已知cos(a-p/2)=-5/13,sin(a/2-p)=4/5且0

根据题意,知a-p/2∈（0,π）,a/2-p∈（0,π/2）,所以sin(a-p/2)=12/13,cos(a/2-p)=3/5.
sin[(a+p)/2]=sin[(a-p/2)-(a/2-p)]
=sin(a-p/2)cos(a/2-p)-cos(a-p/2)sin(a/2-p)
=12/13*3/5-(-5/13)*4/5
=56/65
cos[(a+p)/2]=cos[(a-p/2)-(a/2-p)]
=cos(a-p/2)cos(a/2-p)+sin(a-p/2)sin(a/2-p)
=(-5/13)*3/5+12/13*4/5
=33/65
sin(a+p)=2sin[(a+p)/2]cos[(a+p)/2]
=2*56/65*33/65
=3696/4225