Is the number of tetrahedral voids always equal to double the number of octahedral voids in any crystal?
Note that, in case of tetrahedral voids, each sphere is in contact with three tetrahedral sites above, and three tetrahedral sites below, hence, there are two tetrahedral sites associated with each sphere. Thus,

Number of tetrahedral voids $(N_{th}) = 2 \times$ (Rank of unit cell)



Or alternatively, if you divide a FCC lattice into 8 octants, each octant will have a tetrahedral site at it's body centre.



For octahedral sites, each sphere is associated with one octahedral sites, thus, 

Number of octahedral voids $(N_o) =$ Rank of unit cell

For FCC, besides the body center, there is one site at the center of each of the 12 edges. It is surrounded by 6 atoms, three belonging to the same cell unit cell, and the three belonging to two adjacent unit cells.

For clarity see:



For HCP,