证明：sinasin(a+2b)-sinbsin(b+2a)=sin(a+b)sin(a-b)
<==>sinasin(a+b+b)-sinbsin(a+b+a)=sin(a+b)sin(a-b)
<==>sinasin(a+b)cosb+sinacos(a+b)sinb-sinbsin(a+b)cosa-sinbcos(a+b)sina=sin(a+b)sin(a-b)
<==>sin(a+b)(sinacosb-cosasinb)=sin(a+b)sin(a-b)
<==>sin(a+b)sin(a-b)=sin(a+b)sin(a-b)恒成立
以上各步可逆,证毕