已知向量m=(根号3sinx/4,1),向量n=(cosx/4,cos^2 x/4) 若向量m垂直向量n,求cos(2π/3-X)

问题描述:

已知向量m=(根号3sinx/4,1),向量n=(cosx/4,cos^2 x/4) 若向量m垂直向量n,求cos(2π/3-X)

m•n=√3sin(x/4)cos(x/4)+cos²(x/4)
=(√3/2)sin(x/2)+(1/2)cos(x/2)+1/2
=cos(x/2-π/3)+1/2
因为向量m垂直向量n,所以cos(x/2-π/3)+1/2=0,
cos(x/2-π/3)=-1/2
所以cos(2π/3-X)=cos(x-2π/3)
= cos2(x/2-π/3)=2 cos²(x/2-π/3)-1
=-1/2.