theorem Th5:
  sin(r) = 0 implies r = 2*PI*[\r/(2*PI)/] or r = PI+2*PI*[\r/(2*PI)/]
  proof
    set i = [\r/(2*PI)/];
    assume
A1: sin 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;
    sin w = sin r by A2,COMPLEX2:8;
    then w = 0 or w = PI by A1,A3,A4,COMPTRIG:17;
    hence thesis by A2;
  end;
