sin(a-b)=3/5,sin(a+b)=-3/5,a-b属于(π/2,π)a+b属于(3π/2,2π),则cos2b=?

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sin(a-b)=3/5,sin(a+b)=-3/5,a-b属于(π/2,π)a+b属于(3π/2,2π),则cos2b=?

sin(a-b)=3/5,sin(a+b)=-3/5,a-b属于(π/2,π)a+b属于(3π/2,2π),则cos2b=?
sin(a-b)=3/5,sin(a+b)=-3/5,a-b属于(π/2,π)a+b属于(3π/2,2π),则cos2b=?

sin(a-b)=3/5,sin(a+b)=-3/5,a-b属于(π/2,π)a+b属于(3π/2,2π),则cos2b=?
因为sin(a+b)=-3/5,sin(a-b)=3/5,且a-b∈(π/2,π),a+b∈(3π/2,2π),
则cos(a+b)=4/5,cos(a-b)=-4/5,
所以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.