((b^2-c^2)/a^2 )*sin2A+((c^2-a^2)/b^2)sin2B+((a^2-b^2)/c^2)sin2C=
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((b^2-c^2)/a^2 )*sin2A+((c^2-a^2)/b^2)sin2B+((a^2-b^2)/c^2)sin2C=
((b^2-c^2)/a^2 )*sin2A+((c^2-a^2)/b^2)sin2B+((a^2-b^2)/c^2)sin2C=
((b^2-c^2)/a^2 )*sin2A+((c^2-a^2)/b^2)sin2B+((a^2-b^2)/c^2)sin2C=
题意不明,我默认你的a、b、c为三角形的三边长,而A、B、C为对应的三角,这样由正弦定理有:(b^2-c^2)/a^2=(sin^2B-sin^2c)/sin^2A=(sinB+sinC)(sinB-sinC)/sinA*sinA=[4sin(B+C)/2*sin(B-C)/2*sin(B-c)/2*cos(B-C)/2]/sin^2A=sin(B+C)*sin(B-C)/sin^2A=sin(B-C)/sinA,于是((b^2-c^2)/a^2 )*sin2A=2sinAcosA*sin(B-C)/sinA=sin(B-C)cosA=-sin(B-C)cos(B+C)=sin2C-sin2B;同理可得((c^2-a^2)/b^2)sin2B=sin2A-sin2C;((a^2-b^2)/c^2)sin2C=sin2B-sin2A.于是((b^2-c^2)/a^2 )*sin2A+((c^2-a^2)/b^2)sin2B+((a^2-b^2)/c^2)sin2C=0.
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