cos^4π／6-sin^π／6
= (√3/2)^4- (1/2)^4
= 9/16-1/16
= 8/16
= 1/2
f(x)=log2(x^2-x1),
f(2) = log2(2^2-2^1） = log2 3
【方法二】用平方差公式：
cos^4π／6-sin^π／6
= [cos^2(π／6)+sin^2(π／6)] [cos^2(π／6)-sin^2(π／6)]
= 1*cos(π/3)
= 1/2
f(x)=log2(x^2-x1),
f(2) = log2(2^2-2^1）= log2[(2+1)(2-1)] = log2 3