已知π/2
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已知π/2已知π/2已知π/2∵π/2(1)sin2a=sin[(a+b)+(a-b)]=sin(a+b)cos(a-b)+sin(a-b)cos(a+b)因为π/2sin2a=-56/65(2)si
已知π/2 已知π/2 (1)sin2a=sin[(a+b)+(a-b)]=sin(a+b)cos(a-b)+sin(a-b)cos(a+b) 由a,b的范围知
已知π/2
∵π/2
因为π/2
(2)sin2b=sin[(a+b)-(a-b)]=sin(a+b)cos(a-b)-sin(a-b)cos(a+b)
所以sin2b=-16/65
(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