Then AE⊥CD of BE⊥CD.

∴∠AEB is the plane angle of dihedral angle A-CD-B.

The side length of regular tetrahedron A-BCD is 1.

∴, AE=√3/2 in regular triangle ACD.

In regular triangle BCD, BE=√3/2.

And AB= 1,

∴, cos∠ AEB = (AE 2+BE 2-AB 2)/(2AE * BE) =1/3 in △ABE.

∴ The plane cosine of dihedral angle A-CD-B is 1/3.

Please accept it if you are satisfied, thank you!