Question

在三角形ABC中,已知sinA\sinC=sin(A-B)\sin(B-C),求...

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

Answer 1

解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/dsinB=b/dsinC=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^2Csin^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=1cos^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