《2025年金版教程高考科学复习解决方案数学》

例1　　(1)(2024·海南海口模拟)若$\tan\alpha\tan\beta = 2$，则$\frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)}$的值为（ ）
A. -3
B. $-\frac{1}{3}$
C. $\frac{1}{3}$
D. 3

答案： A [由题意，得$\frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)} = \frac{\cos\alpha\cos\beta + \sin\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta} = \frac{1 + \tan\alpha\tan\beta}{1 - \tan\alpha\tan\beta} = \frac{1 + 2}{1 - 2} = - 3$. 故选A.]

(2)(2024·九省联考)已知$\theta\in(\frac{3\pi}{4},\pi)$，$\tan2\theta = - 4\tan(\theta + \frac{\pi}{4})$，则$\frac{1 + \sin2\theta}{2\cos^{2}\theta + \sin2\theta} =$（ ）
A. $\frac{1}{4}$
B. $\frac{3}{4}$
C. 1
D. $\frac{3}{2}$

答案： A [由$\theta\in(\frac{3\pi}{4},\pi)$，$\tan2\theta = - 4\tan(\theta + \frac{\pi}{4})$，得$\frac{2\tan\theta}{1 - \tan^{2}\theta} = \frac{- 4(\tan\theta + 1)}{1 - \tan\theta}$，则$- 4(\tan\theta + 1)^{2} = 2\tan\theta$，则$(2\tan\theta + 1)(\tan\theta + 2) = 0$，解得$\tan\theta = - 2$或$\tan\theta = - \frac{1}{2}$，因为$\theta\in(\frac{3\pi}{4},\pi)$，所以$\tan\theta\in(- 1,0)$，所以$\tan\theta = - \frac{1}{2}$，则$\frac{1 + \sin2\theta}{2\cos^{2}\theta + \sin2\theta} = \frac{\sin^{2}\theta + \cos^{2}\theta + 2\sin\theta\cos\theta}{2\cos^{2}\theta + 2\sin\theta\cos\theta} = \frac{\tan^{2}\theta + 1 + 2\tan\theta}{2 + 2\tan\theta} = \frac{\frac{1}{4} + 1 - 1}{2 + (- 1)} = \frac{1}{4}$. 故选A.]

1.(2024·安徽亳州模拟)已知$\sin\alpha = \frac{3}{5}$，$\alpha\in(\frac{\pi}{2},\pi)$，若$\frac{\sin(\alpha + \beta)}{\cos\beta} = 4$，则$\tan(\alpha + \beta) =$（ ）
A. $-\frac{16}{7}$
B. $-\frac{7}{8}$
C. $\frac{16}{7}$
D. $\frac{2}{3}$

答案： C [因为$\sin\alpha = \frac{3}{5}$，$\alpha\in(\frac{\pi}{2},\pi)$，所以$\cos\alpha = - \sqrt{1 - \sin^{2}\alpha} = - \frac{4}{5}$，$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = - \frac{3}{4}$，因为$\frac{\sin(\alpha + \beta)}{\cos\beta} = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\beta} = \sin\alpha + \cos\alpha\tan\beta = \frac{3}{5} - \frac{4}{5}\tan\beta = 4$，所以$\tan\beta = - \frac{17}{4}$，所以$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{- \frac{3}{4} - \frac{17}{4}}{1 - ( - \frac{3}{4})\times( - \frac{17}{4})} = \frac{16}{7}$. 故选C.]

2.(2023·河北保定模拟)已知锐角$\theta$满足$2\cos2\theta = 1 + \sin2\theta$，则$\tan\theta =$（ ）
A. $\frac{1}{3}$
B. $\frac{1}{2}$
C. 2
D. 3

答案： A [$\because2\cos2\theta = 1 + \sin2\theta$，$\therefore2(\cos^{2}\theta - \sin^{2}\theta) = (\sin\theta + \cos\theta)^{2}$，即$2(\cos\theta - \sin\theta)(\sin\theta + \cos\theta) = (\sin\theta + \cos\theta)^{2}$，又$\theta$为锐角，$\therefore\sin\theta + \cos\theta＞0$，$\therefore2(\cos\theta - \sin\theta) = \sin\theta + \cos\theta$，即$\cos\theta = 3\sin\theta$，$\therefore\tan\theta = \frac{1}{3}$. 故选A.]

例2　(1)(2023·湖北武汉模拟)$\sin109^{\circ}\cos296^{\circ} + \cos71^{\circ}\sin64^{\circ} =$（ ）
A. $\frac{1}{2}$
B. $\frac{\sqrt{2}}{2}$
C. $\frac{\sqrt{3}}{2}$
D. 1

答案： B [$\sin109^{\circ}\cos296^{\circ} + \cos71^{\circ}\sin64^{\circ} = \sin(180^{\circ} - 71^{\circ})\cdot\cos(360^{\circ} - 64^{\circ}) + \cos71^{\circ}\sin64^{\circ} = \sin71^{\circ}\cos64^{\circ} + \cos71^{\circ}\sin64^{\circ} = \sin(71^{\circ} + 64^{\circ}) = \sin135^{\circ} = \frac{\sqrt{2}}{2}$. 故选B.]

(2)(2024·广西梧州模拟)$\frac{1 + \tan\frac{7\pi}{12}}{1 - \tan\frac{7\pi}{12}} =$（ ）
A. $-\frac{\sqrt{3}}{3}$
B. $\frac{\sqrt{3}}{3}$
C. $-\sqrt{3}$
D. $\sqrt{3}$

答案： A [因为$\frac{1 + \tan\frac{7\pi}{12}}{1 - \tan\frac{7\pi}{12}} = \frac{\tan\frac{\pi}{4} + \tan\frac{7\pi}{12}}{1 - \tan\frac{\pi}{4}\tan\frac{7\pi}{12}} = \tan(\frac{\pi}{4} + \frac{7\pi}{12}) = \tan\frac{10\pi}{12} = \tan\frac{5\pi}{6} = \tan(\pi - \frac{\pi}{6}) = - \tan\frac{\pi}{6} = - \frac{\sqrt{3}}{3}$. 故选A.]

3.(2024·福建永安三中模拟)$\cos(\alpha - 35^{\circ})\cos(25^{\circ} + \alpha) + \sin(\alpha - 35^{\circ})\sin(25^{\circ} + \alpha)$的值为（ ）
A. $-\frac{1}{2}$
B. $\frac{1}{2}$
C. $-\frac{\sqrt{3}}{2}$
D. $\frac{\sqrt{3}}{2}$

答案： B [由两角差的余弦公式，得$\cos(\alpha - 35^{\circ})\cos(25^{\circ} + \alpha) + \sin(\alpha - 35^{\circ})\sin(25^{\circ} + \alpha) = \cos[(\alpha - 35^{\circ}) - (25^{\circ} + \alpha)] = \cos(- 60^{\circ}) = \frac{1}{2}$. 故选B.]