a+b=(cosA+cosB,sinA+sinB)
a-b=(cosA-cosB, sinA-sinB)

|a+b|²=(cosA+cosB)²+(sinA+sinB)²
=2+2(cosAcosB+sinAsinB)
=2+2cos(A-B)

|a-b|²=(cosA-cosB)²+(sinA-sinB)²
=2-2cos(A-B)

由题设|a+b|=2|a-b|可得：
2+2cos(A-B)=8-8cos(A-B)
10cos(A-B)=6
∴cos(A-B)=3/52+2cos(A-B)=8-8cos(A-B)后面为什么是8|a+b|=2|a-b|的两边平方。|a+b|²=4|a-b|²