因为sin(a+b)=-3/5,sin(a-b)=3/5,且a-b∈(π/2,π),a+b∈(3π/2,2π)
所以
cos(a+b)=4/5,cos(a-b)=-4/5,
cos2b=cos[(a+b)-(a-b)]
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=4/5×(-4/5)+(-3/5)×(3/5)
=-16/25-9/25
=-1
因为sin(a+b)=-3/5,sin(a-b)=3/5,且a-b∈(π/2,π),a+b∈(3π/2,2π)
所以
cos(a+b)=4/5,cos(a-b)=-4/5,
cos2b=cos[(a+b)-(a-b)]
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=4/5×(-4/5)+(-3/5)×(3/5)
=-16/25-9/25
=-1