《2026年学易优高考二轮总复习数学》

1. (2025·全国二卷)已知 $0 ＜ \alpha ＜ \pi$，$\cos \frac{\alpha}{2} = \frac{\sqrt{5}}{5}$，则 $\sin(\alpha - \frac{\pi}{4}) =$(D)

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

答案： 1.D 二倍角公式＋两角差的正弦公式＋同角三角函数的基本关系 $\cos \alpha = 2\cos^2 \frac{\alpha}{2} - 1 = 2 × \frac{1}{5} - 1 = -\frac{3}{5}$，因为 $0 ＜ \alpha ＜ \pi$，所以 $\sin \alpha = \frac{4}{5}$，所以 $\sin \left( \alpha - \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} (\sin \alpha - \cos \alpha) = \frac{\sqrt{2}}{2} × \frac{7}{5} = \frac{7\sqrt{2}}{10}$。

2. (2023·新高考Ⅱ卷)已知 $\alpha$ 为锐角，$\cos \alpha = \frac{1 + \sqrt{5}}{4}$，则 $\sin \frac{\alpha}{2} =$(D)

A.$\frac{3 - \sqrt{5}}{8}$
B.$\frac{-1 + \sqrt{5}}{8}$
C.$\frac{3 - \sqrt{5}}{4}$
D.$\frac{-1 + \sqrt{5}}{4}$

答案： 2.D 解法一 由题意，$\cos \alpha = \frac{1 + \sqrt{5}}{4} = 1 - 2\sin^2 \frac{\alpha}{2}$，得 $\sin^2 \frac{\alpha}{2} = \frac{3 - \sqrt{5}}{8} = \frac{6 - 2\sqrt{5}}{16} = \frac{(\sqrt{5} - 1)^2}{16}$，又 $\alpha$ 为锐角，所以 $\sin \frac{\alpha}{2} ＞ 0$，所以 $\sin \frac{\alpha}{2} = \frac{-1 + \sqrt{5}}{4}$，故选 D.
解法二 由题意，$\cos \alpha = \frac{1 + \sqrt{5}}{4} = 1 - 2\sin^2 \frac{\alpha}{2}$，得 $\sin^2 \frac{\alpha}{2} = \frac{3 - \sqrt{5}}{8}$，将选项逐个代入验证可知 D 选项满足，故选 D.

3. (2024·全国甲卷)已知 $\frac{\cos \alpha}{\cos \alpha - \sin \alpha} = \sqrt{3}$，则 $\tan(\alpha + \frac{\pi}{4}) =$(B)

A.$2\sqrt{3} + 1$
B.$2\sqrt{3} - 1$
C.$\frac{\sqrt{3}}{2}$
D.$1 - \sqrt{3}$

答案： 3.B 根据题意有 $\frac{\cos \alpha - \sin \alpha}{\cos \alpha} = \frac{\sqrt{3}}{3}$，即 $1 - \tan \alpha = \frac{\sqrt{3}}{3}$，所以 $\tan \alpha = 1 - \frac{\sqrt{3}}{3} = \frac{2 - \sqrt{3}}{3}$，所以 $\tan \left( \alpha + \frac{\pi}{4} \right) = \frac{\tan \alpha + 1}{1 - \tan \alpha} = \frac{2 - \frac{\sqrt{3}}{3}}{ \frac{\sqrt{3}}{3}} = 2\sqrt{3} - 1$，故选 B.

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. $- \frac{2\sqrt{2}}{3}$ 由题知 $\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha · \tan \beta} = \frac{4}{1 - \sqrt{2} - 1} = -2\sqrt{2}$，即 $\sin (\alpha + \beta) = -2\sqrt{2} \cos (\alpha + \beta)$，又 $\sin^2 (\alpha + \beta) + \cos^2 (\alpha + \beta) = 1$，可得 $\sin (\alpha + \beta) = \pm \frac{2\sqrt{2}}{3}$. 由 $2k\pi ＜ \alpha ＜ 2k\pi + \frac{\pi}{2}, k \in \mathbf{Z}, 2m\pi + \pi ＜ \beta ＜ 2m\pi + \frac{3\pi}{2}, m \in \mathbf{Z}$，得 $2(k + m)\pi + \pi ＜ \alpha + \beta ＜ 2(k + m)\pi + 2\pi, k + m \in \mathbf{Z}$. 又 $\tan (\alpha + \beta) ＜ 0$，所以 $\alpha + \beta$ 是第四象限角，故 $\sin (\alpha + \beta) = - \frac{2\sqrt{2}}{3}$.

(1) (2025·武汉联考) $\frac{\sin 20^{\circ} \sin 70^{\circ}}{1 - 2\sin^{2}25^{\circ}} =$(C)

A.$\frac{\sqrt{3}}{4}$
B.$\frac{\sqrt{3}}{2}$
C.$\frac{1}{2}$
D.$2$

答案：
(1)C $\frac{\sin 20° \sin 70°}{1 - 2\sin^2 25°} = \frac{\sin 20° \cos 20°}{\cos 50°} = \frac{\frac{1}{2} \sin 40°}{\sin 40°} = \frac{1}{2}$.

(2) (2025·西安模拟) $\frac{\cos 55^{\circ} + \sin 25^{\circ} \sin 30^{\circ}}{\cos 25^{\circ}}$等于(C)

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

答案：
(2)C $\frac{\cos 55° + \sin 25° \sin 30°}{\cos 25°} = \frac{\cos (25° + 30°) + \sin 25° \sin 30°}{\cos 25°} = \frac{\cos 25° \cos 30° - \sin 25° \sin 30° + \sin 25° \sin 30°}{\cos 25°} = \frac{\cos 25° \cos 30°}{\cos 25°} = \cos 30° = \frac{\sqrt{3}}{2}$.