《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
(1) 已知$\tan (\alpha+\beta)=\frac{2}{5}, \tan \left(\beta-\frac{\pi}{4}\right)=\frac{1}{4}$, 那么$\tan \left(\alpha+\frac{\pi}{4}\right)$等于 (C)
A.$\frac{13}{18}$
B.$\frac{13}{22}$
C.$\frac{3}{22}$
D.$\frac{1}{6}$
(2) 已知$\cos \left(\alpha+\frac{\pi}{6}\right)=\frac{1}{7}, 0＜\alpha＜\pi$, 则$\sin \alpha$的值为 (C)
A.$\frac{3 \sqrt{3}}{14}$
B.$\frac{5 \sqrt{3}}{14}$
C.$\frac{11}{14}$
D.$\frac{13}{14}$
(3) 若$0＜\alpha＜\frac{\pi}{2}, -\frac{\pi}{2}＜\beta＜0, \cos \left(\frac{\pi}{4}+\alpha\right)=\frac{1}{3}, \cos \left(\frac{\pi}{4}-\frac{\beta}{2}\right)=\frac{\sqrt{3}}{3}$, 求$\cos \left(\alpha+\frac{\beta}{2}\right)$.

答案： 对点训练2:
(1)C
(2)C
(3)见解析
【解析】
(1) $\tan(\alpha + \frac{\pi}{4}) = \tan[(\alpha + \beta) - (\beta - \frac{\pi}{4})] =$
$\frac{\tan(\alpha + \beta) - \tan(\beta - \frac{\pi}{4})}{1 + \tan(\alpha + \beta)\tan(\beta - \frac{\pi}{4})} = \frac{\frac{2}{5} - \frac{1}{4}}{1 + \frac{2}{5} × \frac{1}{4}} = \frac{3}{22}$.故选C.
(2)因为 $\cos(\alpha + \frac{\pi}{6}) = \frac{1}{7}$,且$0 ＜ \alpha ＜ \pi$,则$\frac{\pi}{6} ＜ \alpha + \frac{\pi}{6} ＜$
$\frac{\pi}{2}$,所以 $\sin(\alpha + \frac{\pi}{6}) = \sqrt{1 - \cos^2(\alpha + \frac{\pi}{6})} = \frac{4\sqrt{3}}{7}$,所以 $\sin \alpha =$
$\sin[(\alpha + \frac{\pi}{6}) - \frac{\pi}{6}] = \sin(\alpha + \frac{\pi}{6})\cos \frac{\pi}{6} - \cos(\alpha + \frac{\pi}{6})\sin \frac{\pi}{6}$
=$\frac{4\sqrt{3}}{7} × \frac{\sqrt{3}}{2} - \frac{1}{7} × \frac{1}{2} = \frac{11}{14}$.故选C.
(3)$\because 0 ＜ \alpha ＜ \frac{\pi}{2}$,$\therefore \frac{\pi}{4} ＜ \frac{\pi}{4} + \alpha ＜ \frac{3\pi}{4}$,
$\therefore \sin(\frac{\pi}{4} + \alpha) = \frac{2\sqrt{2}}{3}$,
$\because -\frac{\pi}{2} ＜ \beta ＜ 0$,$\therefore \frac{\pi}{4} ＜ \frac{\pi}{4} - \beta ＜ \frac{\pi}{2}$,$\therefore \sin(\frac{\pi}{4} - \beta) = \frac{\sqrt{6}}{3}$
$\cos(\alpha + \frac{\beta}{2}) = \cos[(\frac{\pi}{4} + \alpha) - (\frac{\pi}{4} - \frac{\beta}{2})]$
=$\cos(\frac{\pi}{4} + \alpha)\cos(\frac{\pi}{4} - \frac{\beta}{2}) + \sin(\frac{\pi}{4} + \alpha)\sin(\frac{\pi}{4} - \frac{\beta}{2})$
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=$\frac{1}{3} × \frac{\sqrt{3}}{3} + \frac{2\sqrt{2}}{3} × \frac{\sqrt{6}}{3} = \frac{5\sqrt{3}}{9}$.

例3. 已知$\sin \alpha=\frac{\sqrt{5}}{5}, \sin \beta=\frac{\sqrt{10}}{10}$, 且$\alpha, \beta$为锐角, 求$\alpha+\beta$的值.
[归纳提升]

答案： 例3:$\because \alpha, \beta$为锐角,$\sin \alpha = \frac{\sqrt{5}}{5}$,$\sin \beta = \frac{\sqrt{10}}{10}$,
$\therefore \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \frac{2\sqrt{5}}{5}$,$\cos \beta = \sqrt{1 - \sin^2 \beta} = \frac{3\sqrt{10}}{10}$,
$\therefore \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \frac{2\sqrt{5}}{5} × \frac{3\sqrt{10}}{10} - \frac{\sqrt{5}}{5} ×$
$\frac{\sqrt{10}}{10} = \frac{\sqrt{2}}{2}$,
$\because \alpha, \beta$为锐角,$\therefore 0° ＜ \alpha + \beta ＜ 180°$,$\therefore \alpha + \beta = 45°$.