PI in ].0 ,4.[ by SIN_COS:def 32;
then A1: 0 < PI by XXREAL_1:4;
then PI / 4 < PI / 1 by XREAL_1:78;
hence arccos ((sqrt 2) / 2) = PI / 4 by A1, Th13, SIN_COS6:94; :: thesis: verum