关于三角函数的我不知道怎么由第二步到第三步

是不是这个
(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))
乘出来
sin^A*sin(A-B)+sin^B*sin(A-B)=sin^A*sin(A+B)-sin^B*sin(A+B)
移项
sin^A*sin(A+B)-sin^A*sin(A-B)=sin^B*sin(A-B)+sin^B*sin(A+B)
所以sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B))

三角形ABC的形状是直角三角形,证明如下:
∵a/simA=b/sinB=2R,
a=sinA*2R,b=sinB*2R,
(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),
等式右边有:
(a^2-b^2)sin(A+B)=sin(A+B)*(a+b)(a-b)
=sin(A+B)*[(sinA+sinB)(sinA-sinB...