《2026年南方凤凰台5A新考案高中数学二轮基础版》

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$\tan \alpha = - \frac{3}{4}$，则$\tan \left( \alpha + \frac{\pi}{4} \right) = \frac{\tan \alpha + 1}{1 - \tan \alpha} = \frac{1}{7}$。

2. (人 A 必一 P217 练习 5 改)设 $0 ＜ \alpha ＜ \dfrac{\pi}{2}, -\dfrac{\pi}{2} ＜ \beta ＜ 0$，且 $\sin \alpha = \dfrac{\sqrt{10}}{10}, \sin \beta = -\dfrac{\sqrt{5}}{5}$，则 $\alpha - \beta =$(B)

A.$\dfrac{\pi}{6}$
B.$\dfrac{\pi}{4}$
C.$\dfrac{\pi}{3}$
D.$\dfrac{3\pi}{4}$

答案： 2.B 【解析】因为$0 ＜ \alpha ＜ \frac{\pi}{2}$，$ - \frac{\pi}{2} ＜ \beta ＜ 0$，所以$\cos \alpha = \sqrt{1 - \sin^{2} \alpha} = \sqrt{1 - \left(\frac{1}{5}\right)} = \frac{2\sqrt{5}}{5}$，$\cos \beta = \sqrt{1 - \sin^{2} \beta} = \sqrt{1 - \left(-\frac{1}{5}\right)} = \frac{2\sqrt{5}}{5}$，$0 ＜ \alpha - \beta ＜ \pi$，$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta = \frac{2\sqrt{5}}{5} × \frac{2\sqrt{5}}{5} + \frac{\sqrt{5}}{5} × \left(-\frac{\sqrt{5}}{5}\right) = \frac{3}{5}$，此部分大模型计算有误，以OCR为准，$0 ＜ \alpha - \beta ＜ \pi$，$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta = \frac{3\sqrt{10}}{10} × \frac{2\sqrt{5}}{5} + \frac{\sqrt{10}}{10} × \left(-\frac{\sqrt{5}}{5}\right) = \frac{\sqrt{2}}{2}$，所以$\alpha - \beta = \frac{\pi}{4}$。

3. (人 A 必一 P223 练习 5 改)下列各式的值为 $\dfrac{1}{2}$ 的是(B)

A.$\sin 15^{\circ} \sin 75^{\circ}$
B.$\cos^2 \dfrac{\pi}{6} - \sin^2 \dfrac{\pi}{6}$
C.$\dfrac{\tan 30^{\circ}}{1 - \tan^2 30^{\circ}}$
D.$\sqrt{\dfrac{1 + \cos 60^{\circ}}{2}}$

答案： 3.B 【解析】对于A，$\sin 15^{\circ} \sin 75^{\circ} = \sin 15^{\circ} \cos 15^{\circ} = \frac{1}{2} × 2 \sin 15^{\circ} \cos 15^{\circ} = \frac{1}{2} \sin 30^{\circ} = \frac{1}{4}$，故A错误；对于B，$\cos^{2} \frac{\pi}{6} - \sin^{2} \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}$，故B正确；对于C，原式$= \frac{1}{2} × \frac{2 \tan 30^{\circ}}{1 - \tan^{2} 30^{\circ}} = \frac{1}{2} \tan 60^{\circ} = \frac{\sqrt{3}}{2}$，故C错误；对于D，原式$= \cos 30^{\circ} = \frac{\sqrt{3}}{2}$，故D错误。

4. (人 A 必一 P223 练习 2 改)若 $\cos \theta = -\dfrac{3}{5}, \theta \in \left(\dfrac{\pi}{2}, \pi\right)$，则 $\sin\left(2\theta + \dfrac{\pi}{4}\right) =$$ - \frac{31 \sqrt{2}}{50}$。

答案： 4.$ - \frac{31 \sqrt{2}}{50}$ 【解析】因为$\cos \theta = - \frac{3}{5}$，$\theta \in \left( \frac{\pi}{2}, \pi \right)$，所以$\sin \theta = \sqrt{1 - \cos^{2} \theta} = \frac{4}{5}$，所以$\sin 2 \theta = 2 \sin \theta \cos \theta = - \frac{24}{25}$，$\cos 2 \theta = 1 - 2 \sin^{2} \theta = - \frac{7}{25}$，所以$\sin \left( 2 \theta + \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \left( \sin 2 \theta + \cos 2 \theta \right) = \frac{\sqrt{2}}{2} × \left( - \frac{24}{25} - \frac{7}{25} \right) = - \frac{31 \sqrt{2}}{50}$。

5. (2025·全国Ⅱ卷)已知 $\alpha \in (0, \pi), \cos \dfrac{\alpha}{2} = \dfrac{\sqrt{5}}{5}$，则 $\sin\left(\alpha - \dfrac{\pi}{4}\right) =$(D)

A.$\dfrac{\sqrt{2}}{10}$
B.$\dfrac{\sqrt{2}}{5}$
C.$\dfrac{3\sqrt{2}}{10}$
D.$\dfrac{7\sqrt{2}}{10}$

答案： 5.D 【解析】由题知$\cos \alpha = 2 \cos^{2} \frac{\alpha}{2} - 1 = 2 × \left( \frac{\sqrt{5}}{5} \right)^{2} - 1 = - \frac{3}{5}$，因为$0 ＜ \alpha ＜ \pi$，所以$\frac{\pi}{2} ＜ \alpha ＜ \pi$，则$\sin \alpha = \sqrt{1 - \cos^{2} \alpha} = \sqrt{1 - \left( - \frac{3}{5} \right)^{2}} = \frac{4}{5}$，则$\sin \left( \alpha - \frac{\pi}{4} \right) = \sin \alpha \cos \frac{\pi}{4} - \cos \alpha \sin \frac{\pi}{4} = \frac{4}{5} × \frac{\sqrt{2}}{2} - \left( - \frac{3}{5} \right) × \frac{\sqrt{2}}{2} = \frac{7 \sqrt{2}}{10}$。