已知0＜β＜π/4＜α＜3π/4,cos(π/4-α)=3/5,sina(3π/4+β)=5/13,求sina(α+β)的值

0＜β＜π/4＜α＜3π/4
可知-
-π/2<π/4-α<0
cos(π/4-α)=3/5，
则sin(π/4-α)<0
则sin(π/4-α)=-根号[1-（cos(π/4-α)）^2]=-4/5
3π/4<3π/4+β<π
sin(3π/4+β)=5/13
则cos(3π/4+β)<0
cos(3π/4+β)=-12/13
而sin(α+β)=-cos（3π/4+β-（π/4-α））=-cos（α+β+π/2）
而
cos（3π/4+β-（π/4-α））
=cos(3π/4+β)cos(π/4-α)+sin(π/4-α)sin(3π/4+β)
=-12/13*3/5-4/5*5/13
=-56/65
所以sin(α+β)=56/65

π/4<α<3π/4 ，cos(π/4+α)=-3/5 …… sin(π/4+α)=4/5

0<β<π/4， sin(3π/4+β)=5/13…… cos(3π/4+β)= - 12/13

sina(α+β)= - sin（α+β+π）=- sin（(π/4+α)+(3π/4+β）)

= - (sin(π/4+α)*cos(3π/4+β) + cos(π/4+α)*sin(3π/4+β) )

=63/65

cos(π/4-α)=cos(α-π/4)=3/5
π/4＜α＜3π/4
所以0＜α-π/4＜π/2
所以sin(α-π/4)>0
由(sin)^2+(cos)^2=1
cos(α-π/4)=3/5
所以sin(α-π/4)=4/5

sina(3π/4+β)=sin[π-(3π/4+β)]=sin(π/4-β)=-sin(β-π/4)=5/13
所以sin(β-π/4)=-5/13
0＜β＜π/4
所以-π/4＜β-π/4＜0
所以cos(β-π/4)>0<