如图,在正方体ABCD-A'B'C'D'中,E,F分别是棱AA',BB'的中点,求A'F与D'E所成角的余弦值.

取CC‘中点G,连结D'G、FG和EG因为在正方形BCC'B'中,点F、G分别是BB'、CC'的中点所以有：FG//B'C',FG=B'C'又A'D'//B'C',A'D'=B'C'那么：FG//A'D',FG=A'D'所以四边形A'D'GF是平行四边形所以：A'F//D'G则可知∠ED'G就是A'F与D'E所成角（或其补角）令正方体棱长为a,那么易得：D'E=D'G=(√5)a/2,EG=A'C'=√2*a所以在△D'EG中,由余弦定理得：cos∠ED'G=(D'E²+D'G²-EG²)/(2*D'E*D'G)=(5a²/4 +5a²/4 - 2a²)/[2*(√5)a/2 *(√5)a/2]=(a²/4)/(5a²/2)=1/10即A'F与D'E所成角的余弦值为1/10 再问： 参考答案里面貌似是五分之一 = = 再答： 抱歉！算错了，更正： cos∠ED'G=(D'E²+D'G²-EG²)/(2*D'E*D'G) =(5a²/4 +5a²/4 - 2a²)/[2*(√5)a/2 *(√5)a/2] =(a²/2)/(5a²/2) =1/5 即A'F与D'E所成角的余弦值为1/5