《2026年南方凤凰台5A新考案数学二轮提高版》

1. (人 A 必一 P218 例 3 改)若 $\alpha \in \left( \dfrac{\pi}{2}, \pi \right)$，$\sin \alpha = \dfrac{3}{5}$，则 $\tan \left( \alpha + \dfrac{\pi}{4} \right) =$ (A)

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

答案： 1.A【解析】由$\alpha\in\left(\frac{\pi}{2},\pi\right),\sin\alpha=\frac{3}{5}$，得$\cos\alpha=-\frac{4}{5}$，所以
$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{3}{-4}=-\frac{3}{4}$，则$\tan\left(\alpha+\frac{\pi}{4}\right)=\frac{\tan\alpha + 1}{1 - \tan\alpha}=\frac{1}{7}$。

2. (2025·武汉 4 月调研)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： 2.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$\cos2\alpha=\frac{\cos^{2}\alpha-\sin^{2}\alpha}{\cos^{2}\alpha+\sin^{2}\alpha}=\frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}=\frac{7}{25}$。
方法二：$\cos2\alpha=\sin\left(2\alpha+\frac{\pi}{2}\right)=2\sin\left(\alpha+\frac{\pi}{4}\right)\cos\left(\alpha+\frac{\pi}{4}\right)=\frac{2\sin\left(\alpha+\frac{\pi}{4}\right)\cos\left(\alpha+\frac{\pi}{4}\right)}{\sin^{2}\left(\alpha+\frac{\pi}{4}\right)+\cos^{2}\left(\alpha+\frac{\pi}{4}\right)}=\frac{2\tan\left(\alpha+\frac{\pi}{4}\right)}{\tan^{2}\left(\alpha+\frac{\pi}{4}\right)+1}=\frac{14}{49 + 1}=\frac{7}{25}$。

3. 已知 $\sin \left( \alpha - \dfrac{\pi}{6} \right) = \dfrac{1}{3}$，则 $\sin \left( 2 \alpha - \dfrac{\pi}{6} \right) + \cos 2 \alpha =$ $\frac{7}{9}$ .

答案： 3.$\frac{7}{9}$【解析】$\sin\left(2\alpha-\frac{\pi}{6}\right)+\cos2\alpha=\frac{\sqrt{3}}{2}\sin2\alpha+\frac{1}{2}\cos2\alpha=\sin\left(2\alpha+\frac{\pi}{6}\right)=\cos\left[\frac{\pi}{2}-\left(2\alpha+\frac{\pi}{6}\right)\right]=\cos\left(\frac{\pi}{3}-2\alpha\right)=\cos\left(2\alpha-\frac{\pi}{3}\right)=1 - 2\sin^{2}\left(\alpha-\frac{\pi}{6}\right)=1 - 2×\left(\frac{1}{3}\right)^{2}=\frac{7}{9}$。

4. (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： 4.$-\frac{2\sqrt{2}}{3}$【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$\alpha\in\left(2k\pi,2k\pi+\frac{\pi}{2}\right),\beta\in\left(2m\pi+\pi,2m\pi+\frac{3\pi}{2}\right),k,m\in\mathbf{Z}$，所以$\alpha+\beta\in\left((2m + 2k)\pi+\pi,(2m+2k)\pi+2\pi\right),k,m\in\mathbf{Z}$。又$\tan(\alpha+\beta)=-2\sqrt{2}＜0$，所以$\alpha+\beta\in\left((2m + 2k)\pi+\frac{3\pi}{2},(2m+2k)\pi+2\pi\right),k,m\in\mathbf{Z}$，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，则$\cos\alpha＞0$，
$\cos\beta＜0$，$\cos\alpha=\frac{\cos\alpha}{\sqrt{\sin^{2}\alpha+\cos^{2}\alpha}}=\frac{1}{\sqrt{1+\tan^{2}\alpha}}$，$\cos\beta=\frac{\cos\beta}{\sqrt{\sin^{2}\beta+\cos^{2}\beta}}=\frac{-1}{\sqrt{1+\tan^{2}\beta}}$，则$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\cos\alpha\cos\beta(\tan\alpha+\tan\beta)=4\cos\alpha\cos\beta=\frac{-4}{\sqrt{1+\tan^{2}\alpha}\sqrt{1+\tan^{2}\beta}}=\frac{-4}{\sqrt{4^{2}+2}}=\frac{2\sqrt{2}}{3}$。

例 1 已知 $2 \cos \left( 2 x + \dfrac{\pi}{12} \right) \cos \left( x - \dfrac{\pi}{12} \right) - \cos 3 x = \dfrac{1}{4}$，则 $\sin \left( \dfrac{\pi}{6} - 2 x \right) =$ (D)

A.$\dfrac{1}{2}$
B.$- \dfrac{1}{2}$
C.$\dfrac{7}{8}$
D.$- \dfrac{7}{8}$

答案： 例1 D【解析】由已知得$2\cos\left(2x+\frac{\pi}{12}\right)\cos\left(x-\frac{\pi}{12}\right)-\cos\left[\left(2x+\frac{\pi}{12}\right)+\left(x-\frac{\pi}{12}\right)\right]=\frac{1}{4}$，化简得$\cos\left(2x+\frac{\pi}{12}\right)\cos\left(x-\frac{\pi}{12}\right)+\sin\left(2x+\frac{\pi}{12}\right)\sin\left(x-\frac{\pi}{12}\right)=\cos\left[\left(2x+\frac{\pi}{12}\right)-\left(x-\frac{\pi}{12}\right)\right]=\cos\left(x+\frac{\pi}{6}\right)=\frac{1}{4}$.令$t=x+\frac{\pi}{6}$，则$x=t-\frac{\pi}{6}$，$\cos t=\frac{1}{4}$，所以
$\sin\left(\frac{\pi}{6}-2x\right)=\sin\left[\frac{\pi}{6}-2\left(t-\frac{\pi}{6}\right)\right]=\sin\left(\frac{\pi}{2}-2t\right)=\cos2t=2\cos^{2}t - 1=2×\left(\frac{1}{4}\right)^{2}-1=-\frac{7}{8}$。