已知长方体的一条对角线与共顶点的三条棱所成的角分别是α β γ则（cos^2)α+（cos^2)β+（cos^2) γ=

在长方体ABCD-A'B'C'D'中,BD'是一条体对角线,不妨设∠BD'A'=α,∠BD'D=β,∠BD'C'=γ
联结BA'、BD、BC'
由于D'A'⊥平面AA'B'B,所以D'A'⊥A'B,则在Rt△A'BD'中,(cosα)^2=(A'D'/BD')^2
由于D'D⊥平面ABCD,所以D'D⊥BD,则在Rt△BDD'中,(cosβ)^2=(DD'/BD')^2
由于D'C'⊥平面BB'C'C,所以D'C'⊥BC',则在Rt△BC'D'中,(cosγ)^2=(C'D'/BD')^2

所以(cosα)^2+(cosβ)^2+(cosγ)^2
=(A'D'/BD')^2+(DD'/BD')^2+(C'D'/BD')^2=(A'D'^2+DD'^2+C'D'^2)/BD'^2
而在Rt△BDD'中,BD'^2=BD^2+DD'^2
在Rt△ABD中,BD^2=AB^2+AD^2,所以BD'^2=AB^2+AD^2+DD'^2
即BD'^2=C'D'^2+A'D'^2+DD'^2
所以(cosα)^2+(cosβ)^2+(cosγ)^2==(A'D'^2+DD'^2+C'D'^2)/BD'^2=BD'^2/BD'^2=1