《2025年高考帮数学》

4.用“五点法”画$y = 2\sin(2x+\frac{\pi}{3})$在一个周期内的简图时，所描的五个点分别是$(-\frac{\pi}{6},0),(\frac{\pi}{12},2),(\frac{\pi}{3},0),(\frac{7\pi}{12},-2)$，________.

答案： 令$2x+\frac{\pi}{3}=2\pi$，则解得$x = \frac{5\pi}{6}$，故最后一个关键点是$(\frac{5\pi}{6},0)$.

例1
(1)[2021全国卷乙]把函数$y = f(x)$图象上所有点的横坐标缩短到原来的$\frac{1}{2}$倍,纵坐标不变,再把所得曲线向右平移$\frac{\pi}{3}$个单位长度,得到函数$y = \sin(x - \frac{\pi}{4})$的图象,则$f(x) =$ （ ）
A. $\sin(\frac{x}{2} - \frac{7\pi}{12})$ B. $\sin(\frac{x}{2} + \frac{\pi}{12})$
C. $\sin(2x - \frac{7\pi}{12})$ D. $\sin(2x + \frac{\pi}{12})$
(2)[2023全国卷甲]函数$y = f(x)$的图象由函数$y = \cos(2x + \frac{\pi}{6})$的图象向左平移$\frac{\pi}{6}$个单位长度得到,则$y = f(x)$的图象与直线$y = \frac{1}{2}x - \frac{1}{2}$的交点个数为 （ ）
A. 1 B. 2 C. 3 D. 4

答案：

(1)B　依题意,将$y = \sin(x - \frac{\pi}{4})$的图象向左平移$\frac{\pi}{3}$个单位长度，再将所得曲线上所有点的横坐标扩大到原来的2倍，得到$f(x)$的图象,所以$y = \sin(x - \frac{\pi}{4})$的图象$\xrightarrow{向左平移\frac{\pi}{3}个单位长度}y = \sin(x + \frac{\pi}{12})$的图象$\xrightarrow{所有点的横坐标扩大到原来的2倍}f(x) = \sin(\frac{x}{2} + \frac{\pi}{12})$的图象。
(2)C　把函数$y = \cos(2x + \frac{\pi}{6})$的图象向左平移$\frac{\pi}{6}$个单位长度后得到函数$f(x) = \cos[2(x + \frac{\pi}{6}) + \frac{\pi}{6}] = \cos(2x + \frac{\pi}{2}) = - \sin 2x$的图象.作出函数$f(x)$的部分图象和直线$y = \frac{1}{2}x - \frac{1}{2}$,如图所示.观察图象知，共有3个交点.故选C.
　

训练1
(1)[2023江西南昌联考]为了得到函数$y = 2\cos(2x - \frac{2\pi}{3})$的图象,只需将函数$y = 2\sin x$ （ ）
A. 图象上所有点的横坐标缩短到原来的$\frac{1}{2}$,纵坐标不变,再向右平移$\frac{\pi}{12}$个单位长度
B. 图象上所有点的横坐标伸长到原来的2倍,纵坐标不变,再向左平移$\frac{\pi}{12}$个单位长度
C. 图象向右平移$\frac{\pi}{3}$个单位长度,再将所有点的横坐标伸长到原来的2倍,纵坐标不变
D. 图象向左平移$\frac{\pi}{6}$个单位长度,再将所有点的横坐标缩短到原来的$\frac{1}{2}$,纵坐标不变
(2)[2023郑州二模]将函数$y = \sin(2x + \frac{\pi}{3})$图象上的点$A(m,n)$向右平移$\frac{1}{4}$个周期(最小正周期)得到点$A'$,若$A'$位于函数$y = \cos 2x$的图象上,则$m$的值可以是 （ ）
A. $\frac{\pi}{12}$ B. $\frac{\pi}{6}$ C. $\frac{\pi}{3}$ D. $\frac{5\pi}{12}$

答案：
(1)A将函数$y = 2\sin x = 2\cos(x - \frac{\pi}{2})$的图象上所有点的横坐标缩短到原来的$\frac{1}{2}$,纵坐标不变,得到函数$y = 2\cos(2x - \frac{\pi}{2})$的图象,再将所得图象向右平移$\frac{\pi}{12}$个单位长度,得到函数$y = 2\cos[2(x - \frac{\pi}{12}) - \frac{\pi}{2}] = 2\cos(2x - \frac{2\pi}{3})$的图象，故A正确，B错误。将函数$y = 2\sin x = 2\cos(x - \frac{\pi}{2})$的图象上所有的点向右平移$\frac{\pi}{6}$个单位长度得到函数$y = 2\cos(x - \frac{\pi}{6} - \frac{\pi}{2}) = 2\cos(x - \frac{2\pi}{3})$的图象，再将所得图象上所有点的横坐标缩短到原来的$\frac{1}{2}$，纵坐标不变，得到函数$y = 2\cos(2x - \frac{2\pi}{3})$的图象，故C,D错误.故选A.
(2)D　函数$y = \sin(2x + \frac{\pi}{3})$的最小正周期$T = \frac{2\pi}{2} = \pi$,设$A'(m',n')$,由题知将点$A(m,n)$向右平移$\frac{\pi}{4}$个单位长度得到点$A'(m',n')$,则$\begin{cases}m' = m + \frac{\pi}{4}\\n' = n\end{cases}$,因为$A'$位于函数$y = \cos 2x$的图象上,所以$n' = \cos 2m'$,所以$n = \cos 2(m + \frac{\pi}{4})$,又$n = \sin(2m + \frac{\pi}{3})$,所以$\sin(2m + \frac{\pi}{3}) = \cos 2(m + \frac{\pi}{4}) = \cos(2m + \frac{\pi}{2}) = - \sin 2m$,即$\sin 2m\cos\frac{\pi}{3} + \cos 2m\sin\frac{\pi}{3} = - \sin 2m$,化简得$\frac{3}{2}\sin 2m = - \frac{\sqrt{3}}{2}\cos 2m$,所以$\tan 2m = - \frac{\sqrt{3}}{3}$,故$2m = - \frac{\pi}{6} + k\pi(k \in \mathbf{Z})$,$m = - \frac{\pi}{12} + \frac{k\pi}{2}(k \in \mathbf{Z})$,当$k = 1$时,$m = \frac{5\pi}{12}$,故选D.

例2
(1)[2023新高考卷Ⅱ]已知函数$f(x) = \sin(\omega x + \varphi)$,如图,$A,B$是直线$y = \frac{1}{2}$与曲线$y = f(x)$的两个交点,若$|AB| = \frac{\pi}{6}$,则$f(\pi) =$________.

(2)[2021全国卷甲]已知函数$f(x) = 2\cos(\omega x + \varphi)$的部分图象如图所示,则满足条件$(f(x) - f(-\frac{7\pi}{4}))\cdot (f(x) - f(\frac{4\pi}{3})) ＞ 0$的最小正整数$x$为________.

答案：
(1)$-\frac{\sqrt{3}}{2}$　对比正弦函数$y = \sin x$的图象，不妨设点$(\frac{2\pi}{3},0)$为“五点作图法”中的第五个点，所以$\frac{2\pi}{3}\omega + \varphi = 2\pi$ ①.设$A,B$的横坐标分别为$x_A,x_B$,由题知$|AB| = x_B - x_A = \frac{\pi}{6}$,$\begin{cases}\omega x_A + \varphi = \frac{\pi}{6}\\\omega x_B + \varphi = \frac{5\pi}{6}\end{cases}$两式相减，得$\omega(x_B - x_A) = \frac{4\pi}{6}$,即$\frac{\pi}{6}\omega = \frac{4\pi}{6}$,解得$\omega = 4$.代入①,得$\varphi = - \frac{2\pi}{3}$,所以$f(\pi) = \sin(4\pi - \frac{2\pi}{3}) = - \sin\frac{2\pi}{3} = - \frac{\sqrt{3}}{2}$.
(2)2　由题图可知，$\frac{3}{4}T = \frac{13\pi}{12} - \frac{\pi}{3} = \frac{3\pi}{4}$（$T$为$f(x)$的最小正周期），解得$T = \pi$,所以$\omega = \pm 2$.
当$\omega = - 2$时,$f(x) = 2\cos(- 2x + \varphi)$,因为$\frac{\pi}{3} + \frac{T}{4} = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$,所以函数$f(x)$的图象经过点$(\frac{7\pi}{12}, - 1)$,所以$2\cos(- 2\times\frac{7\pi}{12} + \varphi) = - 1$,所以$- 2\times\frac{7\pi}{12} + \varphi = \pi + 2k\pi,k \in \mathbf{Z}$,得$\varphi = \frac{13\pi}{6} + 2k\pi,k \in \mathbf{Z}$,令$k = - 1$,则$\varphi = \frac{\pi}{6}$,所以$f(x) = 2\cos(- 2x + \frac{\pi}{6}) = 2\cos(2x - \frac{\pi}{6})$.
当$\omega = 2$时,$f(x) = 2\cos(2x + \varphi)$.点$(\frac{\pi}{3},0)$可看作“五点作图法”中的第二个点，则$2\times\frac{\pi}{3} + \varphi = \frac{\pi}{2}$,得$\varphi = - \frac{\pi}{6}$,所以$f(x) = 2\cos(2x - \frac{\pi}{6})$.
综上$f(-\frac{7\pi}{4}) = 2\cos[2\times(-\frac{7\pi}{4}) - \frac{\pi}{6}] = 2\cos(-\frac{11\pi}{3}) = 2\cos\frac{\pi}{3} = 1$,$f(\frac{4\pi}{3}) = 2\cos(2\times\frac{4\pi}{3} - \frac{\pi}{6}) = 2\cos\frac{5\pi}{2} = 0$,所以$(f(x) - f(-\frac{7\pi}{4}))\cdot(f(x) - f(\frac{4\pi}{3})) ＞ 0$,即$(f(x) - 1)f(x) ＞ 0$,可得$f(x) ＞ 1$或$f(x) ＜ 0$,所以$\cos(2x - \frac{\pi}{6}) ＞ \frac{1}{2}$或$\cos(2x - \frac{\pi}{6}) ＜ 0$.当$\cos(2x - \frac{\pi}{6}) ＜ 0$时，$\frac{\pi}{2} + 2k\pi ＜ 2x - \frac{\pi}{6} ＜ \frac{3\pi}{2} + 2k\pi,k \in \mathbf{Z}$,解得$\frac{\pi}{3} + k\pi ＜ x ＜ \frac{5\pi}{6} + k\pi,k \in \mathbf{Z}$,此时最小正整数$x$为2.当$\cos(2x - \frac{\pi}{6}) ＞ \frac{1}{2}$时，$-\frac{\pi}{3} + 2k\pi ＜ 2x - \frac{\pi}{6} ＜ \frac{\pi}{3} + 2k\pi,k \in \mathbf{Z}$,解得$-\frac{\pi}{12} + k\pi ＜ x ＜ \frac{\pi}{4} + k\pi,k \in \mathbf{Z}$,此时最小正整数$x$为3.综上,最小正整数$x$为2.