sin(pi/6-a)=1/3,则cos(2pi/3+2a)的值为

∵sin(π/6-α)=1/3∴cos[2(π/6-α)]=1-sin²(π/6-α)=7/9即cos(π/3-2α)=7/9cos(2π/3+2α)=cos[2(π/3+α)]=cos[2(π/2-π/6+α)]=cos[π-(π/3-2α)]= -cos(π/3-2α)=-7/9