sinx+siny+sinz=0,cosx+cosy+cosz=0,求cos（y-z）的值.
问题描述：
sinx+siny+sinz=0,cosx+cosy+cosz=0,求cos（y-z）的值.
答
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
答
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