Question

3 在三角形ABC中，已知（a2+b2）sin（A-B）=（a2-b2）sin(A+B) 求证：ABC是等腰或直角三角形

(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),
(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
其中，sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B)) ，也就是第三步怎么来的

Answer 1

他这是合并同类项
(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)
sin^Asin(A-B)+ sin^Bsin(A-B)=sin^Asin(A+B) -sin^Bsin(A+B)
sin^A*sin(A+B)-sin^Asin(A-B))=sin^B*sin(A-B)+sin^Bsin(A+B)
sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B))