已知π/2<b<a<3π/4,cos(a-b)=12/13,sin(a+b)=-3/5,求sin2a,sin2b的值

∵π/2<b<a<3π/4,
∴0<a-b<π/4,π<a+b<3π/2.
∵cos(a-b)=12/13,sin(a+b)=-3/5,
∴sin(a-b)=√(1-cos²(a-b))=5/13,
cos(a+b)=-√(1-sin²(a+b))=-4/5.
故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,
sin2b=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)
=-16/65.

由a,b的范围知
(a-b)属于(0,π/4)
(a+b)属于(π,3π/2)
易得cos(a+b)=-4/5
sin(a-b)=5/13
又知sin2a=sin((a+b)+(a-b))=sin(a+b)cos(a-b)+sin(a-b)cos(a+b)=-56/65
sin2b=sin((a+b)-(a-b))=sin(a+b)cos(a-b)-sin(a-b)cos(a+b)=-16/65

(1)sin2a=sin[(a+b)+(a-b)]=sin(a+b)cos(a-b)+sin(a-b)cos(a+b)
因为π/2<b<a<3π/4，所以sin(a-b)=5/13,cos(a+b)=-4/5.所以
sin2a=-56/65

(2)sin2b=sin[(a+b)-(a-b)]=sin(a+b)cos(a-b)-sin(a-b)cos(a+b)
所以sin2b=-16/65