二面角问题 设E为正方体ABCD-A1B1C1D1的棱CC1的中点，求平面AB1E和底面A1B1C1D1所成的角的余弦值

设棱长为a
延长AE,与A1C1的延长线交于F,连续B1F
则平面AB1E和底面A1B1C1D1所成的角即为二面角A-B1F-A1的角.

过C1作C1M垂直B1F交B1F于M,连结EM
则∠EMC1为二面角A-B1F-A1的角

AF=2A1C1=2√2a
B1F²=A1B1²+A1F²-2A1B1*A1Fcos∠B1A1F
B1F=√5a

A1B1/sin∠A1FB1=B1F/sin∠B1A1F
sin∠A1FB1=a*√2/(2√5a)=√10/10
C1M=C1F*sin∠A1FB1=√5/5

tg∠EMC1=C1E/C1M=√5/2
sin²∠EMC1/cos²∠EMC1=5/4
5cos²∠EMC1=4-4cos²∠EMC1
cos²∠EMC1=4/9
cos∠EMC1=2/3