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.

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