问题描述: 在三角形ABC中,角A,B,C对边分别为a,b,c.证明:(a*2--b*2)/c*2=sin(A--B)/sinC*为平方的意思 1个回答 分类:数学 2014-11-09 问题解答: 我来补答 a/sinA=b/sinB=c/sinC=2R,a=sinA*2R,b=sinB*2R,c=sinC*2R,左边有(a^2-b^2)/c^2=(sin^2A-sin^2B)/sin^2C=[(sinA+sinB)(sinA-sinB)]/sin^2c=[2sin(A+B)/2*cos(A-B)/2*2cos(A+B)/2*sin(A-B)/2]/sin^2C=[sin(A+B)*sin(A-B)]/sin^2C,而,(A+B+C)=180,A+B=180-C,sin(A+B)=sin(180-C)=sinC,则,[sin(A+B)*sin(A-B)]/sin^2C=sin(A-B)/sinC=右边,等式成立. 展开全文阅读