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高1数学`已知2tanA=3tangB,求证tan(A-B)=s

2005-08-08 17:12:33︵***
已知2tanA=3tangB,求证tan(A-B)=sin2B/(5-cos2B)高1数学`已知2tanA=3tangB,求证tan(A-B)=sin2B/(5-cos2B)：已知2tanA=3tangB,求证tan(A-B)=sin2B/(?

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  • 已知2tanA=3tangB,求证tan(A-B)=sin2B/(5-cos2B) tanA=3tanB/2 sin2B/(5-cos2B) =2sinBcosB/(5-(2cosB^2-1)) =2sinBcosB/(6-2cosB^2) =sinBcosB/(3-cosB^2) =sinBcosB/(3sinB^2+2cosB^2) 上下同除以cosB^2 =tanB/3tanB^2+2 tan(A-B)=(tanA-tanB)/(1+tanAtanB) =1/2*tanB/(1+3/2*tanB^2) =tanB/3tanB^2+2 左边等于右边
    2005-08-08 17:19:45
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