若2sin(π/4+α)=sinθ+cosθ,2sin²β=sin2θ,求证:sin2α+1/2cos2β=0
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若2sin(π/4+α)=sinθ+cosθ,2sin²β=sin2θ,求证:sin2α+1/2cos2β=0
若2sin(π/4+α)=sinθ+cosθ,2sin²β=sin2θ,求证:sin2α+1/2cos2β=0
若2sin(π/4+α)=sinθ+cosθ,2sin²β=sin2θ,求证:sin2α+1/2cos2β=0
(2sin(π/4+α))^2=2(sinα+cosα)^2=2+4sinαcosα=2+2sin2α=(sinθ+cosθ)^2=1+sin2θ,所以
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若2sin(π/4+α)=sinθ+cosθ,2sin²β=sin2θ,求证:sin2α+1/2cos2β=0
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