《2025年世纪金榜高中全程复习方略高中数学A版》

[例4]化简：
(1)$3\cos x + 3\sin x$；
(2)$\sin\frac{\pi}{12} - \sqrt{3}\cos\frac{\pi}{12}$；
(3)$3\sqrt{15}\sin x + 3\sqrt{5}\cos x$。

答案： 【解析】
(1)原式$=3\sqrt{2}(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x)=3\sqrt{2}\cos(x-\frac{\pi}{4})$.
(2)方法一:原式$=2(\frac{1}{2}\sin\frac{\pi}{12}-\frac{\sqrt{3}}{2}\cos\frac{\pi}{12})=2(\sin\frac{\pi}{6}\sin\frac{\pi}{12}-\cos\frac{\pi}{6}\cos\frac{\pi}{12})=-2\cos(\frac{\pi}{6}+\frac{\pi}{12})=-2\cos\frac{\pi}{4}=-\sqrt{2}$.
方法二:原式$=2(\frac{1}{2}\sin\frac{\pi}{12}-\frac{\sqrt{3}}{2}\cos\frac{\pi}{12})=2(\cos\frac{\pi}{3}\sin\frac{\pi}{12}-\sin\frac{\pi}{3}\cos\frac{\pi}{12})=-2\sin(\frac{\pi}{3}-\frac{\pi}{12})=-2\sin\frac{\pi}{4}=-\sqrt{2}$.
(3)$3\sqrt{15}\sin x+3\sqrt{5}\cos x=6\sqrt{5}(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x)=6\sqrt{5}(\sin x\cos\frac{\pi}{6}+\cos x\sin\frac{\pi}{6})=6\sqrt{5}\sin(x+\frac{\pi}{6})$.

对点训练
1. $\cos(\alpha - 35^{\circ})\cos(25^{\circ} + \alpha) + \sin(\alpha - 35^{\circ})\sin(25^{\circ} + \alpha)$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$\cos(\alpha - 35^{\circ})\cos(25^{\circ}+\alpha)+\sin(\alpha - 35^{\circ})\sin(25^{\circ}+\alpha)=\cos[(\alpha - 35^{\circ})-(25^{\circ}+\alpha)]=\cos(-60^{\circ})=\frac{1}{2}$.

2. 若$\alpha + \beta = \frac{2\pi}{3}$，则$\sqrt{3}\tan\alpha\tan\beta - \tan\alpha - \tan\beta$的值为________。

答案： 【解析】因为$\alpha+\beta=\frac{2\pi}{3}$,所以$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\tan(\pi-\frac{\pi}{3})=-\sqrt{3}$,所以$\tan\alpha+\tan\beta=-\sqrt{3}(1-\tan\alpha\tan\beta)$,所以$\sqrt{3}\tan\alpha\tan\beta-\tan\alpha-\tan\beta=\sqrt{3}\tan\alpha\tan\beta-(\tan\alpha+\tan\beta)=\sqrt{3}\tan\alpha\tan\beta+\sqrt{3}-\sqrt{3}\tan\alpha\tan\beta=\sqrt{3}$.答案:$\sqrt{3}$

3. (2022·北京高考)若函数$f(x) = A\sin x - \sqrt{3}\cos x$的一个零点为$\frac{\pi}{3}$，则$A =$________；$f(\frac{\pi}{12}) =$________。

答案： 【解析】依题意得$f(\frac{\pi}{3})=A\times\frac{\sqrt{3}}{2}-\sqrt{3}\times\frac{1}{2}=0$,解得$A = 1$,所以$f(x)=\sin x-\sqrt{3}\cos x=2\sin(x-\frac{\pi}{3})$,所以$f(\frac{\pi}{12})=2\sin(\frac{\pi}{12}-\frac{\pi}{3})=-\sqrt{2}$.答案:$1\ -\sqrt{2}$

[例5](1)(2024·株洲模拟)已知$\theta\in(0,\frac{\pi}{2})$，$\sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{5}}{5}$，则$\tan\theta =$ （ ）
A. 2
B. $\frac{1}{2}$
C. 3
D. $\frac{1}{3}$

答案： C 因为$\theta\in(0,\frac{\pi}{2})$,所以$-\frac{\pi}{4}\lt\theta-\frac{\pi}{4}\lt\frac{\pi}{4}$,故$\cos(\theta-\frac{\pi}{4})=\sqrt{1-\sin^{2}(\theta-\frac{\pi}{4})}=\frac{2\sqrt{5}}{5}$,所以$\sin\theta=\sin[(\theta-\frac{\pi}{4})+\frac{\pi}{4}]=\frac{\sqrt{2}}{2}[\sin(\theta-\frac{\pi}{4})+\cos(\theta-\frac{\pi}{4})]=\frac{3\sqrt{10}}{10}$,故$\cos\theta=\sqrt{1-\sin^{2}\theta}=\frac{\sqrt{10}}{10}$,因此$\tan\theta=\frac{\sin\theta}{\cos\theta}=3$.

(2)已知$\alpha,\beta\in(\frac{\pi}{3},\frac{5\pi}{6})$，若$\sin(\alpha + \frac{\pi}{6}) = \frac{4}{5}$，$\cos(\beta - \frac{5\pi}{6}) = \frac{5}{13}$，则$\sin(\alpha - \beta)$的值为 （ ）
A. $\frac{16}{65}$
B. $\frac{33}{65}$
C. $\frac{56}{65}$
D. $\frac{63}{65}$

答案： A 由题意可得$\alpha+\frac{\pi}{6}\in(\frac{\pi}{2},\pi)$,$\beta-\frac{5\pi}{6}\in(-\frac{\pi}{2},0)$,所以$\cos(\alpha+\frac{\pi}{6})=-\frac{3}{5}$,$\sin(\beta-\frac{5\pi}{6})=-\frac{12}{13}$,所以$\sin(\alpha-\beta)=-\sin[(\alpha+\frac{\pi}{6})-(\beta-\frac{5\pi}{6})]=-\frac{4}{5}\times\frac{5}{13}+(-\frac{3}{5})\times(-\frac{12}{13})=\frac{16}{65}$.

对点训练
1. (2024·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 因为$2\alpha=(\alpha+\beta)+(\alpha-\beta)$,所以$\tan2\alpha=\frac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta)\tan(\alpha-\beta)}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}=1$.又$\tan(\pi - 2\alpha)=-\tan2\alpha$,所以$\tan(\pi - 2\alpha)=-1$.

2. 已知$0\lt\alpha\lt\frac{\pi}{2}\lt\beta\lt\pi$，$\tan\alpha = \frac{4}{3}$，$\cos(\beta - \alpha) = \frac{\sqrt{2}}{10}$，则$\sin\alpha =$________，$\cos\beta =$________。

答案： 【解析】因为$0\lt\alpha\lt\frac{\pi}{2}$,且$\tan\alpha=\frac{4}{3}$,所以$\sin\alpha=\frac{4}{5},\cos\alpha=\frac{3}{5}$,由$0\lt\alpha\lt\frac{\pi}{2}\lt\beta\lt\pi$,则$0\lt\beta-\alpha\lt\pi$,又因为$\cos(\beta-\alpha)=\frac{\sqrt{2}}{10}$,则$\sin(\beta-\alpha)=\frac{7\sqrt{2}}{10}$,所以$\cos\beta=\cos[(\beta-\alpha)+\alpha]=\cos(\beta-\alpha)\cos\alpha-\sin(\beta-\alpha)\sin\alpha=\frac{\sqrt{2}}{10}\times\frac{3}{5}-\frac{7\sqrt{2}}{10}\times\frac{4}{5}=-\frac{\sqrt{2}}{2}$.答案:$\frac{4}{5}\ -\frac{\sqrt{2}}{2}$