三角函数单调区间

本小妞迷上赌 提交于 2019-12-10 12:26:45

前言

典例剖析

例1已知函数\(f(x)=\cfrac{\sqrt{3}}{3}[cos(2x+\cfrac{\pi}{6})+4sinxcosx]+1\)\(x\in R\)

(1).求\(f(x)\)的单调区间;

分析:化简为正弦型或者余弦型,

\(f(x)=\cfrac{\sqrt{3}}{3}[cos(2x+\cfrac{\pi}{6})+4sinxcosx]+1\)

\(=\cfrac{\sqrt{3}}{3}[(cos2x\cdot cos\cfrac{\pi}{6}-sin2x\cdot sin\cfrac{\pi}{6})+2sin2x]+1\)

\(=\cfrac{\sqrt{3}}{3}(\cfrac{\sqrt{3}}{2}cos2x-\cfrac{1}{2}sin2x+2sin2x)+1\)

\(=\cfrac{\sqrt{3}}{3}(\cfrac{\sqrt{3}}{2}cos2x+\cfrac{3}{2}sin2x)+1\)

\(=\cfrac{\sqrt{3}}{2}sin2x+\cfrac{1}{2}cos2x+1\)

即函数\(f(x)=sin(2x+\cfrac{\pi}{6})+1\),[化简到此结束]

\(2k\pi-\cfrac{\pi}{2}\leqslant 2x+\cfrac{\pi}{6}\leqslant 2k\pi+\cfrac{\pi}{2}\)\(k\in Z\)

解得单调递增区间为\([k\pi-\cfrac{\pi}{3},k\pi+\cfrac{\pi}{6}]\)\((k\in Z)\)

\(2k\pi+\cfrac{\pi}{2}\leqslant 2x+\cfrac{\pi}{6}\leqslant 2k\pi+\cfrac{3\pi}{2}\)\(k\in Z\)

解得单调递减区间为\([k\pi+\cfrac{\pi}{6},k\pi+\cfrac{2\pi}{3}]\)\((k\in Z)\)

注意:求\(f(x)=Asin(\omega x+\phi)+k\)型函数的单调区间,为减少错误,一般需要将\(\omega\)转化为\(\omega>0\)

(2).令\(g(x)=af(x)+b\),若函数\(g(x)\)在区间\([-\cfrac{\pi}{6},\cfrac{\pi}{4}]\)上的值域为\([-1,1]\),求\(a+b\)的值;

提示:当\(x\in [-\cfrac{\pi}{6},\cfrac{\pi}{4}]\)时,\(sin(2x+\cfrac{\pi}{6})\in [-\cfrac{1}{2},1]\),则\(f(x)\in [\cfrac{1}{2},2]\)

以下针对一次项系数\(a\)分类讨论如下:

①当\(a>0\)时,由\(2a+b=1\)\(\cfrac{1}{2}a+b=-1\),解得\(a=\cfrac{4}{3}\)\(b=-\cfrac{5}{3}\),故\(a+b=-\cfrac{1}{3}\)

②当\(a<0\)时,由\(2a+b=-1\)\(\cfrac{1}{2}a+b=1\),解得\(a=-\cfrac{4}{3}\)\(b=\cfrac{5}{3}\),故\(a+b=\cfrac{1}{3}\)

\(a+b\)的值为\(\cfrac{1}{3}\)\(-\cfrac{1}{3}\)

例2【2016\(\cdot\)天津高考】已知函数\(f(x)=4tanx\cdot sin(\cfrac{\pi}{2}-x)\cdot cos(x-\cfrac{\pi}{3})-\sqrt{3}\)

(1).求函数的定义域;

分析:由函数解析式可知,需要让\(tanx\)有意义,故定义域为\(\{x\mid x\neq k\pi+\cfrac{\pi}{2},k\in Z\}\)

(2).试讨论\(f(x)\)在区间\([-\cfrac{\pi}{4},\cfrac{\pi}{4}]\)上的单调性。

分析:先将所给函数化简为正弦型或者余弦型,

\(f(x)=4tan\cdot cosx(cosx\cdot \cfrac{1}{2}+sinx\cdot \cfrac{\sqrt{3}}{2})-\sqrt{3}\)

\(=4sinx(cosx\cdot \cfrac{1}{2}+sinx\cdot \cfrac{\sqrt{3}}{2})-\sqrt{3}\)

\(=2sinxcosx+2\sqrt{3}sin^2x-\sqrt{3}\)

\(=sin2x+\sqrt{3}(1-cos2x)-\sqrt{3}\)

\(=sin2x-\sqrt{3}cos2x\)

\(=2sin(2x-\cfrac{\pi}{3})\)

法1:先求解函数在\(x\in R\)上的单调区间,

\(2k\pi-\cfrac{\pi}{2}\leq 2x-\cfrac{\pi}{3}\leq 2k\pi+\cfrac{\pi}{2}(k\in Z)\)

得到单调递增区间为\([k\pi-\cfrac{\pi}{12},k\pi+\cfrac{5\pi}{12}](k\in Z)\)

又因为\(x\in [-\cfrac{\pi}{4},\cfrac{\pi}{4}]\)

[说明:求得的单调递增区间和给定区间求交集,即为所求的单调递增区间;剩余的即为单调递减区间]

然后给\(k\)赋值,令\(k=0\)

得到函数在区间\([-\cfrac{\pi}{12},\cfrac{\pi}{4}]\)上单调递增,在区间\([-\cfrac{\pi}{4},-\cfrac{\pi}{12}]\)上单调递减。

法2:由\(-\cfrac{\pi}{4}\leq x\leq \cfrac{\pi}{4}\),求得\(-\cfrac{5\pi}{6}\leq 2x-\cfrac{\pi}{3}\leq \cfrac{\pi}{6}\),结合横轴为\(2x-\cfrac{\pi}{3}\)的图像可知,

\(-\cfrac{5\pi}{6}\leq 2x-\cfrac{\pi}{3}\leq -\cfrac{\pi}{2}\)时,求得函数在区间\([-\cfrac{\pi}{4},-\cfrac{\pi}{12}]\)单调递减;

\(-\cfrac{\pi}{2}\leq 2x-\cfrac{\pi}{3}\leq \cfrac{\pi}{6}\)时,求得函数在区间\([-\cfrac{\pi}{12},\cfrac{\pi}{4}]\)单调递增;

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!