Question

(a2+b2)sin(A-B)=(a2-b2)sinC,判断三角形形状

Answer 1

(a^2+b^2)sin(A-B)=(a^2-b^2)sinC,

(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)

sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B))

sin^A*2cosAsinB=sin^B*2sinAcosB

sin^A*2cosAsinB-sin^B*2sinAcosB=0

sinAsinB(sin2A-sin2B)=0

sin2A=sin2B

2A=2B 或2A+2B=180度

A=B或A+B=90度

故△ABC是等腰三角形或直角三角形