问几道高一数学题~~·~麻烦朋友帮忙解答一下

cos(α-β)=12/13， sin(α+β)=-3/5， π/2<β<α<3π/4
->pai/2<a<3pai/4,-3pai/4<-b<-pai/2,
-pai/4<a-b<pai/4,pai<a+b<3pai/2
sin(a-b)=5/13,cos(a+b)=-4/5

sin2α
=sin(a-b+a+b)
=sin(a-b)cos(a+b)+cos(a-b)sin(a+b)
=5/13*(-4/5)+12/13*(-3/5)
=-56/65

(2).已知0<β<π/4, π/4<α<3π/4, cos(π/4-α)=3/5, sin(β+3π/4)=5/13
求sin(α+β)
0<β<π/4, π/4<α<3π/4
3pai/4<b+3pai/4<pai,-3pai/4<-a<-pai/4,-pai/2<pai/4-a<0
sin(pai/4-a)=-4/5,cos(b+3pai/4)=-12/13

sin(a+b+pai/2)=sin(b+3pai/4-pai/4+a)
=sin(b+3pai/4)cos(pai/4-a)-sin(pai/4-a）cos(b+3pai/4)

=5/13*3/5-(-4/5)*(-12/13)
=-33/65
sin(a+b)>0
sin(a+b)=根号（1－（－33/65)^2)(这一步自己算一下）

1.sin2α =sin(α-β+α+β)

2.sin(α+β)=sin(β+3π/4-π/4+α-π/2)=-cos(β+3π/4-π/4+α)
其他自己算吧