若sin(π/6-a)cos(π/3+b)+sin(π/3+a)cos(π/6-b)=12/13,sin(a+b)=-3/5,且π/2
问题描述：
若sin(π/6-a)cos(π/3+b)+sin(π/3+a)cos(π/6-b)=12/13,sin(a+b)=-3/5,且π/2
答
sin(π/6-a)cos(π/3+b)+sin(π/3+a)cos(π/6-b)=12/13cos[π/2-(π/6-a)]sin[π/2-(π/3+b)]+sin(π/3+a)cos(π/6-b)=12/13cos(π/3+a)sin(π/6-b)+sin(π/3+a)cos(π/6-b)=12/13sin[(π/3+a))+(π/6-b)]=12/13sin(...