已知5siny=sin(2x+y),求证:tan(x+y)=3/2tanx
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已知5siny=sin(2x+y),求证:tan(x+y)=3/2tanx已知5siny=sin(2x+y),求证:tan(x+y)=3/2tanx已知5siny=sin(2x+y),求证:tan(x
已知5siny=sin(2x+y),求证:tan(x+y)=3/2tanx
已知5siny=sin(2x+y),求证:tan(x+y)=3/2tanx
已知5siny=sin(2x+y),求证:tan(x+y)=3/2tanx
sin[(x+y)+x]=5sin[(x+y)-x]
sin(x+y)·cosx+cos(x+y)·sinx=5·sin(x+y)·cosx-5·cos(x+y)·sinx
4·sin(x+y)·cosx=6·cos(x+y)·sinx
两边同除以cos(x+y)·cosx得:2·tan(x+y)=3·tanx
2x+y=(x+y)+x
y=(x+y)-x 代入原式
5sin(x+y)cosx-5sinxcos(x+y)=sin(x+y)cosx+sinxcos(x+y)
sin(x+y)/cos(x+y)=3sinx/2cosx
得证
相信换元是可以的
令A=2X+Y,B=Y
则5sinB=sinA
即证tan[(A+B)/2]=3/2tan[(A-B)/2]
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