《2025年高考帮数学》

例1 (1)[2024山东模拟]若$\tan\theta = 2$,则$1+\sin\theta\cos\theta =$ ( )
A.$\frac{7}{3}$ B.$\frac{7}{5}$ C.$\frac{5}{4}$ D.$\frac{5}{3}$
(2)[2023全国卷乙]若$\theta\in(0,\frac{\pi}{2}),\tan\theta=\frac{1}{2}$,则$\sin\theta - \cos\theta =$______.

答案：
(1)B 易知$\cos\theta\neq0$，则$1 + \sin\theta\cos\theta=\frac{1 + \sin\theta\cos\theta}{1}=\frac{\sin^{2}\theta+\cos^{2}\theta+\sin\theta\cos\theta}{\sin^{2}\theta+\cos^{2}\theta}=\frac{\tan^{2}\theta+\tan\theta + 1}{\tan^{2}\theta + 1}=\frac{2^{2}+2 + 1}{2^{2}+1}=\frac{7}{5}$
(2)$-\frac{\sqrt{5}}{5}$ 由$\begin{cases}\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{1}{2},\\\sin^{2}\theta+\cos^{2}\theta = 1,\end{cases}$且$\theta\in(0,\frac{\pi}{2})$，解得$\begin{cases}\sin\theta=\frac{\sqrt{5}}{5},\\\cos\theta=\frac{2\sqrt{5}}{5},\end{cases}$故$\sin\theta-\cos\theta=-\frac{\sqrt{5}}{5}$

训练1 [多选/2023江西省上饶市第一中学模拟]已知$\theta\in(-\pi,0),\sin\theta+\cos\theta=\frac{7}{13}$,则下列结论正确的是 ( )
A.$\theta\in(-\pi,-\frac{\pi}{2})$
B.$\cos\theta=\frac{12}{13}$
C.$\tan\theta=\frac{5}{12}$
D.$\sin\theta - \cos\theta = -\frac{17}{13}$

答案： BD 由$\sin\theta+\cos\theta=\frac{7}{13}$可得，$\cos\theta=\frac{7}{13}-\sin\theta$，则$(\frac{7}{13}-\sin\theta)^{2}+\sin^{2}\theta = 1$，解得$\sin\theta=\frac{12}{13}$或$\sin\theta=-\frac{5}{13}$。由$\theta\in(-\pi,0)$，可得$\sin\theta=-\frac{5}{13}$，$\cos\theta=\frac{12}{13}$，故B正确；由$\sin\theta=-\frac{5}{13}＜0$，$\cos\theta=\frac{12}{13}＞0$可得$\theta$为第四象限角，又$\theta\in(-\pi,0)$，所以$\theta\in(-\frac{\pi}{2},0)$，故A错误；$\tan\theta=\frac{\sin\theta}{\cos\theta}=-\frac{5}{12}$，故C错误；$\sin\theta-\cos\theta=-\frac{5}{13}-\frac{12}{13}=-\frac{17}{13}$，故D正确。故选BD。

例2 (1)[全国卷Ⅲ]函数$f(x)=\frac{1}{5}\sin(x+\frac{\pi}{3})+\cos(x - \frac{\pi}{6})$的最大值为 ( )
A.$\frac{6}{5}$ B.1 C.$\frac{3}{5}$ D.$\frac{1}{5}$
(2)[北京高考]若函数$f(x)=\sin(x+\varphi)+\cos x$的最大值为2,则常数$\varphi$的一个取值为______.

答案：
(1)A 因为$\cos(x-\frac{\pi}{6})=\cos[(x+\frac{\pi}{3})-\frac{\pi}{2}]=\sin(x+\frac{\pi}{3})$，所以$f(x)=\frac{6}{5}\sin(x+\frac{\pi}{3})$，所以$f(x)$的最大值为$\frac{6}{5}$，故选A。
(2)$\frac{\pi}{2}$(答案不唯一) 易知当$y = \sin(x+\varphi)$，$y = \cos x$同时取得最大值1时，函数$f(x)=\sin(x+\varphi)+\cos x$取得最大值2，故$\sin(x+\varphi)=\cos x$，则$\varphi=\frac{\pi}{2}+2k\pi$，$k\in\mathbf{Z}$，故常数$\varphi$的一个取值为$\frac{\pi}{2}$

训练2 (1)[2023山东省济宁市模拟]已知$\cos(\frac{\pi}{6}-\theta)=\frac{1}{3}$,则$\cos(\frac{5\pi}{6}+\theta)+2\sin(\frac{5\pi}{3}-\theta)$的值为______.
(2)已知$\sin\alpha$是方程$5x^{2}-7x - 6 = 0$的根,且$\alpha$是第三象限角,则$\frac{\sin(-\alpha-\frac{3\pi}{2})\cos(\frac{3\pi}{2}-\alpha)}{\cos(\frac{\pi}{2}-\alpha)\sin(\frac{\pi}{2}+\alpha)}\cdot\tan^{2}(\pi - \alpha)$的值为______.

答案：
(1)−1 原式$=\cos[\pi-(\frac{\pi}{6}-\theta)]+2\sin[\frac{3\pi}{2}+(\frac{\pi}{6}-\theta)]=-\cos(\frac{\pi}{6}-\theta)-2\cos(\frac{\pi}{6}-\theta)=-3\cos(\frac{\pi}{6}-\theta)=-1$。
(2)$-\frac{9}{16}$ 原式$=\frac{-\sin(\frac{3\pi}{2}+\alpha)\cos(\frac{3\pi}{2}-\alpha)}{\sin\alpha\cos\alpha}\cdot\tan^{2}\alpha=\frac{-\cos\alpha\sin\alpha}{\sin\alpha\cos\alpha}\cdot\tan^{2}\alpha=-\tan^{2}\alpha$。解方程$5x^{2}-7x - 6 = 0$，得$x_{1}=-\frac{3}{5}$，$x_{2}=2$。又$\alpha$是第三象限角，$\therefore\sin\alpha=-\frac{3}{5}$，$\therefore\cos\alpha=-\frac{4}{5}$，$\therefore\tan\alpha=\frac{3}{4}$。故原式$=-\tan^{2}\alpha=-\frac{9}{16}$

例3 (1)[2023陕西模拟]已知$0\lt\alpha\lt\frac{\pi}{2},\cos(\alpha+\frac{\pi}{3})=-\frac{2}{3}$,则$\tan(\frac{2\pi}{3}-\alpha)=$ ( )
A.$\frac{\sqrt{5}}{2}$ B.$-\frac{\sqrt{5}}{2}$ C.$\frac{\sqrt{5}}{3}$ D.$-\frac{\sqrt{5}}{3}$
(2)[全国卷Ⅰ]已知$\theta$是第四象限角,且$\sin(\theta+\frac{\pi}{4})=\frac{3}{5}$,则$\tan(\theta - \frac{\pi}{4})=$______.

答案：
(1)A 由$0＜\alpha＜\frac{\pi}{2}$，得$\frac{\pi}{3}＜\alpha+\frac{\pi}{3}＜\frac{5\pi}{6}$，则$\sin(\alpha+\frac{\pi}{3})=\sqrt{1-\cos^{2}(\alpha+\frac{\pi}{3})}=\sqrt{1-(-\frac{2}{3})^{2}}=\frac{\sqrt{5}}{3}$，所以$\tan(\alpha+\frac{\pi}{3})=\frac{\sin(\alpha+\frac{\pi}{3})}{\cos(\alpha+\frac{\pi}{3})}=-\frac{\sqrt{5}}{2}$，所以$\tan(\frac{2\pi}{3}-\alpha)=\tan[\pi-(\alpha+\frac{\pi}{3})]=-\tan(\alpha+\frac{\pi}{3})=\frac{\sqrt{5}}{2}$。故选A。
(2)$-\frac{4}{3}$ 解法一 因为$\sin(\theta+\frac{\pi}{4})=\frac{3}{5}$，所以$\cos(\theta-\frac{\pi}{4})=\sin[\frac{\pi}{2}+(\theta-\frac{\pi}{4})]=\sin(\theta+\frac{\pi}{4})=\frac{3}{5}$。因为$\theta$为第四象限角，所以$-\frac{\pi}{2}+2k\pi＜\theta＜2k\pi$，$k\in\mathbf{Z}$，所以$-\frac{3\pi}{4}+2k\pi＜\theta-\frac{\pi}{4}＜2k\pi-\frac{\pi}{4}$，$k\in\mathbf{Z}$，所以$\sin(\theta-\frac{\pi}{4})=-\sqrt{1-(\frac{3}{5})^{2}}=-\frac{4}{5}$，所以$\tan(\theta-\frac{\pi}{4})=\frac{\sin(\theta-\frac{\pi}{4})}{\cos(\theta-\frac{\pi}{4})}=-\frac{4}{3}$。
解法二 因为$\theta$是第四象限角，且$\sin(\theta+\frac{\pi}{4})=\frac{3}{5}$，所以$\theta+\frac{\pi}{4}$为第一象限角，所以$\cos(\theta+\frac{\pi}{4})=\frac{4}{5}$，所以$\tan(\theta-\frac{\pi}{4})=\frac{\sin(\theta-\frac{\pi}{4})}{\cos(\theta-\frac{\pi}{4})}=\frac{-\cos[\frac{\pi}{2}+(\theta-\frac{\pi}{4})]}{\sin[\frac{\pi}{2}+(\theta-\frac{\pi}{4})]}=-\frac{\cos(\theta+\frac{\pi}{4})}{\sin(\theta+\frac{\pi}{4})}=-\frac{4}{3}$。

训练3 [2024安徽省皖江名校联考]已知在平面直角坐标系中,点$M(2,4)$在角$\alpha$终边上,则$\frac{\sin^{3}(\pi - \alpha)+\cos^{3}(-\alpha)}{\sin^{3}\alpha - 2\cos^{3}\alpha}=$ ( )
A.$\frac{2}{3}$
B.$\frac{3}{2}$
C.$-\frac{3}{5}$
D.$-\frac{5}{3}$

答案： B 由题意可得$\tan\alpha=2$，所以原式$=\frac{\sin^{3}\alpha+\cos^{3}\alpha}{\sin^{3}\alpha-2\cos^{3}\alpha}=\frac{\tan^{3}\alpha + 1}{\tan^{3}\alpha-2}=\frac{8 + 1}{8-2}=\frac{3}{2}$。故选B。