Question

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

Answer 1

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.