《2025年高考帮数学》

3.[全国卷Ⅱ]若$\sin x=-\frac{2}{3}$,则$\cos2x=$________.

答案： 3.$\frac{1}{9}$ $\cos2x = 1 - 2\sin^{2}x = 1 - 2\times(-\frac{2}{3})^{2} = \frac{1}{9}$.

4.[易错题]$\frac{1 + \tan15^{\circ}}{1 - \tan15^{\circ}}=$________.

答案： 4.$\sqrt{3}$ $\frac{1 + \tan15^{\circ}}{1 - \tan15^{\circ}} = \frac{\tan45^{\circ} + \tan15^{\circ}}{1 - \tan45^{\circ}\tan15^{\circ}} = \tan(45^{\circ} + 15^{\circ}) = \tan60^{\circ} = \sqrt{3}$.

5.若$\sin x-\sqrt{3}\cos x=2\sin(x-\varphi),\varphi＞0$,则$\varphi$的最小值为________.

答案： 5.$\frac{\pi}{3}$ 因为$\sin x - \sqrt{3}\cos x = 2(\frac{1}{2}\sin x - \frac{\sqrt{3}}{2}\cos x) = 2(\sin x\cos\varphi - \cos x\sin\varphi)$，所以$\cos\varphi = \frac{1}{2},\sin\varphi = \frac{\sqrt{3}}{2}$. 因为$\varphi＞0$，所以$\varphi$的最小值为$\frac{\pi}{3}$.

例1　(1)[2023新高考卷Ⅰ]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.$-\frac{7}{9}$
(2)[全国卷Ⅲ]已知$2\tan\theta - \tan(\theta + \frac{\pi}{4}) = 7$,则$\tan\theta =$　　(　　)
A.$-2$　　　B.$-1$　　　C.1　　　　D.2

答案：
(1)B 依题意，得$\begin{cases}\sin\alpha\cos\beta - \cos\alpha\sin\beta = \frac{1}{3},\\\cos\alpha\sin\beta = \frac{1}{6},\end{cases}$所以$\sin\alpha\cos\beta = \frac{1}{2}$，所以$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$，所以$\cos(2\alpha + 2\beta) = 1 - 2\sin^{2}(\alpha + \beta) = 1 - 2\times(\frac{2}{3})^{2} = \frac{1}{9}$，故选B.
(2)D 由已知得$2\tan\theta - \frac{\tan\theta + 1}{1 - \tan\theta} = 7$，得$\tan\theta = 2$.

训练1　(1)[全国卷Ⅰ]已知$\alpha\in(0,\pi)$,且$3\cos2\alpha - 8\cos\alpha = 5$,则$\sin\alpha =$　　(　　)
A.$\frac{\sqrt{5}}{3}$　　　B.$\frac{2}{3}$　　　C.$\frac{1}{3}$　　　D.$\frac{\sqrt{5}}{9}$
(2)[2024广西玉林市联考]已知$\cos(\alpha + \beta)=\frac{1}{3},\cos\alpha\cos\beta=\frac{1}{2}$,则$\cos(2\alpha - 2\beta)=$　　(　　)
A.$-\frac{7}{9}$　　B.$-\frac{1}{9}$　　C.$\frac{1}{9}$　　　D.$\frac{7}{9}$

答案：
(1)A $\because 3\cos2\alpha - 8\cos\alpha = 5$，$\therefore 3(2\cos^{2}\alpha - 1) - 8\cos\alpha = 5$，$\therefore 6\cos^{2}\alpha - 8\cos\alpha - 8 = 0$，$\therefore 3\cos^{2}\alpha - 4\cos\alpha - 4 = 0$，解得$\cos\alpha = 2$(舍去)或$\cos\alpha = -\frac{2}{3}$.$\because\alpha\in(0,\pi)$，$\therefore\sin\alpha = \sqrt{1 - \cos^{2}\alpha} = \frac{\sqrt{5}}{3}$. 故选A.
(2)B 由$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$，即$\frac{1}{3} = \frac{1}{2} - \sin\alpha\sin\beta$，可得$\sin\alpha\sin\beta = \frac{1}{6}$，则$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$，所以$\cos(2\alpha - 2\beta) = 2\cos^{2}(\alpha - \beta) - 1 = 2\times(\frac{2}{3})^{2} - 1 = -\frac{1}{9}$. 故选B.

例2　(1)[2023新高考卷Ⅱ]已知$\alpha$为锐角,$\cos\alpha=\frac{1 + \sqrt{5}}{4}$,则$\sin\frac{\alpha}{2}=$　　(　　)
A.$\frac{3 - \sqrt{5}}{8}$　　　　　　　B.$\frac{-1 + \sqrt{5}}{8}$
C.$\frac{3 - \sqrt{5}}{4}$　　　　　　　D.$\frac{-1 + \sqrt{5}}{4}$
(2)[2021全国卷乙]$\cos^{2}\frac{\pi}{12}-\cos^{2}\frac{5\pi}{12}=$　　(　　)
A.$\frac{1}{2}$　　　B.$\frac{\sqrt{3}}{3}$　　　C.$\frac{\sqrt{2}}{2}$　　　D.$\frac{\sqrt{3}}{2}$
(3)[2022新高考卷Ⅱ]若$\sin(\alpha + \beta) + \cos(\alpha + \beta)=2\sqrt{2}\cos(\alpha + \frac{\pi}{4})\sin\beta$,则　　(　　)
A.$\tan(\alpha - \beta)=1$　　　B.$\tan(\alpha + \beta)=1$
C.$\tan(\alpha - \beta)= - 1$　　D.$\tan(\alpha + \beta)= - 1$

答案：
(1)D $\cos\alpha = \frac{1 + \sqrt{5}}{4} = 1 - 2\sin^{2}\frac{\alpha}{2}$，得$\sin^{2}\frac{\alpha}{2} = \frac{3 - \sqrt{5}}{8} = \frac{6 - 2\sqrt{5}}{16} = (\frac{\sqrt{5} - 1}{4})^{2}$，又$\alpha$为锐角，所以$\sin\frac{\alpha}{2} ＞ 0$，所以$\sin\frac{\alpha}{2} = \frac{-1 + \sqrt{5}}{4}$，故选D.
(2)D 解法一 原式$=\frac{1 + \cos\frac{\pi}{6}}{2} - \frac{1 + \cos\frac{5\pi}{6}}{2} = \frac{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})}{2} = \frac{\sqrt{3}}{2}$.
解法二 因为$\cos\frac{5\pi}{12} = \sin(\frac{\pi}{2} - \frac{5\pi}{12}) = \sin\frac{\pi}{12}$，所以$\cos^{2}\frac{\pi}{12} - \cos^{2}\frac{5\pi}{12} = \cos^{2}\frac{\pi}{12} - \sin^{2}\frac{\pi}{12} = \cos(2\times\frac{\pi}{12}) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$. 故选D.
(3)C $\sin(\alpha + \beta) + \cos(\alpha + \beta) = \sqrt{2}\sin(\alpha + \beta + \frac{\pi}{4}) = 2\sqrt{2}\sin\beta\cdot\cos(\alpha + \frac{\pi}{4})$，所以$\sin(\alpha + \frac{\pi}{4})\cos\beta + \sin\beta\cos(\alpha + \frac{\pi}{4}) = 2\sin\beta\cos(\alpha + \frac{\pi}{4})$，整理得$\sin(\alpha + \frac{\pi}{4})\cos\beta - \sin\beta\cos(\alpha + \frac{\pi}{4}) = 0$，即$\sin(\alpha + \frac{\pi}{4} - \beta) = 0$，所以$\alpha - \beta + \frac{\pi}{4} = k\pi,k\in\mathbf{Z}$，所以$\tan(\alpha - \beta) = \tan(k\pi - \frac{\pi}{4}) = -1$.

训练2　(1)在$\triangle ABC$中,$C = 120^{\circ},\tan A+\tan B=\frac{2\sqrt{3}}{3}$,则$\tan A\tan B$的值为　　(　　)
A.$\frac{1}{4}$　　　B.$\frac{1}{3}$　　　C.$\frac{1}{2}$　　　D.$\frac{5}{3}$
(2)[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：
(1)B 解法一 由题意得$\tan(A + B) = -\tan C = -\tan120^{\circ} = \sqrt{3}$，所以$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} = \sqrt{3}$，即$\frac{\frac{2\sqrt{3}}{3}}{1 - \tan A\tan B} = \sqrt{3}$，解得$\tan A\tan B = \frac{1}{3}$，故选B.
解法二 由已知，可取$A = B = 30^{\circ}$，则$\tan A\tan B = \frac{\sqrt{3}}{3} \times \frac{\sqrt{3}}{3} = \frac{1}{3}$，故选B.
(2)$1$ $-\sqrt{2}$ 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$f(\frac{\pi}{12}) = 2\sin(\frac{\pi}{12} - \frac{\pi}{3}) = -\sqrt{2}$.

例3　(1)[2024山东省部分学校联考]已知$\sin(x + \frac{\pi}{12})=-\frac{1}{4}$,则$\cos(\frac{5\pi}{6}-2x)=$　　(　　)
A.$\frac{7}{8}$　　　B.$\frac{1}{8}$　　　C.$-\frac{7}{8}$　　D.$-\frac{1}{8}$
(2)若$\tan(\alpha + 2\beta)=2,\tan\beta = - 3$,则$\tan(\alpha + \beta)=$________,$\tan\alpha =$________。

答案：
(1)C 因为$\sin(x + \frac{\pi}{12}) = -\frac{1}{4}$，所以$\cos(\frac{5\pi}{6} - 2x) = \cos(\pi - \frac{\pi}{6} - 2x) = -\cos(\frac{\pi}{6} + 2x) = -[1 - 2\sin^{2}(x + \frac{\pi}{12})] = -[1 - 2\times(-\frac{1}{4})^{2}] = -\frac{7}{8}$. 故选C.
(2)$-1$ $\frac{1}{2}$ 因为$\tan(\alpha + 2\beta) = 2,\tan\beta = -3$，所以$\tan(\alpha + \beta) = \tan((\alpha + 2\beta) - \beta) = \frac{\tan(\alpha + 2\beta) - \tan\beta}{1 + \tan(\alpha + 2\beta)\tan\beta} = \frac{2 - (-3)}{1 + 2\times(-3)} = -1$，$\tan\alpha = \tan((\alpha + \beta) - \beta) = \frac{-1 - (-3)}{1 + (-1)\times(-3)} = \frac{1}{2}$.