2.因为,2{cos[(π/4)-(x/2)]}^2=cos2[(π/4)-(x/2)] 1=sinx 1xtan[(π/4)-(x/2)]=xsin[(π/4)-(x/2)]/cos[(π/4)-(x/2)]=2xsin[(π/4)-(x/2)]*cos[(π/4)-(x/2)]/2{cos[(π/4)-(x/2)]}^2=xsin2[(π/4)-(x/2)]/(sinx 1)=xcosx/(sinx 1)所以,原式=(sinx 1){3cosx/2{cos[(π/4)-1)]}^2-2tan[(π/4)-1]}=(sinx 1)[3cosx/(sinx 1)-2cosx/(sinx 1)]=3cosx-2cosx=cosx