已知sinA+cosA=2sina,sinA*cosA=(sinb)^2,求证4(sin2a)^2=(cos2b)^2
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![已知sinA+cosA=2sina,sinA*cosA=(sinb)^2,求证4(sin2a)^2=(cos2b)^2](/uploads/image/z/5529315-3-5.jpg?t=%E5%B7%B2%E7%9F%A5sinA%2BcosA%3D2sina%2CsinA%2AcosA%3D%28sinb%29%5E2%2C%E6%B1%82%E8%AF%814%28sin2a%29%5E2%3D%28cos2b%29%5E2)
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已知sinA+cosA=2sina,sinA*cosA=(sinb)^2,求证4(sin2a)^2=(cos2b)^2
已知sinA+cosA=2sina,sinA*cosA=(sinb)^2,求证4(sin2a)^2=(cos2b)^2
已知sinA+cosA=2sina,sinA*cosA=(sinb)^2,求证4(sin2a)^2=(cos2b)^2
LZ,证明的结果好像是错的,应该是sin2a改为cos2a吧.