《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 2. (1) 若$\cos(\frac{\pi}{4} - \alpha) = \frac{4}{5}$, 则$\sin 2\alpha =$$\frac{7}{25}$.
(2) 已知$\alpha$是第二象限角, $\tan(\pi + 2\alpha) = -\frac{4}{3}$, 则$\tan \alpha =$$-\frac{1}{2}$.
[归纳提升]

答案： 例2：
(1)$\frac{7}{25}$
(2)$-\frac{1}{2}$
(1)方法一：由$\cos(\frac{\pi}{4} - \alpha) = \frac{4}{5}$,
得$\frac{\sqrt{2}}{2}(\sin \alpha + \cos \alpha) = \frac{4}{5}$. 两边同时平方,得$\frac{1}{2}(\sin \alpha + \cos \alpha)^2 =$
$\frac{16}{25}$. 故$1 + \sin 2\alpha = \frac{32}{25}$.
所以$\sin 2\alpha = \frac{7}{25}$.
方法二：由二倍角公式,得$\cos^2(\frac{\pi}{4} - \alpha) = \frac{1 + \cos(\frac{\pi}{2} - 2\alpha)}{2} =$
$\frac{1 + \sin 2\alpha}{2} = \frac{16}{25}$,所以$\sin 2\alpha = \frac{7}{25}$.
方法三：因为$\cos(\frac{\pi}{4} - \alpha) = \frac{4}{5}$,所以$\sin 2\alpha =$
$\cos(\frac{\pi}{2} - 2\alpha) = \cos 2(\frac{\pi}{4} - \alpha) = 2\cos^2(\frac{\pi}{4} - \alpha) - 1 = 2 × \frac{16}{25} -$
$1 = \frac{7}{25}$.
(2)由题设得$\tan(\pi + 2\alpha) = \tan 2\alpha = -\frac{4}{3}$. 由二倍角公式,
得$\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^2 \alpha} = -\frac{4}{3}$,整理得$2\tan^2 \alpha - 3\tan \alpha - 2 = 0$,解得
$\tan \alpha = 2$或$\tan \alpha = -\frac{1}{2}$. 因为$\alpha$是第二象限的角,所以$\tan \alpha =$
$-\frac{1}{2}$.

对点训练 2
已知$\tan \alpha + \frac{1}{\tan \alpha} = \frac{5}{2}, \alpha \in (\frac{\pi}{4}, \frac{\pi}{2})$, 求$\cos 2\alpha$和$\sin(2\alpha + \frac{\pi}{4})$的值.

答案： 对点训练2：由$\tan \alpha + \frac{1}{\tan \alpha} = \frac{5}{2}$,得$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} = \frac{5}{2}$,
则$\frac{2}{\sin 2\alpha} = \frac{5}{2}$,即$\sin 2\alpha = \frac{4}{5}$,
因为$\alpha \in (\frac{\pi}{4}, \frac{\pi}{2})$,所以$2\alpha \in (\frac{\pi}{2}, \pi)$,
所以$\cos 2\alpha = -\sqrt{1 - \sin^2 2\alpha} = -\frac{3}{5}$,$\sin(2\alpha + \frac{\pi}{4}) =$
$\sin 2\alpha · \cos \frac{\pi}{4} + \cos 2\alpha · \sin \frac{\pi}{4} = \frac{4}{5} × \frac{\sqrt{2}}{2} - \frac{3}{5} × \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{10}$.

例 3. 已知$\tan(\alpha - \beta) = \frac{1}{2}, \tan \beta = -\frac{1}{7}$, 且$\alpha, \beta \in (0, \pi)$, 求$2\alpha - \beta$的值.
【分析】 本题根据$\tan \beta = -\frac{1}{7} ＜ 0$且$\beta \in (0, \pi)$, 确定$\frac{\pi}{2} ＜ \beta ＜ \pi$, 可求得$\tan \alpha = \frac{1}{3}$且$\alpha \in (0, \pi)$, 确定$0 ＜ \alpha ＜ \frac{\pi}{4}$, 这是求角的范围的关键.
[归纳提升]

答案： 例3：因为$2\alpha - \beta = 2(\alpha - \beta) + \beta$,$\tan(\alpha - \beta) = \frac{1}{2}$,
而$\tan 2(\alpha - \beta) = \frac{2\tan(\alpha - \beta)}{1 - \tan^2(\alpha - \beta)} = \frac{4}{3}$.
从而$\tan(2\alpha - \beta) = \tan[2(\alpha - \beta) + \beta] =$
$\frac{\tan 2(\alpha - \beta) + \tan \beta}{1 - \tan 2(\alpha - \beta) · \tan \beta} = \frac{\frac{4}{3} - \frac{1}{7}}{1 + \frac{4}{3} × \frac{1}{7}} = 1$.
又因为$\tan \alpha = \tan[(\alpha - \beta) + \beta] =$
$\frac{\tan(\alpha - \beta) + \tan \beta}{1 - \tan(\alpha - \beta) · \tan \beta} = \frac{\frac{1}{2} - \frac{1}{7}}{1 - \frac{1}{2} × \frac{1}{7}} ＜ 1$,
且$\alpha \in (0, \pi)$,所以$0 ＜ \alpha ＜ \frac{\pi}{4}$. 所以$0 ＜ 2\alpha ＜ \frac{\pi}{2}$.
又因为$\tan \beta = -\frac{1}{7} ＜ 0$,且$\beta \in (0, \pi)$,
所以$\frac{\pi}{2} ＜ \beta ＜ \pi$,$-\pi ＜ -\beta ＜ -\frac{\pi}{2}$,所以$-\pi ＜ 2\alpha - \beta ＜ 0$.
所以$2\alpha - \beta = -\frac{3\pi}{4}$.

对点训练 3
已知$\tan \alpha = \frac{1}{7}, \tan \beta = \frac{1}{3}$, 并且$\alpha, \beta$均为锐角, 求$\alpha + 2\beta$的值.

答案： 对点训练3：因为$\tan \beta = \frac{1}{3}$,
所以$\tan 2\beta = \frac{2\tan \beta}{1 - \tan^2 \beta} = \frac{2 × \frac{1}{3}}{1 - (\frac{1}{3})^2} = \frac{3}{4}$.
所以$\tan(\alpha + 2\beta) = \frac{\tan \alpha + \tan 2\beta}{1 - \tan \alpha\tan 2\beta} = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7} × \frac{3}{4}} = 1$.
$0 ＜ \tan \alpha = \frac{1}{7} ＜ 1$,$0 ＜ \tan \beta = \frac{1}{3} ＜ 1$,$\alpha,\beta$均为锐角,所以
$0 ＜ \alpha ＜ \frac{\pi}{4}$,$0 ＜ \beta ＜ \frac{\pi}{4}$,$0 ＜ 2\beta ＜ \frac{\pi}{2}$. 所以$0 ＜ \alpha + 2\beta ＜ \frac{3\pi}{4}$,又$\tan(\alpha$
$+ 2\beta) = 1$. 所以$\alpha + 2\beta = \frac{\pi}{4}$.