《2025年高考总复习首选用卷数学人教版》

12.(2023·广东珠海高三月考)已知$\frac{\sin\alpha+2\cos\alpha}{\sin\alpha-2\cos\alpha}=5$,则$\cos^{2}\alpha+\frac{1}{2}\sin2\alpha=$______.

答案： 12.答案 $\frac{2}{5}$
解析 $\because\frac{\sin\alpha+2\cos\alpha}{\sin\alpha-2\cos\alpha}=5$,$\therefore\sin\alpha+2\cos\alpha=5\sin\alpha-10\cos\alpha$,即$12\cos\alpha=4\sin\alpha$,则$\tan\alpha=3$,则$\cos^{2}\alpha+\frac{1}{2}\sin2\alpha=\cos^{2}\alpha+\sin\alpha\cos\alpha=\frac{\cos^{2}\alpha+\sin\alpha\cos\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}=\frac{1+\tan\alpha}{\tan^{2}\alpha+1}=\frac{1 + 3}{9 + 1}=\frac{2}{5}$.

13.(2024·广东佛山高三联考)在$\triangle ABC$中，点$D$是边$BC$上一点，且$AB = 4$,$BD = 2$,$\cos B=\frac{11}{16}$,$\cos C=\frac{\sqrt{6}}{4}$,则$DC =$______.

答案： 13.答案 3
解析 在$\triangle ABD$中,$AD^{2}=AB^{2}+BD^{2}-2AB\cdot BD\cos B = 9$,可得$AD = 3$. 又由余弦定理得,$\cos\angle ADB=\frac{AD^{2}+BD^{2}-AB^{2}}{2AD\cdot BD}=\frac{9 + 4 - 16}{2\times3\times2}=-\frac{1}{4}$,可得$\sin\angle ADB=\frac{\sqrt{15}}{4}$. 在$\triangle ADC$中,$\sin\angle DAC=\sin(\angle ADB-\angle C)$,由此可得$\sin\angle DAC=\sin\angle ADB\cos C-\cos\angle ADB\sin C$,由已知可得$\sin C=\frac{\sqrt{10}}{4}$,代入可得$\sin\angle DAC=\frac{\sqrt{15}}{4}\times\frac{\sqrt{6}}{4}+\frac{1}{4}\times\frac{\sqrt{10}}{4}=\frac{\sqrt{10}}{4}$,所以$AD = DC$,所以$DC = 3$.

14.(2024·重庆高三第一次质量检测)已知函数$f(x)=\sin(\omega x+\varphi)(\omega＞0,|\varphi|＜\frac{\pi}{2})$,如图，直线$y = \frac{\sqrt{3}}{2}$与曲线$y = f(x)$交于$A,B$两点，$|AB|=\frac{\pi}{6}$,则$\varphi =$______. $y = f(x)$在$[t-\frac{\pi}{4},t](t\in\mathbf{R})$上的最大值与最小值之差的取值范围是______.

答案： 14.答案 $\frac{\pi}{12}$ $[1-\frac{\sqrt{2}}{2},\sqrt{2}]$
解析 令$f(x)=\sin(\omega x+\varphi)=\frac{\sqrt{3}}{2}$,得$\omega x+\varphi=2k\pi+\frac{\pi}{3},k\in\mathbf{Z}$或$\omega x+\varphi=2k\pi+\frac{2\pi}{3},k\in\mathbf{Z}$,由题意,知$|AB|=x_{B}-x_{A}=\frac{\pi}{6}$,且$\begin{cases}\omega x_{A}+\varphi=\frac{\pi}{3}\\\omega x_{B}+\varphi=\frac{2\pi}{3}\end{cases}$,两式相减,得$\omega(x_{B}-x_{A})=\frac{\pi}{3}$,所以$\omega = 2$.由题图可知,$x=-\frac{\pi}{24}$是函数的一个零点,则$2\times(-\frac{\pi}{24})+\varphi=k\pi,k\in\mathbf{Z}$,即$\varphi=\frac{\pi}{12}+k\pi,k\in\mathbf{Z}$. 又因为$|\varphi|＜\frac{\pi}{2}$,则$\varphi=\frac{\pi}{12}$,故$f(x)=\sin(2x+\frac{\pi}{12})$. 设函数$f(x)=\sin(2x+\frac{\pi}{12})$在$[t-\frac{\pi}{4},t],t\in\mathbf{R}$上的最大值与最小值之差为$g(t)$,当函数$f(x)$图象的对称轴不在$[t-\frac{\pi}{4},t],t\in\mathbf{R}$上时,函数$f(x)$在$[t-\frac{\pi}{4},t],t\in\mathbf{R}$上单调,不妨设函数$f(x)$在$[t-\frac{\pi}{4},t],t\in\mathbf{R}$上单调递增,则$g(t)=f(t)-f(t-\frac{\pi}{4})=\sin(2t+\frac{\pi}{12})-\sin[2(t-\frac{\pi}{4})+\frac{\pi}{12}]=\sin(2t+\frac{\pi}{12})+\cos(2t+\frac{\pi}{12})=\sqrt{2}\sin(2t+\frac{\pi}{3})\leqslant\sqrt{2}$;当对称轴在$[t-\frac{\pi}{4},t],t\in\mathbf{R}$上时,不妨设$f(x)$在对称轴处取得最大值$1$,则函数$f(x)$的最小值为$f(t-\frac{\pi}{4})$或$f(t)$,显然当对称轴经过$[t-\frac{\pi}{4},t],t\in\mathbf{R}$中点时,$g(t)$有最小值,所以$2\times\frac{(t-\frac{\pi}{4})+t}{2}+\frac{\pi}{12}=\frac{\pi}{2}+2k\pi,k\in\mathbf{Z}$,则$t=\frac{\pi}{3}+k\pi,k\in\mathbf{Z}$,$f(t)=\sin[2(\frac{\pi}{3}+k\pi)+\frac{\pi}{12}]=\sin(\frac{3\pi}{4}+2k\pi)=\frac{\sqrt{2}}{2}(k\in\mathbf{Z})$,所以$g(t)$的最小值为$1-\frac{\sqrt{2}}{2}$. 综上,函数$f(x)=\sin(2x+\frac{\pi}{12})$在$[t-\frac{\pi}{4},t],t\in\mathbf{R}$上的最大值与最小值之差的取值范围是$[1-\frac{\sqrt{2}}{2},\sqrt{2}]$.

15.(本小题满分13分)已知$\alpha\in(\frac{\pi}{2},\pi)$,且$\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2}=\frac{\sqrt{6}}{2}$.
(1)求$\cos\alpha$的值；
(2)若$\sin(\alpha-\beta)=-\frac{3}{5},\beta\in(\frac{\pi}{2},\pi)$,求$\cos\beta$的值.

答案： 15.解
(1)因为$\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2}=\frac{\sqrt{6}}{2}$,两边同时平方,得$\sin\alpha=\frac{1}{2}$.又$\frac{\pi}{2}＜\alpha＜\pi$,所以$\cos\alpha=-\sqrt{1-\sin^{2}\alpha}=-\frac{\sqrt{3}}{2}$.
(2)因为$\frac{\pi}{2}＜\alpha＜\pi$,$\frac{\pi}{2}＜\beta＜\pi$,所以$-\frac{\pi}{2}＜\alpha-\beta＜\frac{\pi}{2}$,又由$\sin(\alpha-\beta)=-\frac{3}{5}$,得$\cos(\alpha-\beta)=\frac{4}{5}$.所以$\cos\beta=\cos[\alpha-(\alpha-\beta)]=\cos\alpha\cos(\alpha-\beta)+\sin\alpha\sin(\alpha-\beta)=-\frac{\sqrt{3}}{2}\times\frac{4}{5}+\frac{1}{2}\times(-\frac{3}{5})=-\frac{4\sqrt{3}+3}{10}$.