sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则si

sin(a-B)cosa-1/2[sin(2a+B)-sinB]=sin(a-B)cosa-1/2[2cos(a+b)sina]=sin(a-b)cosa-cos(a+b)sina=sinacosbcosa-cosasinbcosa-cosacosbsina+sinasinbsina=sinb(sinasina-sinasina)=-sinbcos2a已知cos(a-π/6)+sina=4√3/5展开得则cosacos(π/6)+sinasin(π/6)+sina=4√3/5即(√3/2)cosa+(3/2)sina=4√3/5两边同除以√3得(√3/2)sina+(1/2)cosa=4/5,由于cos(π/6)=(√3/2),sin(π/6)=1/2。最后再合并得sin(a+π/6)=4/5