作业帮已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5.(1)求证:tanA=2tanB(2)tanB的值
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已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5.(1)求证:tanA=2tanB(2)tanB的值
已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5.(1)求证:tanA=2tanB(2)tanB的值
已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5.(1)求证:tanA=2tanB(2)tanB的值
（1）sin(A+B)=3/5,sin(A-B)=1/5
sin(a+b)=sinAcosB+sinBcosA=3/5
sin(a-b)=sinAcosB-sinBcosA=1/5
两式相加相减后可得：
sinAcosB=2/5
sinBcosA=1/5
将两式相除,可得tanA=2tanB
（2）tan(B)=sinB/cosB=sinBcosA/cosAcosB
cos(A)cos(B)=1/2[cos(A+B)+cos(A-B)]=1/2[4/5+2根号6/5]=(根号6-2)/5
tanB=1/(根号6-2)=(根号6+2)/2
sinacosb+sinbcosa=3/5
sinacosb-sinbcosa=1/5
sinacosb=2/5
sinbcosa=1/5
sinacosb/sinbcosa=tana/tanb=2/1
C是锐角，所以A+B是钝角
cos(A+B)=-4/5,cos(A-B)=2√6/5
cosAcosB-sinAsinB=-4...
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sinacosb+sinbcosa=3/5
sinacosb-sinbcosa=1/5
sinacosb=2/5
sinbcosa=1/5
sinacosb/sinbcosa=tana/tanb=2/1
C是锐角，所以A+B是钝角
cos(A+B)=-4/5,cos(A-B)=2√6/5
cosAcosB-sinAsinB=-4/5
cosAcosB+sinAsinB=2√6/5
sinAsinB=(2+√6)/5
sin(A+B)=3/5
tan(A+B)=3/4=(2*tanB+tanB)/(2*tanB*tanB)
tanB=2
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(1)
tan(A)/tan(B)
=(sin(A)/cos(A))/(sin(B)/cos(B))
=sin(A)cos(B)/(sin(B)cos(A))
sin(A)cos(B)=1/2*[sin(A+B)+sin(A-B)]=2/5
sin(B)cos(A)=1/2*[sin(A+B)-sin(A-B)]=1/5
代入即得证
(2)<...
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(1)
tan(A)/tan(B)
=(sin(A)/cos(A))/(sin(B)/cos(B))
=sin(A)cos(B)/(sin(B)cos(A))
sin(A)cos(B)=1/2*[sin(A+B)+sin(A-B)]=2/5
sin(B)cos(A)=1/2*[sin(A+B)-sin(A-B)]=1/5
代入即得证
(2)
tan(B)=sin(A)/cos(A)=sin(A)cos(B)/cos(A)cos(B)
cos(A)cos(B)=1/2[cos(A+B)+cos(A-B)]=1/2[4/5+2根号6/5]=(2+根号6)/5
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1、因为sin(A+B)=3/5,sin(A-B)=1/5
则sin(A+B)=sinAcosB+cosAsinB=3/5 ①
sin(A-B)=sinAcosB-cosAsinB=1/5 ②
①+ ②得：sinAcosB=2/5
①- ②得：cosAsinB=1/5
则两式一比可得：sinAcosB/cosAsinB=2
即tanA=2tanB
2、
(2)AB边上的高CD=x，tanA=2tanB→x/AD=2(x/BD)
→BD=2AD→AD=AB/3=1，BD=2。
sin(A+B)=3/5→tan(A+B)=3/4，(tanA+tanB)/(1-tanAtanB)=3/4。
3tanB/[1-2(tanB)^2]=3/4，
tanA=x，tanB=x/2，代入得x^2+4x=2，正数解x=√6-2