sinx+siny+sinz=0;cosx+cosy+cosz=0;求cos(x-y)

问题描述:

sinx+siny+sinz=0;cosx+cosy+cosz=0;求cos(x-y)

sinx+siny=-sinzcosx+cosy=-cosz平方相加sin²x+cos²x+sin²y+cos²y+2(cosx+cosy+sinxsiny)=sin²z+cos²z1+1+2cos(x-y)=1所以cos(x-y)=-1/2