已知Ａ(cosα,sinα),Ｂ(cos(2π/3+α),sin(2π/3+α)),Ｃ(cos(4π/3+α),sin(4π/3+α)),Ｏ(0,0)求证：向量ＯＡ+向量ＯＢ+向量ＯＣ＝０
热心网友
因为cosα+cos(α+2π/3)+cos(α+4π/3)=cosα+cosαcos(2π/3)-sinαsin(2π/3)+cosαcos(4π/3)-sinαsin(4π/3)=cosα-(1/2)cosα-(√3/2)sinα-(1/2)cosα+(√3/2)sinα=0sinα+sin(α+2π/3)+sin(α+4π/3)=sinα+sinαcos(2π/3)+cosαsin(2π/3)+sinαcos(4π/3)+cosαsin(4π/3)=sinα-(1/2)sinα+(√3/2)cosα-(1/2)sinα-(√3/2)cosα=0所以向量ＯＡ+向量ＯＢ+向量ＯＣ=(cosα,sinα)+(cos(2π/3+α),sin(2π/3+α))+(cos(4π/3+α),sin(4π/3+α))=（cosα+cos(α+2π/3)+cos(α+4π/3)，sinα+sin(α+2π/3)+sin(α+4π/3)）=（0，0）=0（零向量）。