已知cos(π/4-a)=3/5,sin(5π/4+b)=-12/13,其中π/4<a<3π/4,0<b<π/4,求sin(a+b)的值

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

sin(5π/4+b)
=sin(π+π/4+b)
=-sin(π/4+b)=-12/13
sin(π/4+b)=12/13
0<b<π/4
所以π/4<π/4+b<π/2
所以cos(π/4+b)>0
所以cos(π/4+b)=5/13

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

sin（5π/4+b）= -12/13
=>sin(π+(π/4+b))=-12/13
=>sin(π/4+b)=12/13
=>sin((π/4-a)+(a+b))=12/13
=>sin(π/4-a)cos(a+b)+cos(π/4-a)sin(a+b)
已知cos(π/4-a)=3/5=>sin(π/4-a)=4/5
=>4/5cos(a+b)+3/5sin(a+b)=12/13
又因为cos²(a+b)+sin²(a+b)=1
结果应该很简单了。
下面就自己算吧