\sin 20°} = \frac{1}{8}$.

对点训练 1
(1)$\frac{\sin 70^{\circ}\sqrt{1 - \sin 50^{\circ}}}{\sin 40^{\circ}}$的值为 (B)

A. $\frac{1}{2}$
B. $\frac{\sqrt{2}}{2}$
C. $\sqrt{2}$
D. $2$
(2)$\cos \frac{\pi}{9}\cos \frac{2\pi}{9}\sin \frac{3\pi}{9}\cos \frac{4\pi}{9} =$$\frac{\sqrt{3}}{16}$.

答案： 对点训练1：
(1)B
(2)$\frac{\sqrt{3}}{16}$
(1)$\frac{\sin 70°\sqrt{1 - \sin 50°}}{\sin 40°} =$
$\frac{\cos 20°\sqrt{\sin^2 25° + \cos^2 25° - 2\sin 25°\cos 25°}}{2\sin 20°\cos 20°} = \frac{\cos 25° - \sin 25°}{2\sin 20°}$
$=\frac{\sqrt{2}\sin(45° - 25°)}{2\sin 20°} = \frac{\sqrt{2}}{2}$,故选B.
(2)$\cos \frac{\pi}{9}\cos \frac{2\pi}{9}\cos \frac{3\pi}{9}\cos \frac{4\pi}{9}$
$=\sin \frac{\pi}{9}\cos \frac{\pi}{9}\cos \frac{2\pi}{9}\cos \frac{3\pi}{9}\cos \frac{4\pi}{9}$
$=\frac{1}{2}\sin \frac{2\pi}{9}\cos \frac{2\pi}{9}\cos \frac{\pi}{3}\cos \frac{4\pi}{9}$
$=\frac{1}{4}\sin \frac{4\pi}{9}\cos \frac{4\pi}{9}\sin \frac{\pi}{3} = \frac{1}{8}\sin \frac{8\pi}{9}\sin \frac{\pi}{3} = \frac{1}{8}\sin \frac{\pi}{9}\sin \frac{\pi}{3} = \frac{1}{8}\sin \frac{\pi}{3} = \frac{\sqrt{3}}{16}$.