来自沈寒辉的问题
【已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4】
已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4
1回答
2020-08-1016:09
【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