运用tanα/2=sinα/1+cosα=1-cosα/sinα化简1+sin2α-cos2α/1+sin2α+cos2
运用tanα/2=sinα/1+cosα=1-cosα/sinα化简1+sin2α-cos2α/1+sin2α+cos2α
赞 yuanshiqi723 春芽
共回答了13个问题采纳率：92.3% 举报
(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)/(1+sin2α+cos2α)
=2sinα(sinα+cosα)/[(sinα)^2+(cosα)^2+2sinαcosα+(cosα)^2-(sinα)^2]
=2sinα(sinα+cosα)/[2cosα(sinα+cosα)]
=tanα
=(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)/(1+sin2α+cos2α)
=2sinα(sinα+cosα)/[(sinα)^2+(cosα)^2+2sinαcosα+(cosα)^2-(sinα)^2]
=2sinα(sinα+cosα)/[2cosα(sinα+cosα)]
=tanα
1年前
9
可能相似的问题
-
1年前3个回答
-
已知tanα=7.sin²α+sinαcosα+3cos²α
1年前1个回答
你能帮帮他们吗