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,第,2,讲,三角函数,的图象与性质,专题二三角函数与解三角形,第2讲三角函数的图象与性质专题二三角函数与解三角形,1,考情分析,KAO QING FEN XI,1.,高考对此部分内容的命题主要集中于三角函数的定义、图象与性质,,,主要,考查图象的变换、函数的单调性、奇偶性、周期性、对称性,及,最,值,常与三角恒等变换交汇命题,.,2,.,主要以选择题、填空题的形式考查,难度为中等或偏下,.,考情分析KAO QING FEN XI1.高考对此部分内容的,2,内,容,索,引,考点一,考点二,考点三,专题强化练,内考点一考点二考点三专题强化练,3,1,考点一三角函数的定义、诱导公式及基本关系,PART ONE,1考点一三角函数的定义、诱导公式及基本关系PART ONE,4,核心提炼,核心提炼,二级,结论,二级结论,解得,tan,2.,解得tan 2.,2,考点二三角函数的图象与解析式,PART TWO,2考点二三角函数的图象与解析式PART TWO,11,三角函数图象的变换,核心提炼,三角函数图象的变换核心提炼,例,2,(1),已知函数,f,(,x,),A,sin(,x,)(,A,0,,,0,,,|,|0,,解析,f,(,x,),的最小正周期为,,,2.,又,f,(,x,),A,sin(2,x,),是奇函数,,k,(,k,Z,),,,|,|,,,0,,,f,(,x,),A,sin 2,x,,则,g,(,x,),A,sin,x,,,f,(,x,),2sin 2,x,,,解析f(x)的最小正周期为,2.f(x)2s,对于,,根据图象可知,,x,A,2,x,B,,,f,(,x,),在,(0,2),上有,3,个极大值点,,f,(,x,),在,(0,2),上有,2,个或,3,个极小值点,故,不正确;,对于,根据图象可知,xA2xB,f(x)在(0,2,二轮专题复习(2021新高考)专题二-第2讲-三角函数的图象与性质课件,易错提醒,(1),根据零点求,值时注意是在增区间上还是在减区间上,.,(2),注意变换时,“,先平移后伸缩,”,与,“,先伸缩后平移,”,的区别,.,易错提醒(1)根据零点求值时注意是在增区间上还是在减区间上,解析,由图象知,T,2,,,解析由图象知T0,,,0),的性质,核心提炼函数yAsin(x)(A0,0)的性质,二轮专题复习(2021新高考)专题二-第2讲-三角函数的图象与性质课件,二轮专题复习(2021新高考)专题二-第2讲-三角函数的图象与性质课件,因为,y,cos,在,0,,,上是减函数,,因为ycos 在0,上是减函数,,所以,的取值范围是,(1,2).,故选,C.,所以的取值范围是(1,2).故选C.,规律方法,已知三角函数的单调区间求参数取值范围的三种方法,(1),子集法:求出原函数的相应单调区间,由已知区间是所求某区间的子集,列不等式,(,组,),求解,.,(2),反子集法:由所给区间求出整体角的范围,由该范围是某相应正弦、余弦函数的某个单调区间的子集,列不等式,(,组,),求解,.,规律方法已知三角函数的单调区间求参数取值范围的三种方法,解析,函数,f,(,x,),的定义域为,R,,由,f,(,x,),|cos(,x,)|,|sin|,x,|,|cos,x,|,|sin|,x,|,f,(,x,),,知,f,(,x,),是偶函数,故,A,正确,;,f,(,x,),|cos(,x,)|,|sin|,x,|,|cos,x,|,|sin|,x,|,f,(,x,),,所以,f,(,x,),是周期为,的函数,故,B,正确;,解析函数f(x)的定义域为R,由f(x)|cos(x,又,f,(,x,),是周期为,的函数,所以,f,(,x,),的值域为,1,1,,故,D,不正确,.,又f(x)是周期为的函数,所以f(x)的值域为1,1,2,其中,0,,,A,,,B,,,C,是这两个函数图象的交点,.,2其中0,A,B,C是这两个函数图象的交点.,若存在,ABC,是等腰直角三角形,则,的最小值为,_.,解析,若存在,ABC,是等腰直角三角形,利用直角三角形斜边的中线等于斜边的一半,,若存在ABC是等腰直角三角形,则的最小值为_,4,专题强化练,PART,FOUR,4专题强化练PART FOUR,37,一、单项,选择题,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,一、单项选择题12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,由,3,x,y,1,0,得,,y,3,x,1,,,tan,3,,,12345678910111213141516解析由3x,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,平方得,a,2,2,ab,b,2,0,,即,(,a,b,),2,0,,,则,a,b,0,,,b,a,,,且图象关于点,(,,,0),对称,.,12345678910111213141516平方得a22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,依题意得,函数,f,(,x,),tan(,x,),的最小正周期为,3,,,12345678910111213141516解析依题意得,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以,y,f,(,x,),关于点,(2,0),对称,,作出两个函数的图象,(,图略,),,,可知两函数共有,6,个交点,且都关于点,(2,0),对称,,则易知,6,个交点的横坐标之和为,12.,12345678910111213141516所以yf(x,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,故,A,错误;,12345678910111213141516故A错误;,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,而函数,y,2cos 2,x,为偶函数,故,D,正确,.,12345678910111213141516而函数y2c,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,11.(2020,佛山模拟,),已知函数,f,(,x,),sin,x,sin ,x,,下列结论正确的是,A.,f,(,x,),是奇函数,B.,f,(,x,),是周期函数,C.,f,(,x,),在区间,(0,,,),上有三个零点,D.,f,(,x,),的最大值为,2,1234567891011121314151611.(202,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,x,R,,,f,(,x,),sin(,x,),sin(,x,),sin,x,sin ,x,f,(,x,),,,f,(,x,),是奇函数,故,A,正确,;,y,1,sin,x,的周期,T,1,2,k,,,k,Z,,,y,2,sin ,x,的周期,T,2,2,n,,,n,Z,,,T,1,|,T,1,2,k,,,k,Z,T,2,|,T,2,2,n,,,n,Z,,,f,(,x,),不是周期函数,故,B,错误,;,令,f,(,x,),sin,x,sin ,x,0,,得,sin ,x,sin,x,sin(,x,),,,x,x,2,k,,,k,Z,或,x,x,2,k,,,k,Z,,,12345678910111213141516解析xR,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,f,(,x,),在区间,(0,,,),上有三个零点,故,C,正确;,y,sin,x,与,y,sin ,x,不可能同时取得最大值,1,,故,D,错误,.,12345678910111213141516f(x)在区,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,由图象可知,若,f,(,x,),在,0,2,上有且仅有,3,个极小值点,,12345678910111213141516由图象可知,若,f,(,x,),在,(0,2),上可能有,2,或,3,个极大值点,故,B,错误;,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,f(x)在(0,2)上可能有2或3个极大值点,故B错误;1,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,123456789101112131415161,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,由于函数,y,g,(,x,),的图象关于原点对称,.,12345678910111213141516由于函数yg,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,2,123456789101112131415162,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,综上所述,,1,或,3,,故,正确;,f,(,x,),不可能为偶函数,,错误;,12345678910111213141516综上所述,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,本课结束,更多精彩内容请登录,:,本课结束更多精彩内容请登录:www.xinjiaoyu.co,73,
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