已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
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已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
2sin(3π+θ)=cos(π+θ)
-2sinθ=-cosθ
cosθ=2sinθ
代入sin²θ+cos²θ=1
所以sin²θ=1/5
cos²θ=4/5
sinθcosθ
=sinθ(2sinθ)
=2sin²θ
=2/5
所以原式=4/5
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