高考数学 三角函数诱导公式课件.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,同角关系及诱导公式,一、同角三角函数基本关系式,1.,倒数关系,2.,商数关系,3.,平方关系,tan,cot,=1,sin,csc,=1,cos,sec,=1,sin,2,+,cos,2,=1,1,+,tan,2,=sec,2,1,+co,t,2,=csc,2,tan,=cot,=,sin,cos,cos,sin,二、诱导公式,奇变偶不变,符号看象限,.,3.,本质,通过不相等的两个角的同名三角函数或两个互为余函数的三角函数值相等或互为相反数,反映了三角函数的周期性及各种对称性,.,1.,定义,2.,口诀用自变量,的三角函数表示自变量为,(,k,Z,),的三角函数的公式叫诱导公式,.,2,k,1.,已知,cot(,-,),=2,求,sin(,+,),的值,.,3,2,解,:,cot(,-,)=2,又,cot(,-,)=,-,cot,cot,=,-,2,.,是第二或第四象限角,且,tan,=,-,.,1,2,cos,2,=,.,1+tan,2,1,4,5,又,sin(,+,)=,-,cos,3,2,2,5,5,是第二象限角,2,5,5,是第四象限角,.,-,sin(,+,)=,3,2,cos,=,2,5,5,是第四象限角,.,2,5,5,是第二象限角,-,典型例题,2.,已知,cot,=,m,m,0),求,cos,.,解,:,cot,=,m,(,m,0),角,的终边不在坐标轴上,.,若,是第一或第二象限角,则,c,sc,=,1+cot,2,=,1+,m,2,.,sin,=,c,sc,1,1+,m,2,1,=.,cos,=,sin,cot,=,.,m,1+,m,2,1+,m,2,若,是第三或第四象限角,则,c,sc,=,-,1+cot,2,=,-,1+,m,2,.,sin,=,c,sc,1,1+,m,2,1,=,-,.,cos,=,sin,cot,=,-,.,m,1+,m,2,1+,m,2,3.,已知,sin,+cos,=(0,),求,tan,的值,.,2,3,解法,1,将,已知等式两边平方得,sin,cos,=,-,0,18,7,0,0.,由,sin,cos,0,知,cos,0.,sin,-,cos,=(,sin,-,cos,),2,=,1,-,2sin,cos,=.,4,3,sin,+cos,=,sin,-,cos,=,4,3,2,3,解方程组 得,sin,=,cos,=.,2,+4,6,2,-,4,6,tan,=,.,sin,cos,-,9,-,4,2,7,解法,2,将,已知等式两边平方得,sin,cos,=,-,0,18,7,0,0.,由,sin,cos,0,知,cos,0.,tan,=,.,sin,cos,-,9,-,4,2,7,sin,cos,是方程,x,2,-,x,-,=,0,的根,且,cos,为小根,.,18,7,2,3,cos,=,sin,=.,2,+4,6,2,-,4,6,3.,已知,sin,+cos,=(0,),求,tan,的值,.,2,3,解法,3,由,已知,sin,cos,成等差数列,设其公差为,d,则,2,6,sin,=,-,d,2,6,cos,=,+,d,.,2,6,由,sin,2,+cos,2,=1,得,:(,-,d,),2,+(,+,d,),2,=1.,2,6,2,6,解得,d,=,-,或,.,2,3,2,3,tan,=,.,sin,cos,-,9,-,4,2,7,cos,=,sin,=.,2,+4,6,2,-,4,6,当d,=,时,sin,=,0,与,0,0,矛盾,2,3,2,-,4,6,d,=,-,.,2,3,3.,已知,sin,+cos,=(0,),求,tan,的值,.,2,3,4.,已知,f,(,)=,.(1),化简,f,(,);,sin(,-,)cos(2,-,)tan(,-,+,),3,2,cot(,-,-,)sin(,-,-,),(2),若,是第三象限角,且,cos(,-,)=,求,f,(,),的值,;,3,2,1,5,(3),若,=,-,求,f,(,),的值,;,3,31,解,:,(1),f,(,),=,sin,cos,co,t,-,co,t,sin,=,-,co,s,;,(2),cos(,-)=,-,sin,3,2,由已知可得,sin,=,-,.,1,5,是第三象限角,co,s,0.,co,s,=,-,1,-,sin,2,=,-,2,5,6.,2,5,6,f,(,),=,-,co,s,=.,(3),=,-,=,-,6,2,+,3,31,5,3,f,(,-,)=,-,co,s,(,-,),3,31,3,31,5,3,=,-,cos(,-,6,2,+,),=,-,cos,5,3,=,-,cos,=,-,.,1,2,3,5.,已知,0,tan,+cot,=,求,sin(,-,),的值,.,5,2,2,2,2,3,解,:,tan,+cot,=,2,2,sin,2,由已知可得,sin,=4,5,0,2,c,os,=,1,-,sin,2,3,5,=,.,sin(,-,)=,sin,cos,-,cos,sin,3,3,3,=,-,1,2,3,5,4,5,3,2,=(4,-,3 3,).,10,1,6.,已知,为锐角,且,tan,=,求,的值,.,sin2,cos,-,sin,sin2,cos2,1,2,解,:,tan,=,1,2,又,为锐角,1+tan,2,1,cos,2,=,.,4,5,cos,=,.,5,2,原式,=,2sin,cos,2,-,sin,2sin,cos,cos,2,sin,cos,2,2sin,cos,cos,2,=,1,2,cos,=,=.,5,4,7已知,tan(,-,)=2,求,:(1);,(2)2,sin(3,+,)cos(,+,)+sin(,-,)sin(,-,),.,4cos,2,-,3sin,2,+1,sin,2,-,2sin,cos,-,cos,2,3,2,5,2,解,:,(1),tan(,-,)=2,又,tan(,-,)=,-,tan,tan,=,-,2,.,原式,=,5cos,2,-,2sin,2,sin,2,-,2sin,cos,-,cos,2,1+tan,2,2tan,2,-,tan,=,5,-,2tan,2,tan,2,-,2tan,-,1,=,=,-,.,7,3,(2),由,(1),知,tan,=,-,2,原式2(,-,sin,)(,-,sin,)+(,-,cos,)sin,=2sin,2,-,sin,cos,=cos,2,(2tan,2,-,tan,),=2.,8.,角,的终边上的点,P,与,A(,a,b,),关于,x,轴对称,(,a,0,b,0,),角,的终边上的点,Q,与,A,点关于直线,y,=,x,对称,求,sin,sec,+tan,cot,+,sec,csc,的值,.,解法,1,依题意,P(,a,-,b,),Q(,b,a,),设,r,=,a,2,+,b,2,则,:,sin,=,-,sec,=,tan,=,-,cot,=,sec,=,csc,=.,b,r,r,b,b,a,b,a,r,a,r,a,原式,=,-,+(,-,),+,b,r,r,b,b,a,b,a,r,a,r,a,=,-,1,-,+=0.,a,2,+,b,2,a,2,b,2,a,2,解法,2,依题意,-,=2,k,+,(,k,Z,),即,=2,k,+,+,.,2,2,原式,=sin,+tan,cot(2,k,+,+,)+,cos(,2,k,+,+,),2,1,2,cos,1,sin(,2,k,+,+,),2,1,=sin,+tan,(,-,tan,)+,-,sin,1,cos,1,cos,1,=,-,1,-,tan,2,+sec,2,=0.,课后练习,1.,已知,sin+sin,2,=1,求,cos,2,+cos,4的值,.,解,:,由,sin+sin,2,=1,得,sin=1,-,sin,2,=cos,2,.,cos,2,+cos,4,=sin+sin,2,=1,.,2.,已知,cos,=(,m,-,1),求,sin,cot,.,m,2,+1,2,m,解,:,由已知,cos,0,角,的终边在第二或第三象限或为,x,轴的非正半轴,.,当,角,的终边在第二象限或为,x,轴的非正半轴,时,sin,=,1,-,cos,2,=,m,2,+1,m,2,-,1,2,m,tan,=.,sin,cos,m,2,-,1,当,角,的终边在第三象限,时,sin,=,-,1,-,cos,2,=,1+,m,2,1,-,m,2,2m,tan,=.,sin,cos,1,-,m,2,3.,设,sin,cos,是方程,2,x,2,-,(,3,+1),x,+,m,=0,的两根,求,:,(1),+,及,m,的值,;(2),方程两根,sin,cos,及此时,的值,.,1,-,cot,sin,1,-,tan,cos,解,:,(1),由已知,sin,+cos,=,sin,cos,=.,3,+1,2,2,m,1,-,cot,sin,1,-,tan,cos,+=+,1,-,sin,1,-,cos,cos,sin,sin,cos,=,cos,-,sin,cos,2,-,sin,2,=+,cos,-,sin,cos,2,sin,-,cos,sin,2,=,sin,+cos,=,.,3,+1,2,(,sin,+cos,),2,=1+2sin,cos,2,m,(),2,=1+2,.,3,+1,2,解得,m,=.,3,2,解,:,(2),由,(1),知,原方程为,2,x,2,-,(,3,+1),x,+,=0.,3,2,=2,k,+,或,=2,k,+(,k,Z,).,3,6,3.,设,sin,cos,是方程,2,x,2,-,(,3,+1),x,+,m,=0,的两根,求,:,(1),+,及,m,的值,;(2),方程两根,sin,cos,及此时,的值,.,1,-,cot,sin,1,-,tan,cos,解得,x,1,=,x,2,=.,1,23,2,sin,=,cos,=,1,2,sin,=,cos,=,1,2,或,3,2,3,2,5.,已知,tan(,-,)=,a,2,|cos(,-,)|=,-,cos,求,sec(,+,),的值,;,4.,已知,cos(,-,)=,a,(|,a,|,1),求,cos,(,+,)+sin(,-,),的值,;,5,6,6,2,3,解,:,cos(,-,)=,a,(|,a,|,1),6,cos,(,+,)=,cos,-,(,-,),5,6,6,=,-,cos,(,-,)=,-,a,6,=,cos,(,-,)=,a,6,sin(,-,)=sin,+,(,-,),2,3,6,2,cos,(,+,)+sin(,-,),5,6,2,3,=,-,a,+,a,=0.,解,:,tan(,-,)=,a,2,又,tan(,-,)=,-,tan,tan,=,-,a,2,.,|,cos,(,-,)|=,-,cos,又,|,cos,(,-,)|=|,cos,|,|,cos,|=,-,cos,.,cos,0.,sec(,+,)=,-,cos,1,=,1+tan,2,=,1+,a,4,.,6.,若,+,=0,试判断,cos(sin,),sin(cos,),的符号,;,1,-,cos,2,sin,1,-,sin,2,cos,|,cos,|,sin,|sin,|,cos,解,:,由已知,+,=0,sin,与,cos,异号,.,是第二或第四象限角,.,当,是第二象限角时,-,1,cos,0,0sin,1.,sin(cos,),0.,cos(sin,),sin(cos,),0.,故,cos(sin,),sin(cos,),的符号为“,+,”,号,.,-,-,1,1,2,2,-,cos,0,0sin .,2,2,解,:,是第二象限角,7.,已知,sin,=,cos,=,若,是第二象限角,求实数,a,的值,.,1+,a,3,a,-,1,1+,a,1,-,a,0sin,1,-,1,cos,0.,1+,a,3,a,-,1,1+,a,1,-,a,0 1,-,1 0,解得,0,a,.,1,3,又,sin,2,+cos,21,1+,a,3,a,-,1,1+,a,1,-,a,(),2,+(),2,=1.,整理得,9,a,2,-,10,a,+1=0.,解得,a,=,或,a,=1(,舍去,).,1,9,故实数,a,的值为,.,1,9,8.,在,ABC,中,sinA+cosA,=,AC=2,AB=3,求,tanA,的值和,ABC,的面积,.,2,2,解,:,=,sinA+cosA,=2,sin(A+45),2,2,sin(A+45)=,.,1,2,0A180,A=105,tanA,=tan105=tan(45+60)=,1+,3,1,-,3,=,-,2,-,3.,sinA,=sin105=sin(45+60),=sin45cos60+cos45sin60,1,2,ABC,的面积,S,ABC,=,AC,ABsinA,2+,6,4,=,2,3,1,2,=,(,2+,6).,3,4,2+,6,4,=,.,补充例题,1.,已知,cot,x,=,m,x,(2,k,-,2,k,)(,k,Z,),求,cos,x,的值,.,(2),已知,tan,=2,求,sin,cos,的值,;,(3),已知,sin,+,cos,=,求,cos,4,+sin,4,的值,.,1,2,3.,已知,sin,cos,是方程,x,2,+,px,+,p,+1=0,的两根,求实数,p,的值,.,5.,设,f,(,)=,sin(cos,),g(,)=cos(sin,),(1),若,f,(,),g,(,)0,求角,的取值范围,;(2),设,0,若,f,(,),的最大值、最小值分别是,a,、,b,g,(,),的最大值、最小值分别是,c,、,d,试比较,a,b,c,d,的大小,.,1+,m,2,1+,m,2,m,-,2,5,32,23,-,1,2k,-,2,k,(,k,Z),3,2,2,(1)2,k,+,2,k,+,(,k,Z);,(2),b,d,a,c,.,2.(1),已知,tan,=2,求,2sin,2,-,sin,cos,+,cos,2,的值,;,3,5,-,2,4.,若,=cot,-,csc,求,的取值范围,.,1,-,cos,1+cos,