《2025年高考总复习首选用卷数学人教版》

1. 已知$\alpha \in (\frac{\pi}{2},\pi),\cos\alpha = -\frac{3}{5}$，则$\tan(\alpha +\frac{\pi}{4})=$（ ）
A. $\frac{1}{7}$
B. 7
C. $-\frac{1}{7}$
D. -7

答案： C [因为$\alpha\in(\frac{\pi}{2},\pi)$，$\cos\alpha = -\frac{3}{5}$，所以$\sin\alpha = \frac{4}{5}$，$\tan\alpha = -\frac{4}{3}$，所以$\tan(\alpha+\frac{\pi}{4})=\frac{-\frac{4}{3}+1}{1-(-\frac{4}{3})\times1}=-\frac{1}{7}$。]

2. 已知$\cos\theta = \frac{1}{3}$，则$\sin(2\theta +\frac{\pi}{2})=$（ ）
A. $-\frac{7}{9}$
B. $\frac{7}{9}$
C. $\frac{2}{3}$
D. $-\frac{2}{3}$

答案： A [由题意可知，$\sin(2\theta+\frac{\pi}{2})=\cos2\theta = 2\cos^{2}\theta - 1 = 2\times(\frac{1}{3})^{2}-1=-\frac{7}{9}$。故选A。]

3. $\frac{2\cos10^{\circ}-\sin20^{\circ}}{\sin70^{\circ}}$的值是（ ）
A. $\frac{1}{2}$
B. $\frac{\sqrt{3}}{2}$
C. $\sqrt{3}$
D. $\sqrt{2}$

答案： C [原式$=\frac{2\cos(30^{\circ}-20^{\circ})-\sin20^{\circ}}{\sin70^{\circ}}=\frac{2(\cos30^{\circ}\cos20^{\circ}+\sin30^{\circ}\sin20^{\circ})-\sin20^{\circ}}{\sin70^{\circ}}=\frac{\sqrt{3}\cos20^{\circ}}{\cos20^{\circ}}=\sqrt{3}$。]

4. 已知$\alpha \in (-\pi,0)$，且$3\cos2\alpha + 4\cos\alpha + 1 = 0$，则$\tan\alpha =$（ ）
A. $\frac{\sqrt{2}}{4}$
B. $2\sqrt{2}$
C. $-2\sqrt{2}$
D. $-\frac{\sqrt{2}}{4}$

答案： C [$3\cos2\alpha + 4\cos\alpha + 1 = 0$，$3(2\cos^{2}\alpha - 1)+4\cos\alpha + 1 = 0$，整理得$3\cos^{2}\alpha + 2\cos\alpha - 1=(3\cos\alpha - 1)(\cos\alpha + 1)=0$，解得$\cos\alpha=\frac{1}{3}$或$\cos\alpha = -1$。因为$\alpha\in(-\pi,0)$，所以$\cos\alpha=\frac{1}{3}$，$\sin\alpha = -\frac{2\sqrt{2}}{3}$，$\tan\alpha=\frac{-\frac{2\sqrt{2}}{3}}{\frac{1}{3}}=-2\sqrt{2}$。故选C。]

5. 已知$\alpha$为锐角，且$(\sqrt{3}-\tan10^{\circ})\cos\alpha = 1$，则$\alpha$的值为（ ）
A. $40^{\circ}$
B. $50^{\circ}$
C. $70^{\circ}$
D. $80^{\circ}$

答案： B [$\because(\sqrt{3}-\tan10^{\circ})\cos\alpha = 1$，$\therefore(\sqrt{3}-\frac{\sin10^{\circ}}{\cos10^{\circ}})\cos\alpha = 1$，$\therefore(\sqrt{3}\cos10^{\circ}-\sin10^{\circ})\cos\alpha = \cos10^{\circ}$，$\therefore2\cos40^{\circ}\cos\alpha = \sin80^{\circ}$，$\therefore2\cos40^{\circ}\cos\alpha = 2\sin40^{\circ}\cos40^{\circ}$，$\therefore\cos\alpha = \sin40^{\circ}=\cos50^{\circ}$，又$\alpha$为锐角，$\therefore\alpha = 50^{\circ}$。故选B。]

6. 已知$\cos(\alpha - \beta) = \frac{3}{5},\sin\beta = -\frac{5}{13}$，且$\alpha \in (0,\frac{\pi}{2}),\beta \in (-\frac{\pi}{2},0)$，则$\cos\alpha =$（ ）
A. $\frac{33}{65}$
B. $\frac{56}{65}$
C. $-\frac{33}{65}$
D. $-\frac{56}{65}$

答案： B [$\because\begin{cases}0\lt\alpha\lt\frac{\pi}{2}\\-\frac{\pi}{2}\lt\beta\lt0\end{cases}$，$\therefore0\lt\alpha - \beta\lt\pi$，$\cos\beta=\frac{12}{13}$。又$\cos(\alpha - \beta)=\frac{3}{5}$，$\therefore\sin(\alpha - \beta)=\frac{4}{5}$，$\therefore\cos\alpha=\cos[(\alpha - \beta)+\beta]=\cos(\alpha - \beta)\cos\beta-\sin(\alpha - \beta)\sin\beta=\frac{56}{65}$。]

7. (多选)已知$0＜\theta＜\frac{\pi}{4}$，若$\sin2\theta = m,\cos2\theta = n$，且$m\neq n$，则下列选项中与$\tan(\frac{\pi}{4}-\theta)$恒相等的是（ ）
A. $\frac{n}{1 + m}$
B. $\frac{m}{1 + n}$
C. $\frac{1 - n}{m}$
D. $\frac{1 - m}{n}$

答案： AD [$\tan(\frac{\pi}{4}-\theta)=\frac{1 - \tan\theta}{1 + \tan\theta}=\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}=\frac{\cos^{2}\theta - \sin^{2}\theta}{(\cos\theta + \sin\theta)^{2}}=\frac{\cos2\theta}{1 + \sin2\theta}=\frac{n}{1 + m}$或$\tan(\frac{\pi}{4}-\theta)=\frac{1 - \tan\theta}{1 + \tan\theta}=\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}=\frac{(\cos\theta - \sin\theta)^{2}}{\cos^{2}\theta - \sin^{2}\theta}=\frac{1 - \sin2\theta}{\cos2\theta}=\frac{1 - m}{n}$。故选AD。]