sinB=msin(2A+B) tan(A+B)=3tanA 求m
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/20 15:34:48
x)+sFN
%y 1lcSMR>Ql7:0{eOMiI'U APe@ 0 pg%T4XB1,Cۋ<j :ڸ,&bjL0!rь@x<$Ø (kozdRZ6ycHB 1 +
sinB=msin(2A+B) tan(A+B)=3tanA 求m
sinB=msin(2A+B) tan(A+B)=3tanA 求m
sinB=msin(2A+B) tan(A+B)=3tanA 求m
由,tan(A+B)=3tanA
得,sin(A+B)/cos(A+B)=3sinA/cosA
即,sin(A+B)cosA=3cos(A+B)sinA
sinB=sin[(A+B)-A]
=sin(A+B)cosA-cos(A+B)sinA
=3cos(A+B)sinA-cos(A+B)sinA
=2cos(A+B)sinA
sin(2A+B)=sin[(A+B)+A]
=sin(A+B)cosA+cos(A+B)sinA
=3cos(A+B)sinA+cos(A+B)sinA
=4cos(A+B)sinA
由题知,sinB=msin(2A+B)
即,2cos(A+B)sinA=4mcos(A+B)sinA
解得,m=1/2
所以,m的值为1/2
sinB=msin(2A+B) tan(A+B)=3tanA 求m
已知sinB=msin(2A+B),求证:tan(A+B)=(1+m)tanA/1-m6页
sinb=msin(2a+b)求证:tan(a+b)=(1+m/1—m)*tanaxiexie
已知sinB=msin(2A+B),且tan(A+B)=3tanA 则实数m的值为
sinA=msin(A+b)求证:tan(A+B)=sinB/cosb-m
一道高一三角函数恒等变换题已知sinb=msin(2a+b),m不等于0,2a+b不等于k派(k属于Z).求证tan(a+b)=[(1+m)/(1-m)]tan(a).
在斜三角形ABC中,sinB=msin(2A+B),求证:tanC=m+1/m-1·tanA
求证:sina+sinb/(cosa+cosb)=tan[(a+b)/2]
已知tan(a+b)=2tan a 证明 3sinb=sin(2a+b)
求证:2sinB/(cosA+cosB)=tan(A+B)/2-tan(A-B)/2
已知锐角a,b满足,sinb=2cos(a+b)sina,求证:tan(a+b)=3tan(a)
已知3sinB=sin(2A+B),求证:tan(A+B)=2tan A.
sin(2a+b)+2sinb=0 ,求证:tan a =3 tan (a+b)``
已知sin(2a+B)=5sinB 求证2tan(a+B)=3tan a
若tan(A+B)=2tanA,求证3sinB=sin(2A+B)
已知3sinB=sin(2A+B),求证tan(A+B)=2tanA
已知5sinb=sin(2a+b) 求证 tan(a+b)/tana=3/2
若tan(A+B)=2tanA.求证3sinB=sin(2A+B),