Question

请求知道数学题的详细解答过程，谢谢！

Answer 1

π/2<A<3π/4
π/2<B<3π/4
所以π<A+B<3π/2
所以cos(A+B)<0
sin(A+B)=-3/5
所以cos(A+B)=-4/5

-3π/4<-B<-π/2
π/2<A<3π/4
所以-π/4<A-B<π/4
A>B,
所以0<A-B<π/4
sin(A-B)>0
cos(A-B)=12/13
所以sin(A-B)=5/13

sin2A=sin[(A+B)+(A-B)]
=sin(A+B)cos(A-B)+cos(A+B)sin(A-B)
=-56/65

Answer 2

解：
∵π/2<β<α<3π/4
∴π<β+α<3π/2
∴cos(β+α)<0
故cos(β+α)=-√(1-9/25)=-4/5
又∵0<α-β<π/4
故sin(α-β)=√(1-144/169)=5/13
则有：
sin2α=sin[(α+β)+(α-β)]
=sin(α+β)cos(α-β)+sin(α-β)cos(α+β)
=(-3/5)(12/13)+(5/13)(-4/5)
=-56/65

Answer 3

π/2<β<α<3π/4,第二象限
π<α+β<3π/2第三象限
0<α-β<π/4第一象限

cos( α-β)=12/13==> sin( α-β)=5/13,
sin(α+β)=-3/5==>cos(α+β)=-4/5

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

cos2β
=cos(α+β - α-β)
=cos(α+β)cos(α-β)+sin(α+β)sin(α-β)
=-4/5*12/13 -3/5*5/13
=-63/65