《2026年南方凤凰台5A新考案高中数学二轮基础版》

例 1 (1) (2025·深圳一调)已知 $\dfrac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} = 3$，则 $\dfrac{\tan \alpha}{\tan \beta} =$(C)

A.$\dfrac{1}{3}$
B.$\dfrac{1}{2}$
C.$2$
D.$3$

答案： 例1
(1)C 【解析】方法一：因为$\frac{\sin (\alpha + \beta)}{\sin (\alpha - \beta)} = 3$，所以$4 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta$，所以$\frac{\tan \alpha}{\tan \beta} = 2$。

(2) (2023·新高考Ⅰ卷)若 $\sin(\alpha - \beta) = \dfrac{1}{3}, \cos \alpha \sin \beta = \dfrac{1}{6}$，则 $\cos(2\alpha + 2\beta) =$(B)

A.$\dfrac{7}{9}$
B.$\dfrac{1}{9}$
C.$-\dfrac{1}{9}$
D.$-\dfrac{7}{9}$

答案：
(2)B 【解析】因为$\frac{\tan \alpha}{\tan \beta} + 1 = 3$，可得$\frac{\tan \alpha}{\tan \beta} = 2$。因为$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \frac{1}{3}$，而$\cos \alpha \sin \beta = \frac{1}{6}$，所以$\sin \alpha \cos \beta = \frac{1}{2}$，则$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{2}{3}$，所以$\cos (2 \alpha + 2 \beta) = \cos [2 (\alpha + \beta)] = 1 - 2 \sin^{2} (\alpha + \beta) = 1 - 2 × \left( \frac{2}{3} \right)^{2} = \frac{1}{9}$。

(3) (2024·新高考Ⅱ卷)已知 $\alpha$ 为第一象限角，$\beta$ 为第三象限角，$\tan \alpha + \tan \beta = 4, \tan \alpha \tan \beta = \sqrt{2} + 1$，则 $\sin(\alpha + \beta) =$$ - \frac{2 \sqrt{2}}{3}$。

答案：
(3)$ - \frac{2 \sqrt{2}}{3}$ 【解析】方法一：由题意得$\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{4}{1 - (\sqrt{2} + 1)} = - 2 \sqrt{2}$。因为$\alpha \in \left( 2k \pi, 2k \pi + \frac{\pi}{2} \right)$，$\beta \in \left( 2m \pi + \pi, 2m \pi + \frac{3 \pi}{2} \right)$，$k, m \in \mathbb{Z}$，所以$\alpha + \beta \in \left( (2m + 2k) \pi + \pi, (2m + 2k) \pi + 2 \pi \right)$，$k, m \in \mathbb{Z}$，又$\tan (\alpha + \beta) = - 2 \sqrt{2} ＜ 0$，所以$\alpha + \beta \in \left( (2m + 2k) \pi + \frac{3 \pi}{2}, (2m + 2k) \pi + 2 \pi \right)$，$k, m \in \mathbb{Z}$，则$\sin (\alpha + \beta) ＜ 0$。由$\frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} = - 2 \sqrt{2}$，联立$\sin^{2} (\alpha + \beta) + \cos^{2} (\alpha + \beta) = 1$，解得$\sin (\alpha + \beta) = - \frac{2 \sqrt{2}}{3}$。
方法二：因为$\alpha$为第一象限角，$\beta$为第三象限角，所以$\cos \alpha ＞ 0$，$\cos \beta ＜ 0$，$\cos \alpha = \frac{\cos \alpha}{\sqrt{\sin^{2} \alpha + \cos^{2} \alpha}} = \frac{1}{\sqrt{1 + \tan^{2} \alpha}}$，$\cos \beta = \frac{\cos \beta}{\sqrt{\sin^{2} \beta + \cos^{2} \beta}} = \frac{- 1}{\sqrt{1 + \tan^{2} \beta}}$，则$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \cos \alpha \cos \beta · (\tan \alpha + \tan \beta) = 4 \cos \alpha \cos \beta = \frac{4}{\sqrt{1 + \tan^{2} \alpha} \sqrt{1 + \tan^{2} \beta}} = \frac{4}{\sqrt{(\tan \alpha + \tan \beta)^{2} + (\tan \alpha \tan \beta - 1)^{2}}} = \frac{4}{\sqrt{4^{2} + 2}} = - \frac{2 \sqrt{2}}{3}$。

变式 1
(1) (2025·武汉调研)若 $\tan\left(\alpha + \dfrac{\pi}{4}\right) = 7$，则 $\cos 2\alpha$ 的值为(A)
A.$\dfrac{7}{25}$
B.$\dfrac{3}{4}$
C.$\dfrac{12}{25}$
D.$\dfrac{4}{5}$

答案： 变式1
(1)A 【解析】方法一：由$\tan \left( \alpha + \frac{\pi}{4} \right) = 7$，可得$\frac{\tan \alpha + \tan \frac{\pi}{4}}{1 - \tan \alpha \tan \frac{\pi}{4}} = 7$，即$\frac{\tan \alpha + 1}{1 - \tan \alpha} = 7$，解得$\tan \alpha = \frac{3}{4}$，所以$\cos 2 \alpha = \frac{\cos^{2} \alpha - \sin^{2} \alpha}{\cos^{2} \alpha + \sin^{2} \alpha} = \frac{1 - \tan^{2} \alpha}{1 + \tan^{2} \alpha} = \frac{1 - \left( \frac{3}{4} \right)^{2}}{1 + \left( \frac{3}{4} \right)^{2}} = \frac{7}{25}$。
方法二：$\cos 2 \alpha = \sin \left( 2 \alpha + \frac{\pi}{2} \right) = 2 \sin \left( \alpha + \frac{\pi}{4} \right) \cos \left( \alpha + \frac{\pi}{4} \right) = \frac{2 \tan \left( \alpha + \frac{\pi}{4} \right)}{\tan^{2} \left( \alpha + \frac{\pi}{4} \right)+1} = \frac{14}{49 + 1} = \frac{7}{25}$。

(2) (2022·浙江卷)若 $3\sin \alpha - \sin \beta = \sqrt{10}, \alpha + \beta = \dfrac{\pi}{2}$，则 $\sin \alpha =$$\frac{3 \sqrt{10}}{10}$，$\cos 2\beta =$$\frac{4}{5}$。

答案：
(2)$\frac{3 \sqrt{10}}{10}$ 【解析】方法一：因为$\alpha + \beta = \frac{\pi}{2}$，所以$\cos \beta = \sin \alpha$，$\sin \beta = \cos \alpha$，即$3 \sin \alpha - \cos \alpha = \sqrt{10}$，即$\sqrt{10} \left( \frac{3 \sqrt{10}}{10} \sin \alpha - \frac{\sqrt{10}}{10} \cos \alpha \right) = \sqrt{10}$。令$\sin \theta = \frac{\sqrt{10}}{10}$，$\cos \theta = \frac{3 \sqrt{10}}{10}$，则$\sqrt{10} \sin (\alpha - \theta) = \sqrt{10}$，所以$\alpha - \theta = \frac{\pi}{2} + 2k \pi$，$k \in \mathbb{Z}$，即$\alpha = \theta + \frac{\pi}{2} + 2k \pi$，$k \in \mathbb{Z}$，所以$\sin \alpha = \sin \left( \theta + \frac{\pi}{2} + 2k \pi \right) = \cos \theta = \frac{3 \sqrt{10}}{10}$，则$\cos 2 \beta = 2 \cos^{2} \beta - 1 = 2 \sin^{2} \alpha - 1 = \frac{4}{5}$。
方法二：因为$\alpha + \beta = \frac{\pi}{2}$，所以$\cos \beta = \sin \alpha$，$\sin \beta = \cos \alpha$，即$3 \sin \alpha - \cos \alpha = \sqrt{10}$。将$\cos \alpha = 3 \sin \alpha - \sqrt{10}$代入$\sin^{2} \alpha + \cos^{2} \alpha = 1$，得$10 \sin^{2} \alpha - 6 \sqrt{10} \sin \alpha + 9 = 0$，解得$\sin \alpha = \frac{3 \sqrt{10}}{10}$，则$\cos 2 \beta = 2 \cos^{2} \beta - 1 = 2 \sin^{2} \alpha - 1 = \frac{4}{5}$。

例 2 (1) (2025·杭州期末)已知 $\alpha, \beta \in \left(0, \dfrac{\pi}{2}\right), \sin(\alpha - \beta) = \dfrac{1}{2}, \tan \alpha = 3\tan \beta$，则 $\alpha + \beta =$(C)
A.$\dfrac{\pi}{4}$
B.$\dfrac{\pi}{3}$
C.$\dfrac{\pi}{2}$
D.$\dfrac{2\pi}{3}$

答案： 例2
(1)C 【解析】由已知得$\sin \alpha \cos \beta - \cos \alpha \sin \beta = \frac{1}{2}$ ①。由$\tan \alpha = 3 \tan \beta$，得$\frac{\sin \alpha}{\cos \alpha} = 3 \frac{\sin \beta}{\cos \beta}$，即$\sin \alpha \cos \beta = 3 \cos \alpha \sin \beta$ ②。联立①②，解得$\begin{cases} \sin \alpha \cos \beta = \frac{3}{4} \\ \cos \alpha \sin \beta = \frac{1}{4} \end{cases}$，所以$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = 1$，因为$\alpha, \beta \in \left( 0, \frac{\pi}{2} \right)$，所以$\alpha + \beta \in (0, \pi)$，故$\alpha + \beta = \frac{\pi}{2}$。

(2) (2025·宜春二模)若 $\cos \alpha = 2\sin \dfrac{\pi}{9} · \sin\left(\alpha - \dfrac{2\pi}{9}\right), 0 ＜ \alpha ＜ \pi$，则 $\alpha =$$\frac{7 \pi}{18}$。

答案：
(2)$\frac{7 \pi}{18}$ 【解析】因为$2 \sin \frac{\pi}{9} \sin \left( \alpha - \frac{2 \pi}{9} \right) = \cos \left[ \frac{\pi}{9} - \left( \alpha - \frac{2 \pi}{9} \right) \right] - \cos \left[ \frac{\pi}{9} + \left( \alpha - \frac{2 \pi}{9} \right) \right] = \cos \left( \frac{\pi}{3} - \alpha \right) - \cos \left( \alpha - \frac{\pi}{9} \right)$，所以$\cos \alpha = \cos \left( \frac{\pi}{3} - \alpha \right) - \cos \left( \alpha - \frac{\pi}{9} \right)$，则$\cos \alpha = \frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha - \cos \left( \alpha - \frac{\pi}{9} \right)$，整理得$\frac{1}{2} \cos \alpha - \frac{\sqrt{3}}{2} \sin \alpha = \cos \left( \alpha + \frac{\pi}{3} \right) = - \cos \left( \alpha - \frac{\pi}{9} \right)$。当$0 ＜ \alpha ＜ \frac{2 \pi}{9}$时，$\cos \alpha ＞ 0$，$2 \sin \frac{\pi}{9} · \sin \left( \alpha - \frac{2 \pi}{9} \right) ＜ 0$，不合题意。当$\frac{2 \pi}{9} ＜ \alpha ＜ \pi$时，$2 \sin \frac{\pi}{9} · \sin \left( \alpha - \frac{2 \pi}{9} \right) ＞ 0$，由$\cos \alpha ＞ 0$，得$\frac{2 \pi}{9} ＜ \alpha ＜ \frac{\pi}{2}$，所以$\frac{\pi}{2} ＜ \frac{5 \pi}{9} ＜ \alpha + \frac{\pi}{3} ＜ \frac{5 \pi}{6} ＜ \pi$，$\frac{\pi}{9} ＜ \alpha - \frac{\pi}{9} ＜ \frac{7 \pi}{18} ＜ \frac{\pi}{2}$。由$\cos \left( \alpha - \frac{\pi}{9} \right) = - \cos \left( \alpha + \frac{\pi}{3} \right)$，得到$\alpha - \frac{\pi}{9} + \alpha + \frac{\pi}{3} = \pi$，解得$\alpha = \frac{7 \pi}{18}$。