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(tanαtan2α)/(tan2α-tanα)化简急
人气:243 ℃ 时间:2020-04-06 10:49:47
解答
因tanα=tan(2α-α)=(tan2α-tanα)/(1+tan2αtanα)
故(tanαtan2α)/(tan2α-tanα)=(tanαtan2α)/[(1+tan2αtanα)*tanα]=tan2α/(1+tan2αtanα)
=sin2αcosα/(cos2αcosα+sin2αsinα)=sin2αcosα/cosα=sin2α
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