已知sinx+siny+sinz=0,cosx+cosy+cosz=0,求证cos(x-y)=cos(y-z)=cos(z-x)

问题描述:

已知sinx+siny+sinz=0,cosx+cosy+cosz=0,求证cos(x-y)=cos(y-z)=cos(z-x)

由sinx+siny+sinz=0可得(sinx+siny)²=(-sinz)²,展开为sin²x+sin²y+2sinxsiny=sin²z (1)同理cosx+cosy+cosz=0可得cos²x+cos²y+2cosxcosy=cos²z (2)(1)(2)式相加可得2+2(cosx...