Question

已知向量a=(cosa,sina),b=(cosb .sinb),|a -b|=(4根号13)/13.

1.求cos（a-b)的值 2.若0<a< π/2，-π/2<b<0,且sinb=-4/5，求sina

Answer 1

1. a-b=(cosa-cosb,sina-sinb)
|a-b|=√[(cosa-cosb)^2+(sina-sinb)^2]=√(2-2(cosacosb+sinasinb))
=√(2-2cos（a-b))
=4√13/13
2-2cos(a-b)=16/13
1-cos(a-b)=8/13
cos(a-b)=5/13
2. 若0<a< π/2，-π/2<b<0,且sinb=-4/5， cosb=3/5
0<a-b<π sin(a-b)=12/13
sina=sin[(a-b)+b]
=sin(a-b)cosb+cos(a-b)sinb
=12/13*3/5-5/13*(-4/5)
=16/65

Answer 2

1 (a-b)^2=16/13
故a^2-2ab+b^2=16/13
又a^2=1,b^2=1 ab=cosacosb+sinasinb=cos(a-b)
代入上式得cos(a-b)=5/13
2 因为0<a<π/2，-π/2<b<0
故0<a-b<π，又cos(a-b)>0
所以0<a-b<π/2 则sin(a-b)=12/13
由-π/2<b<0得cosb>0
cosb=3/5
sina=sin(a-b+b)=sin(a-b)cosb+cos(a-b)sinb
=12/13*3/5+5/13*(-4/5)=16/65

Answer 3

1、|a -b|=(4√13)/13,|a -b|²=16/13=|a|²+|b|²-2|a||b|cosθ=1+1-2|a||b|cosθ
|a||b|cosθ=5/13，cos（a-b)=cosacosb+sinasinb=a*b=|a||b|cosθ=5/13
2、容易求的sin(a-b)=12/13，cosb=3/5
sina=sin(a-b+b)=sin(a-b)cosb+cos(a-b)sinb=12/13*3/5+5/13*(-4/5)=16/65