sinx+siny+sinz=0,cosx+cosy+cosz=0,求cos（y-z）的值.

• .1.参考答案-1
  siny+sinz=-sinx①
  cosy+cosz=-cosx②
  ①²+②²得:sin²y+sin²z+2sinysinz+cos²y+cos²z+2cosycosz=sin²x+cos²x
  1+1+2sinysinz+2cosycosz=1
  2cos(y-z)=-1
  cos(y-z)=-½
• .2.参考答案-2
  关键是把x消去
  sinx=-siny-sinz
  cosx=-cosy-cosz
  cos²x+sin²x=(cosy+cosz)²+(siny+sinz)²
  =cos²y+cos²z+2cosycosz+sin²y+sin²z+2sinysinz
  =2+2(cosycosz+sinysinz)
  =2+2cos(y-z)=1
  cos(y-z)=-1/2