《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》

对点训练2
已知$\sin \alpha=\frac{\sqrt{5}}{5},0＜\alpha＜\frac{\pi}{2}$,且$\cos \beta=\frac{\sqrt{10}}{10},0＜\beta＜\frac{\pi}{2}$,则$\cos (\alpha+\beta)=$$- \frac{\sqrt{2}}{10}$.

答案： 对点训练2:$- \frac{\sqrt{2}}{10}$ 因为$\sin \alpha = \frac{\sqrt{5}}{5}$,$0 ＜ \alpha ＜ \frac{\pi}{2}$,且$\cos \beta = \frac{\sqrt{10}}{10}$,$0 ＜ \beta ＜ \frac{\pi}{2}$,所以$\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \frac{2\sqrt{5}}{5}$,$\sin \beta = \sqrt{1 - \cos^2 \beta} = \frac{3\sqrt{10}}{10}$,所以$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \frac{2\sqrt{5}}{5} × \frac{\sqrt{10}}{10} - \frac{\sqrt{5}}{5} × \frac{3\sqrt{10}}{10} = - \frac{\sqrt{2}}{10}$.

例3. (1)已知$\alpha$为三角形的内角且$\frac{1}{2} \cos \alpha+\frac{\sqrt{3}}{2} \sin \alpha=\frac{1}{2}$,则$\alpha=$$\frac{2}{3}\pi$;
(2)已知$\cos (\alpha-\beta)=-\frac{12}{13},\cos (\alpha+\beta)=\frac{12}{13}$,且$\alpha-\beta \in(\frac{\pi}{2},\pi),\alpha+\beta \in(\frac{3 \pi}{2},2 \pi)$,求角$\beta$的值.
【分析】 (1)由公式可求出$\cos (\alpha-\frac{\pi}{3})$的值，再根据$\alpha$的范围确定$\alpha-\frac{\pi}{3}$的值.
(2)由条件可发现角与角之间的关系:$2 \beta=(\alpha+\beta)-(\alpha-\beta)$,所以应先求出$2 \beta$的值，再求$\beta$的值.
▶[归纳提升]

答案： 例3:
(1)$\frac{2}{3}\pi$
(2)见解析
【解析】
(1)$\because \frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha = \cos(\alpha - \frac{\pi}{3}) = \frac{1}{2}$
又$\because 0 ＜ \alpha ＜ \pi$,$\therefore - \frac{\pi}{3} ＜ \alpha - \frac{\pi}{3} ＜ \frac{2\pi}{3}$,
$\therefore \alpha - \frac{\pi}{3} = \frac{\pi}{3}$,$\alpha = \frac{2}{3} \pi$.
(2)由$\alpha - \beta \in (\frac{\pi}{2}, \pi)$,且$\cos(\alpha - \beta) = - \frac{12}{13}$,得$\sin(\alpha - \beta) = - \frac{5}{13}$
由$\alpha + \beta \in (\frac{3\pi}{2}, 2\pi)$,且$\cos(\alpha + \beta) = \frac{12}{13}$,得$\sin(\alpha + \beta) = - \frac{5}{13}$
$\cos 2\beta = \cos[(\alpha + \beta) - (\alpha - \beta)]$
$ = \cos(\alpha + \beta) \cos(\alpha - \beta) + \sin(\alpha + \beta) \sin(\alpha - \beta)$
$ = - \frac{12}{13} × \frac{12}{13} + ( - \frac{5}{13}) × \frac{5}{13} = - 1$.
又因为$\alpha + \beta \in (\frac{3\pi}{2}, 2\pi)$,$\alpha - \beta \in (\frac{\pi}{2}, \pi)$,所以$2\beta \in (\frac{\pi}{2}, \frac{3\pi}{2})$.所以$2\beta = \pi$,所以$\beta = \frac{\pi}{2}$.

对点训练3
已知$\sin (\pi-\alpha)=\frac{4 \sqrt{3}}{7},\cos (\alpha-\beta)=\frac{13}{14},0＜\beta＜\alpha＜\frac{\pi}{2}$,求$\beta$的大小.

答案： 对点训练3:$\because \sin(\pi - \alpha) = \sin \alpha = \frac{4\sqrt{3}}{7}$,$0 ＜ \alpha ＜ \frac{\pi}{2}$,$\therefore \cos \alpha = \frac{1}{7}$,
又$\because 0 ＜ \beta ＜ \alpha ＜ \frac{\pi}{2}$,$\therefore 0 ＜ \alpha - \beta ＜ \frac{\pi}{2}$,又$\cos(\alpha - \beta) = \frac{13}{14}$,
$\therefore \sin(\alpha - \beta) = \frac{3\sqrt{3}}{14}$.
$\therefore \cos \beta = \cos[\alpha - (\alpha - \beta)] = \cos \alpha \cos(\alpha - \beta) + \sin \alpha \sin(\alpha - \beta) = \frac{1}{7} × \frac{13}{14} + \frac{4\sqrt{3}}{7} × \frac{3\sqrt{3}}{14} = \frac{13 + 36}{98} = \frac{1}{2}$.
又$\because 0 ＜ \beta ＜ \frac{\pi}{2}$,$\therefore \beta = \frac{\pi}{3}$.