已知sin[2A-B]=3/5，sinB=-12/13,且A∈[π/2,π],B∈[-π/2，0],则sinA=？

解:
由于A∈[π/2,π],B∈[-π/2，0]
则:2A-B属于[π,5π/2]
又sin(2A-B)=3/5>0
则:2A-B属于[2π,5π/2]
则:cos(2A-B)>0
则:cos(2A-B)
=根号[1-sin^2(2A-B)]
=4/5

又sinB=-12/13,B∈[-π/2，0]
则:cosB
=根号[1-sin^2(B)]
=5/13

则:
sin2A
=sin[(2A-B)+B]
=sin(2A-B)cosB+sinBcos(2A-B)
=-33/65

由于
sin(2A-B)
=sin2AcosB-sinBcos2A
=3/5
则:cos2A=56/65

又A∈[π/2,π]
则:sinA>0
则:
1-2sin^2(A)=cos2A=56/65
解得:
sinA=(3根号130)/130