《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. (1)求值:$\cos 75^{\circ}=$$\frac{\sqrt{6} - \sqrt{2}}{4}$;
(2)求值:$\sin 7^{\circ} \cos 23^{\circ}+\sin 83^{\circ} \cos 67^{\circ}=$$\frac{1}{2}$;
(3)计算:$\cos (\alpha-35^{\circ}) \cos (25^{\circ}+\alpha)+\sin (\alpha-35^{\circ}) \sin (25^{\circ}+\alpha)=$$\frac{1}{2}$.
【分析】 尝试逆用公式求解，非特殊角转化为特殊角的差，然后正用$\mathrm{C}_{\alpha \pm \beta}$进行求值.
▶[归纳提升]

答案： 例1:
(1)$\frac{\sqrt{6} - \sqrt{2}}{4}$
(2)$\frac{1}{2}$
(3)$\frac{1}{2}$
(1)$\cos 75° = \cos(45° + 30°) = \cos 45° · \cos 30° - \sin 45° · \sin 30° = \frac{\sqrt{2}}{2} × \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} × \frac{1}{2}$
$ = \frac{\sqrt{6} - \sqrt{2}}{4}$
(2)原式$ = \cos 83° \cos 23° + \sin 83° \sin 23° = \cos(83° - 23°)$
$ = \cos 60° = \frac{1}{2}$
(3)原式$ = \cos[(\alpha - 35°) - (25° + \alpha)] = \cos(-60°) = \cos 60° = \frac{1}{2}$.

对点训练1
求下列各式的值.
(1)$\cos 40^{\circ} \cos 20^{\circ}-\sin 40^{\circ} \sin 20^{\circ}$;
(2)$\cos \frac{7}{3} \pi \cos \frac{11}{6} \pi+\sin \frac{2}{3} \pi \sin \frac{5}{6} \pi$.

答案： 对点训练1:
(1)原式$ = \cos(40° + 20°) = \cos 60° = \frac{1}{2}$
(2)原式$ = \cos(2\pi + \frac{\pi}{3}) \cos(2\pi - \frac{\pi}{6}) + \sin(\pi - \frac{\pi}{3}) \sin(\pi - \frac{\pi}{6})$
$ = \cos \frac{\pi}{3} \cos \frac{\pi}{6} + \sin \frac{\pi}{3} \sin \frac{\pi}{6} = \cos(\frac{\pi}{3} - \frac{\pi}{6})$
$ = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$.

例2. (1)已知$\sin \alpha=-\frac{4}{5},\sin \beta=\frac{5}{13}$,且$180^{\circ}＜\alpha＜270^{\circ},90^{\circ}＜\beta＜180^{\circ}$,则$\cos (\alpha-\beta)=$$\frac{16}{65}$;
(2)已知$\sin (\alpha+\frac{\pi}{4})=\frac{4}{5}$,且$\frac{\pi}{4}＜\alpha＜\frac{3 \pi}{4}$,求$\cos \alpha$的值.
【分析】 (1)求出$\cos \alpha,\cos \beta$,利用公式进行求解;
(2)利用$\cos \alpha=\cos [(\alpha+\frac{\pi}{4})-\frac{\pi}{4}]$进行凑角.
▶[归纳提升]

答案： 例2:
(1)$\frac{16}{65}$
(2)见解析
【解析】
(1)$\because 180° ＜ \alpha ＜ 270°$,$\therefore \cos \alpha = - \frac{3}{5}$;
又$\because 90° ＜ \beta ＜ 180°$,$\therefore \cos \beta = - \frac{12}{13}$;
$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
$ = ( - \frac{3}{5}) × ( - \frac{12}{13}) + ( - \frac{4}{5}) × \frac{5}{13} = \frac{16}{65}$
(2)$\because \frac{\pi}{4} ＜ \alpha ＜ \frac{3\pi}{4}$,$\therefore \frac{\pi}{2} ＜ \alpha + \frac{\pi}{4} ＜ \pi$,
$\therefore \cos(\alpha + \frac{\pi}{4}) = - \frac{3}{5}$,
$\cos \alpha = \cos[(\alpha + \frac{\pi}{4}) - \frac{\pi}{4}]$
$ = \cos(\alpha + \frac{\pi}{4}) · \cos \frac{\pi}{4} + \sin(\alpha + \frac{\pi}{4}) \sin \frac{\pi}{4}$
$ = - \frac{3}{5} × \frac{\sqrt{2}}{2} + \frac{4}{5} × \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{10}$.