已知平面向量A=（cosa,sina),B=(cosb,sinb),|A-B|=2根号5/5（1）求cos(a-b)的值（2）0＜a＜π/2,-π/2＜b＜0,且sinb=-5/13,求sina的值

Answer 1

A-B=(cosa-cosb, sina-sinb)
|A-B|^2=(cosa-cosb)^2+(sina-sinb)^2=2-2(cosacosb+sinasinb)=4/5
cosacosb+sinasinb=3/5
cos(a-b)=cosacosb+sinasinb=3/5
-π/20sin(a-b)=√(1-[sin(a-b)]^2)=4/5
sinb=-5/13,
cosb=√[1-(sinb)^2]=12/13
sina=sin[(a-b)+b]=sin(a-b)cosb+cos(a-b)sinb=33/65