Question

在△ABC中,角A,B,C所对应的边分别为a,b,c,

证明：a²-b²/c²=sin(A-B)/sinC

Answer 1

y证明 右边=sin(A-B)/sinC
=(sinAcosB-cosAsinB)/ sinC
= sinA/ sinC *cosB-cosA*sinB/ sinC
=a/c*cosB-b/c* cosA
=a/c*(a^2+c^2-b^2)/(2ac)- b/c*(b^2+c^2-a^2)/(2bc)
=(a^2+c^2-b^2)/(2c^2)- (b^2+c^2-a^2)/ (2c^2)
=(2a^2-2b^2)/(2c^2)
=(a^2-b^2)/(c^2)=左边

Answer 2

从右边开始简化，展开带公式，内角和180 就可以出结果了