theorem Th6:
  cos(r) = 0 implies r = PI/2+2*PI*[\r/(2*PI)/] or r = 3*PI/2+2*PI*[\r/(2*PI)/]
  proof
    set i = [\r/(2*PI)/];
    assume
A1: cos r = 0;
    consider w being Real such that
A2: w = (2*PI)*(-i)+r and
A3: 0 <= w and
A4: w < 2*PI by COMPLEX2:1;
    cos w = cos r by A2,COMPLEX2:9;
    then w = PI/2 or w = 3*PI/2 by A1,A3,A4,COMPTRIG:18;
    hence thesis by A2;
  end;
