Question

已知向量a=(cosa.sina).向量b=(cosB.sinB).绝对值向量a-向量b=（二倍根号5）除以5，求cos(a-B)的值

Answer 1

因为a=(cosa,sina),b=(cosB,sinB)
所以a-b=(cosa-cosB,sina-sinB)

所以|a-b|^2=(cosa-cosB)^2+(sina-sinB)^2
=2-2(cosacosB+sinasinB)
=2-2cos(a-B)
=(2√5/5)^2
=4/5

所以cos(a-B)=(2-4/5)/2=3/5

Answer 2

a-b=(cosA-cosB sinA-sinB)
|a-b|²=(cosA-coaB)²+(sinA-sinB)²
=2-2cos(A-B)
=4/5
cos(A-B)=3/5.

Answer 3

向量a-向量b=(cosa-cosb,sina-sinb)
| 向量a-向量b |=(cosa-cosb)^2+(sina-sinb)^2=2-2*cosa*cosb-2*sina*sinb=2-2*cos(a-b)=2√5/5
cos(a-b)=1-√5/5

Answer 4

（向量a-向量b)²=a²+b²-2a·b=(绝对值向量a-向量b)²=4/5
∴1+1-2（cosacosB+sinasinB)=4/5
∴cosacosB+sinasinB=3/5
cos(a-B)=3/5

Answer 5

（cosa-cosb)^2+(sina-sinb)^2=4/5
(cosa)^2-2cosacosb+(cosb)^2+(sina)^2-2sinasinb+(sinb)^2=4/5
2-2cos(a-b)=4/5
cos(a-b)=3/5得解

Answer 6

解：（a，b为向量，α，β为角）|a|-|b|=2 √5 /5
两边平方得a²-2ab+b²=4/5
√cos²α+sin²α -2（cosα×cosβ+sinα×sinβ）+ √cos²β+sin²β =4/5
1-2cos(α-β)+1=4/5
cos(α-β)=3/5