解∵cosA=4/5且A∈(0.π)∴sinA=3/5由正弦a=bsinA/sinB=(√3*3/5)/(1/2)=6√3/5 sin2A=2sinAcosA=2*3/5*4/5=24/25cos2A=2cos²A-1=2*16/25-1=7/25∴sin(2A-B)=sin2AcosB-cos2AsinB=24/25*√3/2-7/25*1/2=(24√3-7)/50
