|a-b|=1,故：|a-b|^2=(a-b)·(a-b)=|a|^2+|b|^2-2a·b=4+|b|^2-2a·b=1即：a·b=(|b|^2+3)/2,而：a·b=|a|*|b|*cos,故：cos=a·b/(2|b|)=(1/4)(3/|b|+|b|)≥sqrt(3)/2,故：cos∈[0,π/6]