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

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/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