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运用tanα/2=sinα/1+cosα=1-cosα/sinα化简1+sin2α-cos2α/1+sin2α+cos2α
人气:336 ℃ 时间:2020-04-27 13:11:27
解答
(1+sin2α-cos2α)/(1+sin2α+cos2α)=(1+sin2α-cos2α)/(1+sin2α+cos2α)=[(sinα)^2+(cosα)^2+2sinαcosα-(cosα)^2+(sinα)^2]/(1+sin2α+cos2α)=2sinα(sinα+cosα)/(1+sin2α+cos2α)=2sinα(sinα+cosα...
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