已知sin(π-α)=-2sin(π/2+α)则cos²α+sin2α

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已知sin(π-α)=-2sin(π/2+α)则cos²α+sin2α
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已知sin(π-α)=-2sin(π/2+α)则cos²α+sin2α
已知sin(π-α)=-2sin(π/2+α)则cos²α+sin2α

已知sin(π-α)=-2sin(π/2+α)则cos²α+sin2α

sin(π-a)=sina
sin(π/2+a)=cosa
∵sin(π-a)=-2sin(π/2+a)
∴sina=-2cosa
∴tana=sina/cosa=-2
∴cos²a+sin2a
=(cos²a+2sinacosa)/(sin²a+cos²a)——除以1=sin²a+cos²a,值不变
=(1+2tana)/(tan²a+1)
=(1-4)/(4+1)
=-3/5