已知sin(a-β)=4/5,sin(a+β)=-12/13,a-β∈(π/2,π),a+β∈(3π/2,2π),求sin2α,cos2β.

α-β∈(π/2,π) cos(α-β)0
cos(α+β)=√[1-sin²(α+β)]=√[1-(-12/13)²]=5/13
cos(2α)=cos[(α+β)+(α-β)]
=cos(α+β)cos(α-β)-sin(α+β)sin(α-β)
=(5/13)(-3/5)-(-12/13)(4/5)
=33/65
cos(2β)=cos[(α+β)-(α-β)]
=cos(α+β)cos(α-β)+sin(α+β)sin(α-β)
=(5/13)(-3/5)+(-12/13)(4/5)
=-63/65�����sin2����cos2��sin(2��)=sin[(��+��)+(��-��)]=sin(��+��)cos(��-��)+cos(��+��)sin(��-��)=(-12/13)(-3/5)+(5/13)(4/5)=56/65