1、4.7,三角函数中,有,关,的,范围,问题,培优,课,第,四章,三角函数与解三角形,在三角函数的图象与性质中,,的求解是近几年高考的一个热点内容,但因其求法复杂,涉及的知识点多,历来是我们复习中的难点,.,例,1,题型一,三角函数的单调性与,的关系,确定函数的单调区间,根据区间之间的包含关系,建立不等式,即可求,的取值范围,.,思维升华,跟踪训练,1,(2023,宜昌模拟,),已知函数,f,(,x,),3sin(,x,),,,0,,若,f,3,,,f,(),0,,,f,(,x,),在,上,单调递减,那么,的取值共有,A.2,个,B.3,个,C.4,个,D.5,个,n,1,2,3,4,5,,,
2、即周期,T,有,5,个不同取值,,的取值共有,5,个,.,题型二,三角函数的对称性与,的关系,例,2,又因为,0,,所以,的最小值为,1.,三角函数两条相邻对称轴或两个相邻对称中心之间的,“,水平间隔,”,为,,,相邻的对称轴和对称中心之间的,“,水平间隔,”,为,,,这就说明,我们可根据三角函数的对称性来研究其周期性,解决问题的关键在于运用整体代换的思想,建立关于,的不等式组,进而可以研究,“,”,的取值范围,.,思维升华,跟踪训练,2,已知函数,f,(,x,),,,若,f,(,x,),的图象的任何一条对称轴与,x,轴交点的横坐标均不属于区间,(3,,,4),,则,的取值范围是,三角函数的最
3、值与,的关系,例,3,将函数,f,(,x,),sin(2,x,)(,0,,,0,2),图象上每点的横坐标变为原来的,2,倍,得到函数,g,(,x,),,函数,g,(,x,),的部分图象如图所示,且,g,(,x,),在,0,2,上恰有一个最大值和一个最小值,(,其中最大值为,1,,最小值为,1),,则,的取值范围是,题型,三,由已知得函数,g,(,x,),sin(,x,),,由,g,(,x,),图象过,点,以及,点在图象上的位置,,由,g,(,x,),在,0,2,上恰有一个最大值和一个最小值,,利用三角函数的最值与对称轴或周期的关系,可以列出关于,的不等式,(,组,),,进而求出,的值或取值范围
4、思维升华,跟踪训练,3,A.11,B.13,C.15,D.17,由,,得,2(,k,1,k,2,),1,,,k,1,,,k,2,Z,,,综上,先检验,15,,,故,的最大值为,15.,三角函数的零点与,的关系,例,4,题型四,三角函数两个零点之间的,“,水平间隔,”,为,,,根据三角函数的零点个数,可以研究,“,”,的取值,.,思维升华,跟踪训练,4,(,2022,全国甲卷,),设函数,f,(,x,),sin,在区间,(0,,,),上恰有三个极值点、两个零点,则,的取值范围是,课时精练,1,2,3,4,5,6,7,8,9,10,11,12,A.,最小值,2,B,.,最大值,2,C.,最小
5、值,1,D,.,最大值,1,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,又,0,,,k,0,0,1,2,3,,,1,2,3,4,5,6,7,8,9,10,11,12,A.1,B.2 C.3 D.4,1,2,3,4,5,6,7,8,9,10,11,12,又,g,(,x,),的图象关于坐标原点对称,,12,k,2(,k,Z
6、),,,0,,,当,k,0,时,,min,2.,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,6.,(2023,银川模拟,),已知函数,f,(,x,),2sin (,0),,若方程,|,f,(,x,)|,1,在区间,(0,2),上恰有,5,个实根,则,的取值范围是,1,2,3,4,5,6,7,8,9,10,11,12,因为原方程在区间,(0,2),上恰有,5,个实根,,1,2,3,4,5,6,7,8,9,10,11,12,1
7、2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,故对任意整数,k,,,(0,2),,所以,错误;,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,A,,,B,,,C,为连续三个交点,不妨设,B,在,x,轴下方,,D,为,AC,的中点,.,由对称性可得,ABC,是以,B,为顶角的等腰三角形,,AC,T,2,CD,,,1,2,3,4,5,6,7,8,9,10,11,12,要使,ABC,为钝角三角形,只需,ACB,0),在,0,,,上有且仅有,3,个零点,则实数,的,取值,范围,是,
8、1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,10.(,2022,全国乙卷,),记函数,f,(,x,),cos(,x,)(,0,0,0,,所以当,k,0,时,,取得最小值,,且最小值为,3.,1,2,3,4,5,6,7,8,9,10,11,12,11.(,2023,黄冈模拟,),已知函数,y,f,(,x,),的图象是由函数,y,cos,x,(,0),的图象向左,平,移,个单位长度所得,若函数,y,f,(,x,),在区间,(,,,2),上单调,,则,的取值范围是,.,当,x,(,,,2),,,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,又,k,Z,,得,k,0,或,k,1,,,1,2,3,4,5,6,7,8,9,10,11,12,0,1,2,3,4,5,6,7,8,9,10,11,12,令,t,x,,则函数,y,sin,t,在区间,,,2,上存在两个极大值点,,则,,可得,2,,,当,2,T,2,时,即,4,,显然符合题意;,1,2,3,4,5,6,7,8,9,10,11,12,