在三角形ABC中,已知sinA\sinC=sin(A-B)\sin(B-C),求证:a^2+c^2=2b^2

1个回答

  • 假设三角形的外接圆的直径为d,对应角的边分别为a,b,c

    则a/sinA=b/sinB=c/sinC=d

    所以

    sinA=a/d

    sinB=b/d

    sinC=c/d

    因为

    sinA=sin(B+C)

    sinC=sin(A+B)

    所以

    sin(B+C)/sin(A+B)=sin(A-B)/sin(B-C)

    即sin(B+C)sin(B-C)=sin(A-B)sin(A+B)

    (sinBcosC+cosBsinC)(sinBcosC-cosBsinC)=(sinAcosB-cosAsinB)(sinAcosB+cosAsinB)

    sin^2Bcos^2C-cos^2Bsin^2C=sin^2Acos^2B-cos^2Asin^2B

    将系数为负的移项

    sin^2Bcos^2C+cos^2Asin^2B=sin^2Acos^2B+cos^2Bsin^2C

    sin^2B(cos^2C+cos^2A)=cos^2B(sin^2A+sin^2C)

    根据sin^2B+cos^2B=1

    可以得到

    cos^2B=1-sin^2B

    所以

    sin^2B(cos^2C+cos^2A)=(1-sin^2B)(sin^2A+sin^2C)

    将上式拆分

    sin^2B(cos^2C+cos^2A)=(sin^2A+sin^2C)-sin^2B(sin^2A+sin^2C)

    将系数为负的移项

    sin^2B(cos^2C+cos^2A+sin^2A+sin^2C)=sin^2A+sin^2C

    因为

    cos^2A+sin^2A=1

    cos^2C+sin^2C=1

    所以

    2sin^2B=sin^2A+sin^2C

    所以

    2(b/d)^2=(a/d)^2+(c/d)^2

    即2b^2=a^2+c^2

    所以命题得证