《2025年世纪金榜高中全程复习方略高中数学A版》

(1)函数$f(x)=\sin ^{2}x+\sqrt{3}\sin x\cos x-\frac{1}{2}$可以化简为 (　　　)
A. $f(x)=\sin(2x - \frac{\pi }{3})$
B. $f(x)=\sin(2x - \frac{\pi }{6})$
C. $f(x)=\sin(2x + \frac{\pi }{3})$
D. $f(x)=\sin(2x + \frac{\pi }{6})$

答案： B $f(x)=\sin^{2}x+\sqrt{3}\sin x\cos x-\frac{1}{2}=\frac{1 - \cos2x}{2}+\frac{\sqrt{3}}{2}\sin2x-\frac{1}{2}=\frac{\sqrt{3}}{2}\sin2x-\frac{1}{2}\cos2x=\sin(2x-\frac{\pi}{6})$.

(2)若$\alpha \in (0,\frac{\pi }{2})$，$\tan 2\alpha =\frac{\cos \alpha }{2 - \sin \alpha }$，则$\tan \alpha$等于 (　　　)
A. $\frac{\sqrt{15}}{15}$
B. $\frac{\sqrt{5}}{5}$
C. $\frac{\sqrt{5}}{3}$
D. $\frac{\sqrt{15}}{3}$

答案： 解法一:因为$\tan2\alpha=\frac{\sin2\alpha}{\cos2\alpha}=\frac{2\sin\alpha\cos\alpha}{1 - 2\sin^{2}\alpha}$,且$\tan2\alpha=\frac{\cos\alpha}{2 - \sin\alpha}$,所以$\frac{2\sin\alpha\cos\alpha}{1 - 2\sin^{2}\alpha}=\frac{\cos\alpha}{2 - \sin\alpha}$,解得$\sin\alpha=\frac{1}{4}$.因为$\alpha\in(0,\frac{\pi}{2})$,所以$\cos\alpha=\frac{\sqrt{15}}{4}$,$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\sqrt{15}}{15}$.
解法二:因为$\tan2\alpha=\frac{2\tan\alpha}{1 - \tan^{2}\alpha}=\frac{\frac{2\sin\alpha}{\cos\alpha}}{1-\frac{\sin^{2}\alpha}{\cos^{2}\alpha}}=\frac{2\sin\alpha\cos\alpha}{\cos^{2}\alpha-\sin^{2}\alpha}=\frac{2\sin\alpha\cos\alpha}{1 - 2\sin^{2}\alpha}$,且$\tan2\alpha=\frac{\cos\alpha}{2 - \sin\alpha}$,所以$\frac{2\sin\alpha\cos\alpha}{1 - 2\sin^{2}\alpha}=\frac{\cos\alpha}{2 - \sin\alpha}$,解得$\sin\alpha=\frac{1}{4}$.因为$\alpha\in(0,\frac{\pi}{2})$,所以$\cos\alpha=\frac{\sqrt{15}}{4}$,$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\sqrt{15}}{15}$.

(3)已知$\sin \alpha + \cos \alpha =\frac{2\sqrt{3}}{3}$，则$\sin ^{2}(\alpha - \frac{\pi }{4})=$____________________.

答案： 【解析】因为$\sin\alpha+\cos\alpha=\frac{2\sqrt{3}}{3}$,两边同时平方得$\sin^{2}\alpha + 2\sin\alpha\cos\alpha+\cos^{2}\alpha=\frac{4}{3}$,即$\sin2\alpha=\frac{1}{3}$,由降幂公式可知$\sin^{2}(\alpha-\frac{\pi}{4})=\frac{1 - \cos(2\alpha-\frac{\pi}{2})}{2}=\frac{1 - \sin2\alpha}{2}=\frac{1}{2}-\frac{1}{2}\sin2\alpha=\frac{1}{3}$.
答案:$\frac{1}{3}$

对点训练
1. 化简：$\frac{2\cos ^{4}x - 2\cos ^{2}x + \frac{1}{2}}{2\tan(\frac{\pi }{4} - x)\sin ^{2}(\frac{\pi }{4} + x)}=$__________.

答案： 【解析】原式$=\frac{\frac{1}{2}(4\cos^{4}x - 4\cos^{2}x + 1)}{2\times\sin(\frac{\pi}{4}-x)\cdot\cos^{2}(\frac{\pi}{4}-x)\div\cos(\frac{\pi}{4}-x)}=\frac{(2\cos^{2}x - 1)^{2}}{4\sin(\frac{\pi}{4}-x)\cos(\frac{\pi}{4}-x)}=\frac{\cos^{2}2x}{2\sin(\frac{\pi}{2}-2x)}=\frac{\cos^{2}2x}{2\cos2x}=\frac{1}{2}\cos2x$.
答案:$\frac{1}{2}\cos2x$

2. 化简：$\sin ^{2}\alpha \sin ^{2}\beta + \cos ^{2}\alpha \cos ^{2}\beta - \frac{1}{2}\cos 2\alpha \cos 2\beta =$__________.

答案： 【解析】原式$=\frac{1 - \cos2\alpha}{2}\cdot\frac{1 - \cos2\beta}{2}+\frac{1 + \cos2\alpha}{2}\cdot\frac{1 + \cos2\beta}{2}-\frac{1}{2}\cos2\alpha\cos2\beta=\frac{1 - \cos2\beta-\cos2\alpha+\cos2\alpha\cos2\beta}{4}+\frac{1 + \cos2\beta+\cos2\alpha+\cos2\alpha\cos2\beta}{4}-\frac{1}{2}\cos2\alpha\cdot\cos2\beta=\frac{1}{2}+\frac{1}{2}\cos2\alpha\cos2\beta-\frac{1}{2}\cos2\alpha\cos2\beta=\frac{1}{2}$.
答案:$\frac{1}{2}$

(1)$\sin 18^{\circ}\cos 36^{\circ}=$__________.

答案： 【解析】原式$=\frac{2\sin18^{\circ}\cos18^{\circ}\cos36^{\circ}}{2\cos18^{\circ}}=\frac{2\sin36^{\circ}\cos36^{\circ}}{4\cos18^{\circ}}=\frac{\sin72^{\circ}}{4\cos18^{\circ}}=\frac{1}{4}$.
答案:$\frac{1}{4}$

(2)$\cos ^{4}\frac{\pi }{12}-\sin ^{4}\frac{\pi }{12}=$__________.

答案： 原式$=(\cos^{2}\frac{\pi}{12}-\sin^{2}\frac{\pi}{12})(\cos^{2}\frac{\pi}{12}+\sin^{2}\frac{\pi}{12})=\cos^{2}\frac{\pi}{12}-\sin^{2}\frac{\pi}{12}=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$.
答案:$\frac{\sqrt{3}}{2}$

(3)$\frac{1}{2\sin 10^{\circ}}-\frac{\sqrt{3}}{2\cos 10^{\circ}}=$__________.

答案： 原式$=\frac{\cos10^{\circ}-\sqrt{3}\sin10^{\circ}}{2\sin10^{\circ}\cdot\cos10^{\circ}}=\frac{2(\frac{1}{2}\cos10^{\circ}-\frac{\sqrt{3}}{2}\sin10^{\circ})}{\sin20^{\circ}}=\frac{2\sin(30^{\circ}-10^{\circ})}{\sin20^{\circ}}=2$.
答案:2

(1)(2023·新高考Ⅰ卷)已知$\sin(\alpha - \beta )=\frac{1}{3},\cos \alpha \sin \beta =\frac{1}{6}$，则$\cos(2\alpha + 2\beta )=$ (　　　)
A. $\frac{7}{9}$
B. $\frac{1}{9}$
C. $-\frac{1}{9}$
D. $-\frac{7}{9}$

答案： B 因为$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta=\frac{1}{3}$,$\cos\alpha\sin\beta=\frac{1}{6}$,所以$\sin\alpha\cos\beta=\frac{1}{2}$,所以$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac{2}{3}$,所以$\cos(2\alpha + 2\beta)=\cos2(\alpha+\beta)=1 - 2\sin^{2}(\alpha+\beta)=1 - 2\times(\frac{2}{3})^{2}=\frac{1}{9}$.

(2)(一题多法)已知$\sin(\alpha - \frac{\pi }{4})=\frac{3}{5},\pi ＜\alpha ＜\frac{5\pi }{4}$，则$\cos(2\alpha - \frac{\pi }{4})=$__________.

答案： 【解析】因为$\sin(\alpha-\frac{\pi}{4})=\frac{3}{5}$,$\frac{3\pi}{4}＜\alpha-\frac{\pi}{4}＜\pi$,所以$\cos(\alpha-\frac{\pi}{4})=-\frac{4}{5}$.
解法一:因为$\cos\alpha=\cos[(\alpha-\frac{\pi}{4})+\frac{\pi}{4}]=\cos(\alpha-\frac{\pi}{4})\cos\frac{\pi}{4}-\sin(\alpha-\frac{\pi}{4})\sin\frac{\pi}{4}=-\frac{4}{5}\times\frac{\sqrt{2}}{2}-\frac{3}{5}\times\frac{\sqrt{2}}{2}=-\frac{7\sqrt{2}}{10}$,$\sin\alpha=\sin[(\alpha-\frac{\pi}{4})+\frac{\pi}{4}]=\sin(\alpha-\frac{\pi}{4})\cos\frac{\pi}{4}+\cos(\alpha-\frac{\pi}{4})\sin\frac{\pi}{4}=\frac{3}{5}\times\frac{\sqrt{2}}{2}-\frac{4}{5}\times\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}$,所以$\cos(2\alpha-\frac{\pi}{4})=\cos[(\alpha-\frac{\pi}{4})+\alpha]=\cos(\alpha-\frac{\pi}{4})\cos\alpha-\sin(\alpha-\frac{\pi}{4})\sin\alpha=-\frac{4}{5}\times(-\frac{7\sqrt{2}}{10})-\frac{3}{5}\times(-\frac{\sqrt{2}}{10})=\frac{31\sqrt{2}}{50}$.
解法二:$\cos(2\alpha-\frac{\pi}{4})=\cos2\alpha\cos\frac{\pi}{4}+\sin2\alpha\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}(\cos2\alpha+\sin2\alpha)$,$\cos2\alpha=-\sin(2\alpha-\frac{\pi}{2})=-\sin2(\alpha-\frac{\pi}{4})=-2\sin(\alpha-\frac{\pi}{4})\cos(\alpha-\frac{\pi}{4})=-2\times\frac{3}{5}\times(-\frac{4}{5})=\frac{24}{25}$,$\sin2\alpha=\cos(2\alpha-\frac{\pi}{2})=1 - 2\sin^{2}(\alpha-\frac{\pi}{4})=1 - 2\times\frac{9}{25}=\frac{7}{25}$,所以原式$=\frac{\sqrt{2}}{2}\times(\frac{24}{25}+\frac{7}{25})=\frac{31\sqrt{2}}{50}$.
答案:$\frac{31\sqrt{2}}{50}$

(1)已知$\alpha$为锐角，且$\sin \alpha \cdot (\sqrt{3}-\tan 10^{\circ}) = 1$，则$\alpha =$__________.

答案： 【解析】由已知得$\sin\alpha=\frac{1}{\sqrt{3}-\tan10^{\circ}}=\frac{1}{\sqrt{3}-\frac{\sin10^{\circ}}{\cos10^{\circ}}}=\frac{\cos10^{\circ}}{\sqrt{3}\cos10^{\circ}-\sin10^{\circ}}=\frac{\cos10^{\circ}}{2\sin50^{\circ}}=\frac{\sin80^{\circ}}{2\sin50^{\circ}}=\frac{2\sin40^{\circ}\cos40^{\circ}}{2\cos40^{\circ}}=\sin40^{\circ}$. 由于$\alpha$为锐角,所以$\alpha = 40^{\circ}$.
答案:$40^{\circ}$

(2)已知$\sin \alpha =\frac{\sqrt{2}}{10},\cos \beta =\frac{3\sqrt{10}}{10}$，且$\alpha,\beta$为锐角，则$\alpha + 2\beta =$__________.

答案： 【解析】因为$\sin\alpha=\frac{\sqrt{2}}{10}$,且$\alpha$为锐角,所以$\cos\alpha=\sqrt{1 - \sin^{2}\alpha}=\sqrt{1-\frac{2}{100}}=\frac{7\sqrt{2}}{10}$,因为$\cos\beta=\frac{3\sqrt{10}}{10}$,且$\beta$为锐角,所以$\sin\beta=\sqrt{1 - \cos^{2}\beta}=\sqrt{1-\frac{90}{100}}=\frac{\sqrt{10}}{10}$,那么$\sin2\beta=2\sin\beta\cos\beta=2\times\frac{\sqrt{10}}{10}\times\frac{3\sqrt{10}}{10}=\frac{3}{5}$,$\cos2\beta=1 - 2\sin^{2}\beta=1 - 2\times(\frac{\sqrt{10}}{10})^{2}=\frac{4}{5}$,所以$\cos(\alpha + 2\beta)=\cos\alpha\cos2\beta-\sin\alpha\sin2\beta=\frac{7\sqrt{2}}{10}\times\frac{4}{5}-\frac{\sqrt{2}}{10}\times\frac{3}{5}=\frac{\sqrt{2}}{2}$,因为$\alpha\in(0,\frac{\pi}{2})$,$\beta\in(0,\frac{\pi}{2})$,所以$2\beta\in(0,\pi)$.所以$\alpha + 2\beta\in(0,\frac{3\pi}{2})$,故$\alpha + 2\beta=\frac{\pi}{4}$.
答案:$\frac{\pi}{4}$