已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4

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已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4k

已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4
已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4

已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4
k=(2sin^2α+2sinαcosα)/(1+sinα/cosα)
=2sinα(sinα+cosα)/[(sinα+cosα)/cosα]
=2sinαcosα
(sinα-cosα)^2
=sin^2α-2sinαcosα+cos^2α
=1-k
π/4