∵(a-b)²=a²+b²-2|a||b|cos
=4+1+2*2*1*cos60°
=7
∴|a-b|=√7
又∵(a+2b)²=a²+4b²+4|a||b|cos
=4+4*1+4*2*1cos60°
=12
∴|a+2b|=2√3
又∵向量(a-b)(a+2b)
=|a|²-2|b|²+|a||b|cos
=4-2*1+2*1*cos60°
=3
∴cos=向量(a-b)(a+2b)/(|a-b|*|a+2b|)
=3/(√7*2√3)
=√21/14
则向量a-b与a+2b的夹角为：arccos√21/14