sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=______

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sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=______
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sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=______
sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=______

sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=______
[sin(π-θ)+cos(2π-θ)]/[cos(5π/2-θ)+sin(3π/2+θ)]=2=(sinθ+cosθ)/(sinθ-cosθ)=2
sinθ=3cosθ
tanθ=3
sinθ*cosθ=sinθcosθ/[sin^θ+cos^θ]=tanθ/[tan^θ+1]=3/10

3 /10

原式=sin(θ)+cos(θ)/sin(θ)-sin(θ)=2
所以con(θ)=2sin(θ) 因为两者平方和为1
cos(θ)=±2√5/5 sin(θ)=±√5/5 (两者同正同负)
cos(θ)*sin(θ)=±2/5

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