设sina=-3/5,sinb=12/13,且a∈(∏,3∏/2),b∈(∏/2,∏),求sin(a-b),cos2a,tanb/2的值

∵sina=-3/5,a∈(∏,3∏/2),
∴cosa=-√(1-sin²a)=-4/5
∵sinb=12/13,b∈(∏/2,∏),
∴cosb=-√(1-sin²b)=-5/13
∴sin(a-b)
=sinacosb-cosasinb
=-3/5*(-5/13)-(-4/5)*12/13
=63/65
cos2a
=1-2sin²a
=1-2*(-3/5)²
=7/25
tan(b/2)
=sin(b/2)/cos(b/2)
=2sin²(b/2)/[2sin(b/2)cos(b/2)]
=(1-cosb)/sinb
=(1+5/13)/(12/13)
=3/2