来自宋向军的问题
已知2sin^α+sin2α/1+tanα=k(π/4
已知2sin^α+sin2α/1+tanα=k(π/4
1回答
2020-08-1016:16
已知2sin^α+sin2α/1+tanα=k(π/4
已知2sin^α+sin2α/1+tanα=k(π/4
(2sin^2α+2sinα*cosα)/(1+tanα)=k(2sin^2α+2sinα*cosα)/(1+tanα)=2sinacosa(sina+cosa)/(sina+cosa)=2sinacosa=k(sinα-cosα)^2=1-2sinacosa=1-ksinα-cosα=±√(1-k)