《2025年高考帮数学》

训练3 (1)[2024江苏省南通市学情检测]已知$\sin(\alpha + \frac{\pi}{6})=\frac{\sqrt{6}}{3}$,则$\sin(\frac{\pi}{6}-2\alpha)=$　　(　　)
A.$-\frac{2\sqrt{2}}{3}$　B.$\frac{2\sqrt{2}}{3}$　　C.$-\frac{1}{3}$　　D.$\frac{1}{3}$
(2)[2024辽宁省辽东南协作体联考]已知$\frac{\pi}{4}＜\alpha＜\frac{3\pi}{4},0＜\beta＜\frac{\pi}{4},\cos(\frac{\pi}{4}-\alpha)=\frac{3}{5},\sin(\frac{3\pi}{4}+\beta)=\frac{5}{13}$,则$\sin(\alpha + \beta)$的值为________。

答案：
(1)C 设$\alpha + \frac{\pi}{6} = t$，则$\alpha = t - \frac{\pi}{6},\sin t = \frac{\sqrt{6}}{3}$，$\therefore\sin(\frac{\pi}{6} - 2\alpha) = \sin[\frac{\pi}{6} - 2(t - \frac{\pi}{6})] = \sin(\frac{\pi}{2} - 2t) = \cos2t = 1 - 2\sin^{2}t = 1 - 2\times(\frac{\sqrt{6}}{3})^{2} = -\frac{1}{3}$，故选C.
(2)$\frac{56}{65}$ $\because\frac{\pi}{4} ＜ \alpha ＜ \frac{3\pi}{4},0 ＜ \beta ＜ \frac{\pi}{4}$，$\therefore -\frac{\pi}{2} ＜ \frac{\pi}{4} - \alpha ＜ 0,\frac{3\pi}{4} ＜ \frac{3\pi}{4} + \beta ＜ \pi$，$\therefore\sin(\frac{\pi}{4} - \alpha) = -\sqrt{1 - \cos^{2}(\frac{\pi}{4} - \alpha)} = -\frac{4}{5},\cos(\frac{3\pi}{4} + \beta) = -\sqrt{1 - \sin^{2}(\frac{3\pi}{4} + \beta)} = -\frac{12}{13}$，$\therefore\sin(\alpha + \beta) = -\cos[\frac{\pi}{2} + (\alpha + \beta)] = -\cos[(\frac{3\pi}{4} + \beta) - (\frac{\pi}{4} - \alpha)] = -\cos(\frac{3\pi}{4} + \beta)\cos(\frac{\pi}{4} - \alpha) - \sin(\frac{3\pi}{4} + \beta)\sin(\frac{\pi}{4} - \alpha) = \frac{12}{13} \times \frac{3}{5} - \frac{5}{13} \times (-\frac{4}{5}) = \frac{56}{65}$.

例1　(1)[2021全国卷甲]若α∈(0,$\frac{\pi}{2}$),tan2α=$\frac{\cos\alpha}{2 - \sin\alpha}$,则tanα=　　　　　　　　　　(　　　)
A.$\frac{\sqrt{15}}{15}$　　　　　　　　B.$\frac{\sqrt{5}}{5}$
C.$\frac{\sqrt{5}}{3}$　　　　　　　　D.$\frac{\sqrt{15}}{3}$
(2)化简:$\frac{2\cos^{2}\alpha - 1}{2\tan(\frac{\pi}{4} - \alpha)\sin^{2}(\frac{\pi}{4} + \alpha)}=$________.

答案：
(1)A　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$\frac{2\sin\alpha\cos\alpha}{1 - 2\sin^{2}\alpha}=\frac{\cos\alpha}{2 - \sin\alpha}$，由$\alpha\in(0,\frac{\pi}{2})$得$\cos\alpha\neq0$，解得$\sin\alpha=\frac{1}{4}$，$\cos\alpha=\frac{\sqrt{15}}{4}$，$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\sqrt{15}}{15}$. 故选A.
(2)1　原式$=\frac{\cos2\alpha}{2\tan(\frac{\pi}{4}-\alpha)\cos^{2}(\frac{\pi}{4}-\alpha)}=\frac{\cos2\alpha}{2\sin(\frac{\pi}{4}-\alpha)\cos(\frac{\pi}{4}-\alpha)}=\frac{\cos2\alpha}{\sin(\frac{\pi}{2}-2\alpha)}=\frac{\cos2\alpha}{\cos2\alpha}=1$.

训练1　[2021新高考卷I]若tanθ=−2,则$\frac{\sin\theta(1 + \sin2\theta)}{\sin\theta + \cos\theta}$=　　　　　　　　　　(　　　)
A.−$\frac{6}{5}$　
　B.−$\frac{2}{5}$　
　C.$\frac{2}{5}$　　
　D.$\frac{6}{5}$

答案： 解法一　因为$\tan\theta=-2$，所以$\frac{\sin\theta(1 + \sin2\theta)}{\sin\theta+\cos\theta}=\frac{\sin\theta(\sin\theta+\cos\theta)^{2}}{\sin\theta+\cos\theta}=\sin\theta(\sin\theta+\cos\theta)=\frac{\sin^{2}\theta+\sin\theta\cos\theta}{\sin^{2}\theta+\cos^{2}\theta}=\frac{\tan^{2}\theta+\tan\theta}{\tan^{2}\theta + 1}=\frac{4 - 2}{4 + 1}=\frac{2}{5}$. 故选C.
解法二　因为$\tan\theta=-2$，所以角$\theta$的终边在第二或第四象限，所以$\begin{cases}\sin\theta=\frac{2}{\sqrt{5}}\\\cos\theta=-\frac{1}{\sqrt{5}}\end{cases}$或$\begin{cases}\sin\theta=-\frac{2}{\sqrt{5}}\\\cos\theta=\frac{1}{\sqrt{5}}\end{cases}$，所以$\frac{\sin\theta(1 + \sin2\theta)}{\sin\theta+\cos\theta}=\frac{\sin\theta(\sin\theta+\cos\theta)^{2}}{\sin\theta+\cos\theta}=\sin\theta(\sin\theta+\cos\theta)=\sin^{2}\theta+\sin\theta\cos\theta=\frac{4}{5}-\frac{2}{5}=\frac{2}{5}$. 故选C.

例2　(1)sin50°(1 + √3tan10°)=________.
　(2)sin10°.sin30°.sin50°.sin70°=________.

答案：
(1)1　$\sin50^{\circ}(1+\sqrt{3}\tan10^{\circ})=\sin50^{\circ}(1 + \tan60^{\circ}\tan10^{\circ})=\sin50^{\circ}\times\frac{\cos60^{\circ}\cos10^{\circ}+\sin60^{\circ}\sin10^{\circ}}{\cos60^{\circ}\cos10^{\circ}}=\sin50^{\circ}\times\frac{\cos(60^{\circ}-10^{\circ})}{\cos60^{\circ}\cos10^{\circ}}=\frac{2\sin50^{\circ}\cos50^{\circ}}{\cos10^{\circ}}=\frac{\sin100^{\circ}}{\cos10^{\circ}}=\frac{\cos10^{\circ}}{\cos10^{\circ}}=1$.
(2)$\frac{1}{16}$　原式$=\frac{1}{2}\cos20^{\circ}\cdot\cos40^{\circ}\cdot\cos80^{\circ}=\frac{\sin20^{\circ}\cdot\cos20^{\circ}\cdot\cos40^{\circ}\cdot\cos80^{\circ}}{2\sin20^{\circ}}=\frac{\sin160^{\circ}}{16\sin20^{\circ}}=\frac{1}{16}$.

例3　(1)[2022浙江高考]若3sinα−sinβ=$\sqrt{10}$,α + β=$\frac{\pi}{2}$,则sinα=________,cos2β=________.
　(2)[江苏高考]已知$\frac{\tan\alpha}{\tan(\alpha + \frac{\pi}{4})}=-\frac{2}{3}$,则sin(2α + $\frac{\pi}{4}$)的值是________.

答案：
(1)$\frac{3\sqrt{10}}{10}$　$\frac{4}{5}$　因为$\alpha+\beta=\frac{\pi}{2}$，所以$\beta=\frac{\pi}{2}-\alpha$，所以$3\sin\alpha-\sin\beta=3\sin\alpha-\sin(\frac{\pi}{2}-\alpha)=3\sin\alpha-\cos\alpha=\sqrt{10}\sin(\alpha-\varphi)=\sqrt{10}$，其中$\sin\varphi=\frac{\sqrt{10}}{10}$，$\cos\varphi=\frac{3\sqrt{10}}{10}$，所以$\alpha-\varphi=\frac{\pi}{2}+2k\pi,k\in\mathbf{Z}$，所以$\alpha=\frac{\pi}{2}+\varphi+2k\pi,k\in\mathbf{Z}$，所以$\sin\alpha=\sin(\frac{\pi}{2}+\varphi+2k\pi)=\cos\varphi=\frac{3\sqrt{10}}{10},k\in\mathbf{Z}$. 因为$\sin\beta=3\sin\alpha-\sqrt{10}=-\frac{\sqrt{10}}{10}$，所以$\cos2\beta=1 - 2\sin^{2}\beta=1-\frac{1}{5}=\frac{4}{5}$.
(2)$\frac{\sqrt{2}}{10}$　解法一　$\frac{\tan\alpha}{\tan(\alpha+\frac{\pi}{4})}=\frac{\tan\alpha}{\frac{\tan\alpha + 1}{1 - \tan\alpha}}=\frac{\tan\alpha(1 - \tan\alpha)}{\tan\alpha + 1}=-\frac{2}{3}$，解得$\tan\alpha=2$或$\tan\alpha=-\frac{1}{3}$. 当$\tan\alpha=2$时，$\sin2\alpha=\frac{2\sin\alpha\cos\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}=\frac{2\tan\alpha}{\tan^{2}\alpha + 1}=\frac{4}{5}$，$\cos2\alpha=\frac{\cos^{2}\alpha-\sin^{2}\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}=\frac{1 - \tan^{2}\alpha}{\tan^{2}\alpha + 1}=-\frac{3}{5}$，此时$\sin2\alpha+\cos2\alpha=\frac{1}{5}$. 同理当$\tan\alpha=-\frac{1}{3}$时，$\sin2\alpha=-\frac{3}{5}$，$\cos2\alpha=\frac{4}{5}$，此时$\sin2\alpha+\cos2\alpha=\frac{1}{5}$，所以$\sin(2\alpha+\frac{\pi}{4})=\frac{\sqrt{2}}{2}(\sin2\alpha+\cos2\alpha)=\frac{\sqrt{2}}{10}$.
解法二　$\frac{\tan\alpha}{\tan(\alpha+\frac{\pi}{4})}=\frac{\sin\alpha\cos(\alpha+\frac{\pi}{4})}{\cos\alpha\sin(\alpha+\frac{\pi}{4})}=-\frac{2}{3}$，则$\sin\alpha\cos(\alpha+\frac{\pi}{4})=-\frac{2}{3}\cos\alpha\sin(\alpha+\frac{\pi}{4})$，又$\frac{\sqrt{2}}{2}=\sin[(\alpha+\frac{\pi}{4})-\alpha]=\sin(\alpha+\frac{\pi}{4})\cos\alpha-\cos(\alpha+\frac{\pi}{4})\sin\alpha=\frac{5}{3}\sin(\alpha+\frac{\pi}{4})\cos\alpha$，则$\sin(\alpha+\frac{\pi}{4})\cos\alpha=\frac{3\sqrt{2}}{10}$，则$\sin(2\alpha+\frac{\pi}{4})=\sin[(\alpha+\frac{\pi}{4})+\alpha]=\sin(\alpha+\frac{\pi}{4})\cos\alpha+\cos(\alpha+\frac{\pi}{4})\sin\alpha=\frac{1}{3}\sin(\alpha+\frac{\pi}{4})\cdot\cos\alpha=\frac{1}{3}\times\frac{3\sqrt{2}}{10}=\frac{\sqrt{2}}{10}$.