求证a^2sin2B+b^2sin2A=2absinc

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求证a^2*sin2B+b^2*sin2A=2absinc求证a^2*sin2B+b^2*sin2A=2absinc求证a^2*sin2B+b^2*sin2A=2absinc因为A+B=π-C,所以s

求证a^2*sin2B+b^2*sin2A=2absinc
求证a^2*sin2B+b^2*sin2A=2absinc

求证a^2*sin2B+b^2*sin2A=2absinc
因为A+B=π-C,所以sin(A+B)=sinC,展开,sinAcosB+cosAsinB=sinC,乘以2sinAsinB,得2sin^2AsinBcosB+2sinAcosAsin^2B=2sinCsinAsinB,即
sin^2Asin2B+sin2Asin^2B=2sinCsinAsinB,由正弦定理得sinA=a/2R,sinB=b/2R,sinC=c/2R,
所以sin^2A=a^2/4R^2,sin^2B=b^2/4R^2,sin^2C=c^2/4R^2,代入,整理得a^2*sin2B+b^2*sin2A=2absinc