《2025年高考总复习首选用卷数学人教版》

1. (2024·黑龙江哈尔滨三中第二次验收考试)已知$\cos2\alpha = \frac{4}{5},\tan\beta = -\frac{1}{7}$，其中$0＜\alpha＜\frac{\pi}{4},\frac{5\pi}{6}＜\beta＜\pi$.
(1)求$\sin(\alpha + \frac{\pi}{4})$的值；
(2)求$\beta - 2\alpha$的值.

答案： 解
(1)由$\cos2\alpha = 2\cos^{2}\alpha - 1 = 1 - 2\sin^{2}\alpha=\frac{4}{5}$，可得$\cos^{2}\alpha=\frac{9}{10}$，$\sin^{2}\alpha=\frac{1}{10}$，又因为$0\lt\alpha\lt\frac{\pi}{4}$，则$\cos\alpha\gt0$，$\sin\alpha\gt0$，可得$\cos\alpha=\frac{3\sqrt{10}}{10}$，$\sin\alpha=\frac{\sqrt{10}}{10}$，所以$\sin(\alpha+\frac{\pi}{4})=\sin\alpha\cos\frac{\pi}{4}+\cos\alpha\sin\frac{\pi}{4}=\frac{\sqrt{10}}{10}\times\frac{\sqrt{2}}{2}+\frac{3\sqrt{10}}{10}\times\frac{\sqrt{2}}{2}=\frac{2\sqrt{5}}{5}$。
(2)因为$0\lt\alpha\lt\frac{\pi}{4}$，则$0\lt2\alpha\lt\frac{\pi}{2}$，且$\cos2\alpha=\frac{4}{5}$，可得$\sin2\alpha=\sqrt{1 - (\cos2\alpha)^{2}}=\frac{3}{5}$，所以$\tan2\alpha=\frac{\sin2\alpha}{\cos2\alpha}=\frac{3}{4}$，可得$\tan(\beta - 2\alpha)=\frac{\tan\beta-\tan2\alpha}{1+\tan\beta\tan2\alpha}=\frac{(-\frac{1}{7})-\frac{3}{4}}{1+(-\frac{1}{7})\times\frac{3}{4}}=-1$，又因为$\frac{5\pi}{6}\lt\beta\lt\pi$，可得$\frac{\pi}{3}\lt\beta - 2\alpha\lt\pi$，所以$\beta - 2\alpha=\frac{3\pi}{4}$。

2. (2024·河南南阳六校高三月考)已知$\frac{\pi}{4}＜\alpha＜\frac{3\pi}{4},\sin(\frac{\pi}{4}-\alpha) = -\frac{1}{2}$.
(1)求$\cos\alpha$的值；
(2)若$0＜\beta＜\frac{\pi}{4},\cos(\frac{\pi}{4}+\beta) = \frac{3}{5}$，求$\cos(2\alpha + \beta)$的值.

答案： 解
(1)因为$\frac{\pi}{4}\lt\alpha\lt\frac{3\pi}{4}$，所以$-\frac{\pi}{2}\lt\frac{\pi}{4}-\alpha\lt0$，又$\sin(\frac{\pi}{4}-\alpha)=-\frac{1}{2}$，所以$\cos(\frac{\pi}{4}-\alpha)=\frac{\sqrt{3}}{2}$，所以$\cos\alpha=\cos[\frac{\pi}{4}-(\frac{\pi}{4}-\alpha)]=\cos\frac{\pi}{4}\cos(\frac{\pi}{4}-\alpha)+\sin\frac{\pi}{4}\sin(\frac{\pi}{4}-\alpha)=\frac{\sqrt{6}-\sqrt{2}}{4}$。
(2)由
(1)，得$\sin\alpha=\sin[\frac{\pi}{4}-(\frac{\pi}{4}-\alpha)]=\sin\frac{\pi}{4}\cos(\frac{\pi}{4}-\alpha)-\cos\frac{\pi}{4}\sin(\frac{\pi}{4}-\alpha)=\frac{\sqrt{6}+\sqrt{2}}{4}$，所以$\cos2\alpha = 2\cos^{2}\alpha - 1 = 2\times(\frac{\sqrt{6}-\sqrt{2}}{4})^{2}-1=-\frac{\sqrt{3}}{2}$，$\sin2\alpha = 2\sin\alpha\cos\alpha = 2\times\frac{\sqrt{6}+\sqrt{2}}{4}\times\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{1}{2}$，又$0\lt\beta\lt\frac{\pi}{4}$，所以$\frac{\pi}{4}+\beta\in(\frac{\pi}{4},\frac{\pi}{2})$，又$\cos(\frac{\pi}{4}+\beta)=\frac{3}{5}$，所以$\sin(\frac{\pi}{4}+\beta)=\frac{4}{5}$，所以$\cos\beta=\cos[(\frac{\pi}{4}+\beta)-\frac{\pi}{4}]=\cos(\frac{\pi}{4}+\beta)\cos\frac{\pi}{4}+\sin(\frac{\pi}{4}+\beta)\sin\frac{\pi}{4}=\frac{7\sqrt{2}}{10}$，$\sin\beta=\sqrt{1 - (\frac{7\sqrt{2}}{10})^{2}}=\frac{\sqrt{2}}{10}$，所以$\cos(2\alpha+\beta)=\cos2\alpha\cos\beta-\sin2\alpha\sin\beta=(-\frac{\sqrt{3}}{2})\times\frac{7\sqrt{2}}{10}-\frac{1}{2}\times\frac{\sqrt{2}}{10}=-\frac{7\sqrt{6}+\sqrt{2}}{20}$。