已知2tanA=tanB,求证:tan(A-B)=sin2B/(5-cos2B)

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已知2tanA=tanB,求证:tan(A-B)=sin2B/(5-cos2B)已知2tanA=tanB,求证:tan(A-B)=sin2B/(5-cos2B)已知2tanA=tanB,求证:tan(

已知2tanA=tanB,求证:tan(A-B)=sin2B/(5-cos2B)
已知2tanA=tanB,求证:tan(A-B)=sin2B/(5-cos2B)

已知2tanA=tanB,求证:tan(A-B)=sin2B/(5-cos2B)
可能我算错了 我算的是tan(A-B)=sin2B/(3-cos2B) 下面是我算的
tan(A-B)=(tanA-tanB)/(1+tanAtanB)=tanB/[1+(tanB)^2]=sinAcosA/[(cosB)^2+2(sinB)^2]
(3-cos2B)=3-[1-2(sinA)^2]=2+2(sinB)^2
sin2B=2sinBcosB
sin2B/(3-cos2B)=2sinBcosB/[2+2(sinB)^2]=sinBcosB/[1+(sinB)^2]=sinBcosB/[(cosB)^2+2(sinB)^2]