sin[(π/4+a)-(3π/4+B)]
=sin(-π/2+a-B)
=-cos(a-B)
所以
原式=-sin[(π/4+a)-(3π/4+B)]
=sin(3π/4+B)cos(π/4+a)-cos(3π/4+B)sin(π/4+a)
因为
兀/4原式=-sin[(π/4+a)-(3π/4+B)]=sin(3π/4+B)cos(π/4+a)-cos(3π/4+B)sin(π/4+a)=5/13×(-3/5)-(-12/13)×(4/5)=33/65