Question

已知cos(a+b)=4/5,cos(a-b)=-4/5,a+b∈(3π/2,2π),a-b∈(2/π,π),求sin2a,sin2b

Answer 1

cos(a-b)=-4/5,a-b∈(π/2,π)
sin(a-b)=3/5
cos(a+b)=4/5
a+b∈(3π/2,2π)
sin(a+b)=-3/5

sin2a
=sin[(a+b)+(a-b)]
=sin(a+b)cos(a-b)+cos(a+b)sin(a-b)
=-3/5*(-4/5)+4/5*3/5
=12/25+12/25
=24/25

sin2b
=[(a+b)-(a-b)]
=sin(a+b)cos(a-b)-cos(a+b)sin(a-b)
=-3/5*(-4/5)-4/5*3/5
=12/25-12/25
=0

Answer 2

∵cos(a+b)=4/5，a+b∈(3π/2,2π),cos(a-b)=-4/5,a-b∈(2/π,π),
∴sin﹙a＋b﹚＝－3／5,sin﹙a－b﹚＝3／5
∴sin2a＝sin[﹙a＋b﹚＋﹙a－b﹚]
＝sin﹙a＋b﹚cos(a-b)＋cos(a+b)sin﹙a－b﹚
＝﹙－3／5﹚×﹙－4/5﹚＋﹙4/5﹚×﹙3／5﹚
＝24／25
sin2b＝sin[﹙a＋b﹚－﹙a－b﹚]
＝sin﹙a＋b﹚cos(a-b)－cos(a+b)sin﹙a－b﹚
＝﹙－3／5﹚×﹙－4/5﹚－﹙4/5﹚×﹙3／5﹚
＝0