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

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

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