[cos^4(π/3+α)]-[cos(π/6-α)]^2=[cos(π/3+α)]^4-[sin(π/3+α)]^4=[cos(π/3+α)]^2-[sin(π/3+α)]^2=cos(2π/3+2α)(sinα-cosα)/(sinα+cosα)=1/3(1-tanα)/(1+tanα)=1/3(1-tanα)/(1+tanα)=2/(1+tanα...額，答案好像是（3-4√3）/10疏忽了一下，改正如下(sinα-cosα)/(sinα+cosα)=1/3(tanα-1)/(tanα+1)=1/3, 1-2/(tanα+1)=1/3, tanα+1=3 tanα=2(cosα)^2=1/5, cos2α=-3/5tan2α=2tanα/[1-(tanα)^2]=-4/3sin2α=4/5cos2π/3=-1/2所以cos(2π/3+2α)=)=(3/10)-(4√3/10)