《2025年金版教程高考科学复习解决方案数学》

4.(2023·江苏常州二模)已知$\sin\alpha - \sqrt{3}\cos\alpha = 1$，则$\sin(\frac{7\pi}{6} - 2\alpha)$的值为________.

答案： 答案 $\frac{1}{2}$
解析 已知$\sin\alpha - \sqrt{3}\cos\alpha = 1$，则$2(\frac{1}{2}\sin\alpha - \frac{\sqrt{3}}{2}\cos\alpha) = 2\sin(\alpha - \frac{\pi}{3}) = 1$，所以$\sin(\alpha - \frac{\pi}{3}) = \frac{1}{2}$，令$\beta = \alpha - \frac{\pi}{3}$，则$\alpha = \beta + \frac{\pi}{3}$，即$\sin\beta = \frac{1}{2}$，所以$\sin(\frac{7\pi}{6} - 2\alpha) = \sin(\frac{7\pi}{6} - 2\beta - \frac{2\pi}{3}) = \sin(\frac{\pi}{2} - 2\beta) = \cos2\beta = 1 - 2\sin^{2}\beta = \frac{1}{2}$.

5. $\tan50^{\circ} - \tan20^{\circ} - \frac{\sqrt{3}}{3}\tan50^{\circ}\tan20^{\circ} =$________.

答案： 答案 $\frac{\sqrt{3}}{3}$
解析 $\tan50^{\circ} - \tan20^{\circ} - \frac{\sqrt{3}}{3}\tan50^{\circ}\tan20^{\circ} = \tan(50^{\circ} - 20^{\circ})(1 + \tan50^{\circ}\tan20^{\circ}) - \frac{\sqrt{3}}{3}\tan50^{\circ}\tan20^{\circ} = \tan30^{\circ}(1 + \tan50^{\circ}\tan20^{\circ}) - \frac{\sqrt{3}}{3}\tan50^{\circ}\tan20^{\circ} = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}\tan50^{\circ}\tan20^{\circ} - \frac{\sqrt{3}}{3}\tan50^{\circ}\tan20^{\circ} = \frac{\sqrt{3}}{3}$.

例3　(1)(2024·四川绵阳模拟)已知$\sin(\frac{\pi}{6} - \alpha) = \frac{\sqrt{2}}{3}$，则$\cos(2\alpha - \frac{4\pi}{3}) =$（ ）
A. $-\frac{5}{9}$
B. $\frac{5}{9}$
C. $-\frac{1}{3}$
D. $\frac{1}{3}$

答案： A [$\cos(2\alpha - \frac{4\pi}{3}) = \cos(-\pi + 2\alpha - \frac{\pi}{3}) = - \cos(2\alpha - \frac{\pi}{3}) = - \cos(\frac{\pi}{3} - 2\alpha) = - [1 - 2\sin^{2}(\frac{\pi}{6} - \alpha)] = - (1 - 2\times\frac{2}{9}) = - \frac{5}{9}$. 故选A.]

(2)已知$\alpha,\beta\in(\frac{3\pi}{4},\pi)$，$\sin(\alpha + \beta) = -\frac{3}{5}$，$\sin(\beta - \frac{\pi}{4}) = \frac{12}{13}$，则$\cos(\alpha + \frac{\pi}{4}) =$________.

答案： 答案 $-\frac{56}{65}$
解析 因为$\alpha,\beta\in(\frac{3\pi}{4},\pi)$，所以$\frac{3\pi}{2}＜\alpha + \beta＜2\pi$，$\frac{\pi}{2}＜\beta - \frac{\pi}{4}＜\frac{3\pi}{4}$，因为$\sin(\alpha + \beta) = - \frac{3}{5}$，$\sin(\beta - \frac{\pi}{4}) = \frac{12}{13}$，所以$\cos(\alpha + \beta) = \frac{4}{5}$，$\cos(\beta - \frac{\pi}{4}) = - \frac{5}{13}$，所以$\cos(\alpha + \frac{\pi}{4}) = \cos[\alpha + \beta - (\beta - \frac{\pi}{4})] = \cos(\alpha + \beta)\cos(\beta - \frac{\pi}{4}) + \sin(\alpha + \beta)\sin(\beta - \frac{\pi}{4}) = \frac{4}{5}\times(- \frac{5}{13}) + ( - \frac{3}{5})\times\frac{12}{13} = - \frac{56}{65}$.

6.(2023·山东烟台模拟)已知$\tan(\alpha + \beta) = \frac{1}{2}$，$\tan(\alpha - \beta) = \frac{1}{3}$，则$\tan(\pi - 2\alpha) =$（ ）
A. 1
B. -1
C. 2
D. -2

答案： B [$\because2\alpha = (\alpha + \beta) + (\alpha - \beta)$，$\therefore\tan2\alpha = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta)\tan(\alpha - \beta)} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}\times\frac{1}{3}} = 1$. 又$\tan(\pi - 2\alpha) = - \tan2\alpha$，$\therefore\tan(\pi - 2\alpha) = - 1$. 故选B.]

7. 已知$0 ＜ x ＜ \frac{\pi}{4}$，$\sin(\frac{\pi}{4} - x) = \frac{5}{13}$，则$\frac{\cos2x}{\cos(\frac{\pi}{4} + x)} =$________.

答案： 答案 $\frac{24}{13}$
解析 $\frac{\cos2x}{\cos(\frac{\pi}{4} + x)} = \frac{\cos^{2}x - \sin^{2}x}{\frac{\sqrt{2}}{2}(\cos x - \sin x)} = \sqrt{2}(\cos x + \sin x) = 2\cos(\frac{\pi}{4} - x)$. 由$0＜x＜\frac{\pi}{4}$得$0＜\frac{\pi}{4} - x＜\frac{\pi}{4}$，$\therefore\cos(\frac{\pi}{4} - x) = \sqrt{1 - \sin^{2}(\frac{\pi}{4} - x)} = \sqrt{1 - (\frac{5}{13})^{2}} = \frac{12}{13}$，所以原式$ = 2\times\frac{12}{13} = \frac{24}{13}$.

例1　$2\sqrt{1 + \sin4}+\sqrt{2 + 2\cos4}=$（ ）
A. $2\cos2$ B. $2\sin2$
C. $4\sin2 + 2\cos2$ D. $2\sin2 + 4\cos2$
[课堂笔记]　____________________________

答案： B [2$\sqrt{1 + \sin4}$ + $\sqrt{2 + 2\cos4}$ = 2$\sqrt{\sin^{2}2 + 2\sin2\cos2 + \cos^{2}2}$ + $\sqrt{2 + 2(2\cos^{2}2 - 1)}$ = 2$\sqrt{(\sin2 + \cos2)^{2}}$ + $\sqrt{4\cos^{2}2}$ = 2|$\sin2 + \cos2$| + 2|$\cos2$|.
∵$\frac{\pi}{2}$ ＜ 2 ＜ $\pi$,
∴$\cos2$ ＜ 0,
∵$\sin2 + \cos2$ = $\sqrt{2}\sin(2 + \frac{\pi}{4})$, 0 ＜ 2 + $\frac{\pi}{4}$ ＜ $\pi$,
∴$\sin2 + \cos2$ ＞ 0,
∴原式 = 2($\sin2 + \cos2$) - 2$\cos2$ = 2$\sin2$. 故选B.]