已知向量m=(根号3sinx/4,1),向量n=(cosx/4,cos^ x/4) 若向量m*n=1,求cos（2π/3-x)的值
问题描述：
已知向量m=(根号3sinx/4,1),向量n=(cosx/4,cos^ x/4) 若向量m*n=1,求cos（2π/3-x)的值
答
m.n=1
(√3sin(x/4),1). (cos(x/4),(cos(x/4))^2)=1
√3sin(x/4). (cos(x/4)+ (cos(x/4))^2 =1
(√3/2)sin(x/2) +( cos(x/2) +1) /2=1
((√3/2)sin(x/2)+(1/2)cosx/2) = 1/2
cos(π/3 -x/2) = 1/2
[cos(π/3 -x/2)]^2 = 1/4
(cos(2π/3 -x) +1)/2 = 1/4
cos(2π/3 -x) = -1/2