《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》

例1. 求值：
(1)$\sin 20^{\circ} \cos 40^{\circ}+\cos 20^{\circ} \sin 40^{\circ}=$$\frac{\sqrt{3}}{2}$；
(2)计算:$\frac{1+\tan 15^{\circ}}{\sqrt{3}-\tan 60^{\circ} \tan 15^{\circ}}=$1 [归纳提升]

答案： 例1:
(1)$\frac{\sqrt{3}}{2}$
(2)1
(1) $\sin 20° \cos 40° + \cos 20° \sin 40° =$
$\sin(20° + 40°) = \sin 60° = \frac{\sqrt{3}}{2}$.
(2)原式=$\frac{\tan 45° + \tan 15°}{\sqrt{3}(1 - \tan 45° \tan 15°)}$
=$\frac{1}{\sqrt{3}} \tan(45° + 15°)$
=$\frac{1}{\sqrt{3}} \tan 60° = \frac{1}{\sqrt{3}} × \sqrt{3} = 1$.

对点训练1
求下列各式的值：
(1)$\sin 347^{\circ} \cos 148^{\circ}+\sin 77^{\circ} \cos 58^{\circ}$；
(2)$\sqrt{3} \sin \frac{\pi}{12}+\cos \frac{\pi}{12}$；
(3)$\frac{1+\tan 105^{\circ}}{1-\tan 105^{\circ}}$.

答案： 对点训练1:
(1)原式=$\sin(360° - 13°)\cos(180° - 32°) +$
$\sin(90° - 13°)\cos(90° - 32°)$
=$\sin 13° \cos 32° + \cos 13° \sin 32°$
=$\sin(13° + 32°) = \sin 45° = \frac{\sqrt{2}}{2}$.
(2)原式=$2(\frac{\sqrt{3}}{2} \sin \frac{\pi}{12} + \frac{1}{2} \cos \frac{\pi}{12})$
=$2(\sin \frac{\pi}{12} \cos \frac{\pi}{6} + \sin \frac{\pi}{6} \cos \frac{\pi}{12})$
=$2\sin(\frac{\pi}{12} + \frac{\pi}{6}) = 2\sin \frac{\pi}{4} = \sqrt{2}$.
(3)原式=$\frac{\tan 45° + \tan 105°}{1 - \tan 45° \tan 105°}$
=$\tan(45° + 105°) = \tan 150° = -\frac{\sqrt{3}}{3}$.

例2. (1)已知$\cos \alpha=-\frac{4}{5}$, 且$\alpha \in\left(\frac{\pi}{2}, \pi\right)$, 则$\tan \left(\frac{\pi}{4}-\alpha\right)=$ (D)
A.$-\frac{1}{7}$
B.$-7$
C.$\frac{1}{7}$
D.$7$
(2)已知$\alpha$为锐角, $\sin \alpha=\frac{3}{5}$, $\beta$是第四象限角, $\cos \beta=\frac{4}{5}$, 则$\sin (\alpha+\beta)=$0；
(3)已知$\frac{\pi}{2}＜\beta＜\alpha＜\frac{3 \pi}{4}, \cos (\alpha-\beta)=\frac{12}{13}, \sin (\alpha+\beta)=-\frac{3}{5}$, 求$\sin 2 \alpha$的值.
【分析】(2) 先求出$\cos \alpha, \sin \beta$的值, 再代入公式$S_{\alpha+\beta}$.
(3) 由$\alpha, \beta$的范围, 确定$\alpha-\beta, \alpha+\beta$的范围, 求出$\sin (\alpha-\beta)、\cos (\alpha+\beta)$的值, 再由$2 \alpha=(\alpha-\beta)+(\alpha+\beta)$变形求值.
[归纳提升]

答案： 例2:
(1)D
(2)0
(3)见解析
【解析】
(1)由 $\cos \alpha = -\frac{4}{5}$,且 $\alpha \in (\frac{\pi}{2}, \pi)$,得 $\sin \alpha$
=$\frac{3}{5}$,
所以 $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = -\frac{3}{4}$,
所以 $\tan(\frac{\pi}{4} - \alpha) = \frac{\tan \frac{\pi}{4} - \tan \alpha}{1 + \tan \frac{\pi}{4} \tan \alpha} = 7$,故
选D.
(2)$\because \alpha$为锐角,$\sin \alpha = \frac{3}{5}$,$\therefore \cos \alpha = \frac{4}{5}$
$\because \beta$为第四象限角,$\cos \beta = \frac{4}{5}$,$\therefore \sin \beta = -\frac{3}{5}$,
$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{3}{5} × \frac{4}{5} + \frac{4}{5} ×$
($-\frac{3}{5}) = 0$.
(3)因为 $\frac{\pi}{2} ＜ \beta ＜ \alpha ＜ \frac{3\pi}{4}$,
所以$0 ＜ \alpha - \beta ＜ \frac{\pi}{4}$,$\pi ＜ \alpha + \beta ＜ \frac{3\pi}{2}$.
又 $\cos(\alpha - \beta) = \frac{12}{13}$,$\sin(\alpha + \beta) = -\frac{3}{5}$,
所以 $\sin(\alpha - \beta) = \sqrt{1 - \cos^2(\alpha - \beta)} = \sqrt{1 - (\frac{12}{13})^2} = \frac{5}{13}$,
$\cos(\alpha + \beta) = -\sqrt{1 - \sin^2(\alpha + \beta)} = -\sqrt{1 - (-\frac{3}{5})^2} =$
$-\frac{4}{5}$.
所以 $\sin 2\alpha = \sin[(\alpha - \beta) + (\alpha + \beta)]$
=$\sin(\alpha - \beta)\cos(\alpha + \beta) + \cos(\alpha - \beta)\sin(\alpha + \beta)$
=$\frac{5}{13} × (-\frac{4}{5}) + \frac{12}{13} × (-\frac{3}{5}) = -\frac{56}{65}$.