《2025年高考总复习首选用卷数学人教版》

18. (2023·江苏姜堰中学高三月考)若$\tan\alpha = 2\tan10^{\circ}$，则$\frac{\cos(\alpha - 80^{\circ})}{\sin(\alpha - 10^{\circ})}=$（ ）
A. 1
B. 2
C. 3
D. 4

答案： C [$\because\tan\alpha = 2\tan10^{\circ}$，$\therefore\frac{\cos(\alpha - 80^{\circ})}{\sin(\alpha - 10^{\circ})}=\frac{\cos(\alpha + 10^{\circ}-90^{\circ})}{\sin(\alpha - 10^{\circ})}=\frac{\sin(\alpha + 10^{\circ})}{\sin(\alpha - 10^{\circ})}=\frac{\sin\alpha\cos10^{\circ}+\cos\alpha\sin10^{\circ}}{\sin\alpha\cos10^{\circ}-\cos\alpha\sin10^{\circ}}=\frac{\tan\alpha+\tan10^{\circ}}{\tan\alpha - \tan10^{\circ}}=\frac{3\tan10^{\circ}}{\tan10^{\circ}} = 3$。故选C。]

19. (2023·山西运城景胜中学高三考试)已知$\tan(\alpha + \frac{\pi}{6}) = \frac{1}{2},\tan(\frac{\pi}{12}+\beta) = \frac{1}{3}$，则$\tan(\alpha - 2\beta)=$（ ）
A. $-\frac{9}{13}$
B. $-\frac{2}{11}$
C. $\frac{10}{11}$
D. $\frac{2}{5}$

答案： B [由$\tan(\frac{\pi}{12}+\beta)=\frac{1}{3}$，得$\tan(\frac{\pi}{6}+2\beta)=\frac{2\tan(\frac{\pi}{12}+\beta)}{1 - \tan^{2}(\frac{\pi}{12}+\beta)}=\frac{2\times\frac{1}{3}}{1 - (\frac{1}{3})^{2}}=\frac{3}{4}$，而$\tan(\alpha+\frac{\pi}{6})=\frac{1}{2}$，故$\tan(\alpha - 2\beta)=\tan[(\alpha+\frac{\pi}{6})-(2\beta+\frac{\pi}{6})]=\frac{\tan(\alpha+\frac{\pi}{6})-\tan(2\beta+\frac{\pi}{6})}{1+\tan(\alpha+\frac{\pi}{6})\tan(2\beta+\frac{\pi}{6})}=\frac{\frac{1}{2}-\frac{3}{4}}{1+\frac{1}{2}\times\frac{3}{4}}=-\frac{2}{11}$。故选B。]

20. (2024·浙江名校联盟高三研究卷)已知$\alpha \in (0,\pi)$，若$\sqrt{3}(\sin\alpha + \sin2\alpha) + \cos\alpha - \cos2\alpha = 0$，则$\sin(\alpha - \frac{\pi}{12}) =$（ ）
A. $\frac{\sqrt{2}}{2}$
B. $\frac{\sqrt{3}}{2}$
C. $\frac{\sqrt{6}+\sqrt{2}}{4}$
D. $\frac{\sqrt{6}-\sqrt{2}}{4}$

答案： C [$\because\sqrt{3}(\sin\alpha+\sin2\alpha)+\cos\alpha-\cos2\alpha = 0$，$\therefore\frac{\sqrt{3}}{2}\sin\alpha+\frac{1}{2}\cos\alpha=\frac{1}{2}\cos2\alpha-\frac{\sqrt{3}}{2}\sin2\alpha$，$\therefore\sin(\alpha+\frac{\pi}{6})=\sin(\frac{\pi}{6}-2\alpha)$，$\therefore\alpha+\frac{\pi}{6}=\frac{\pi}{6}-2\alpha+2k\pi$，$k\in\mathbf{Z}$或$\alpha+\frac{\pi}{6}+\frac{\pi}{6}-2\alpha=2k\pi+\pi$，$k\in\mathbf{Z}$，又$\alpha\in(0,\pi)$，$\therefore\alpha=\frac{2\pi}{3}$，$\sin(\alpha-\frac{\pi}{12})=\sin(\frac{2\pi}{3}-\frac{\pi}{12})=\sin\frac{7\pi}{12}=\sin(\frac{\pi}{3}+\frac{\pi}{4})=\sin\frac{\pi}{3}\cos\frac{\pi}{4}+\cos\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}$。故选C。]

21. (2024·湖南长沙一中高三月考)已知角$\alpha,\beta\in(0,\pi)$，且$\sin(\alpha + \beta) + \cos(\alpha - \beta) = 0,\sin\alpha\sin\beta - 3\cos\alpha\cos\beta = 0$，则$\tan(\alpha + \beta) =$（ ）
A. -2
B. $-\frac{1}{2}$
C. $\frac{1}{2}$
D. 2

答案： D [由$\sin(\alpha + \beta)+\cos(\alpha - \beta)=0$，可得$\sin\alpha\cos\beta+\cos\alpha\sin\beta+\cos\alpha\cos\beta+\sin\alpha\sin\beta = 0$，$\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}=-1$，故$\frac{\tan\alpha+\tan\beta}{1+\tan\alpha\tan\beta}=-1$。又$\sin\alpha\sin\beta - 3\cos\alpha\cos\beta = 0$，故$\sin\alpha\sin\beta = 3\cos\alpha\cos\beta$，即$\tan\alpha\tan\beta = 3$，代入$\frac{\tan\alpha+\tan\beta}{1+\tan\alpha\tan\beta}=-1$可得$\tan\alpha+\tan\beta=-4$。故$\tan(\alpha + \beta)=\frac{\tan\alpha+\tan\beta}{1 - \tan\alpha\tan\beta}=2$。故选D。]

22. (多选)(2023·江苏盐城市伍佑中学高三月考)以下式子均有意义，则下列等式恒成立的是（ ）
A. $\cos\alpha\sin\beta = \frac{\sin(\alpha + \beta) + \sin(\alpha - \beta)}{2}$
B. $\frac{\sin\alpha}{1 - \cos\alpha} = \frac{1 + \cos\alpha}{\sin\alpha}$
C. $\frac{1 - 2\cos x\sin x}{\cos^{2}x - \sin^{2}x} = \frac{1 - \tan x}{1 + \tan x}$
D. $\frac{\sin(2\alpha + \beta)}{\sin\alpha} = 2\cos(\alpha + \beta) + \frac{\sin\beta}{\sin\alpha}$

答案： BCD [对于A，因为$\sin(\alpha + \beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$，$\sin(\alpha - \beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$，所以$\sin\alpha\cos\beta=\frac{\sin(\alpha + \beta)+\sin(\alpha - \beta)}{2}$，故A错误；对于B，因为$\sin^{2}\alpha = 1 - \cos^{2}\alpha=(1+\cos\alpha)(1 - \cos\alpha)$，所以$\frac{\sin\alpha}{1 - \cos\alpha}=\frac{1+\cos\alpha}{\sin\alpha}$，故B正确；对于C，$\frac{1 - \tan x}{1 + \tan x}=\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}=\frac{\cos x - \sin x}{\cos x + \sin x}$，$\frac{1 - 2\sin x\cos x}{\cos^{2}x - \sin^{2}x}=\frac{(\cos x - \sin x)^{2}}{(\cos x + \sin x)(\cos x - \sin x)}=\frac{\cos x - \sin x}{\cos x + \sin x}$，所以$\frac{1 - 2\cos x\sin x}{\cos^{2}x - \sin^{2}x}=\frac{1 - \tan x}{1 + \tan x}$，故C正确；对于D，$\frac{\sin(2\alpha+\beta)}{\sin\alpha}-2\cos(\alpha + \beta)=\frac{\sin[\alpha+(\alpha + \beta)]}{\sin\alpha}-2\cos(\alpha + \beta)=\frac{\sin\alpha\cos(\alpha + \beta)+\cos\alpha\sin(\alpha + \beta)}{\sin\alpha}-2\cos(\alpha + \beta)=\frac{\cos\alpha\sin(\alpha + \beta)-\sin\alpha\cos(\alpha + \beta)}{\sin\alpha}=\frac{\sin[(\alpha + \beta)-\alpha]}{\sin\alpha}=\frac{\sin\beta}{\sin\alpha}$，所以$\frac{\sin(2\alpha+\beta)}{\sin\alpha}=2\cos(\alpha + \beta)+\frac{\sin\beta}{\sin\alpha}$，故D正确。故选BCD。]

23. (2023·重庆七校高三月考)已知$\alpha,\beta$是锐角，且$\sin\alpha = \frac{4\sqrt{3}}{7},\cos(\alpha + \beta) = -\frac{11}{14}$，则$\sin\beta =$________.

答案： 答案 $\frac{\sqrt{3}}{2}$
解析 $\because\alpha,\beta\in(0,\frac{\pi}{2})$，$\therefore0\lt\alpha+\beta\lt\pi$，$\because\sin\alpha=\frac{4\sqrt{3}}{7}$，$\cos(\alpha + \beta)=-\frac{11}{14}$，$\therefore\cos\alpha=\frac{1}{7}$，$\sin(\alpha + \beta)=\frac{5\sqrt{3}}{14}$，$\therefore\sin\beta=\sin[(\alpha + \beta)-\alpha]=\sin(\alpha + \beta)\cos\alpha-\cos(\alpha + \beta)\sin\alpha=\frac{5\sqrt{3}}{14}\times\frac{1}{7}-(-\frac{11}{14})\times\frac{4\sqrt{3}}{7}=\frac{\sqrt{3}}{2}$。

24. (2023·四川绵阳盐亭中学三诊模拟)已知$\cos(\alpha + \frac{\pi}{6}) = \frac{3}{5}$，则$\cos(\frac{2\pi}{3}-2\alpha) =$________.

答案： 答案 $\frac{7}{25}$
解析 $\sin(\frac{\pi}{3}-\alpha)=\cos[\frac{\pi}{2}-(\frac{\pi}{3}-\alpha)]=\cos(\alpha+\frac{\pi}{6})=\frac{3}{5}$，$\cos(\frac{2\pi}{3}-2\alpha)=\cos[2(\frac{\pi}{3}-\alpha)]=1 - 2\sin^{2}(\frac{\pi}{3}-\alpha)=1-2\times(\frac{3}{5})^{2}=\frac{7}{25}$。