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

Question

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

Answer 1

sin(π/6-a)cos(π/3+b)+sin(π/3+a)cos(π/6-b)=12/13
cos[π/2-(π/6-a)]sin[π/2-(π/3+b)]+sin(π/3+a)cos(π/6-b)=12/13
cos(π/3+a)sin(π/6-b)+sin(π/3+a)cos(π/6-b)=12/13
sin[(π/3+a))+(π/6-b)]=12/13
sin(a-b+π/2)=12/13
cos(a-b)=12/13
π/2a-b>0
sin(a-b)=√(1-(cos(a-b))^2)=5/13
π/2cos(a+b)=-√(1-(sin(a+b))^2)=-4/5
sin2a
=sin(a+b+a-b)
=sin(a+b)cos(a-b)+cos(a+b)sin(a-b)
=(-3/5)*(12/13)+(-4/5)*(5/13)
=56/65