theorem Th5:
  -2 * PI < r < 0 implies sin (r/2) <> 0
  proof
    assume
A1: -2 * PI < r < 0;
    assume
A2: sin(r/2) = 0;
    0 * (-1) < r * (-1) & r * (-1) < (-2 * PI) * (-1) by A1,XREAL_1:69;
    then
A3: sin ((-r)/2) <> 0 by Th4;
    reconsider r0 = r/2 as Real;
    sin (-r0) = - sin (r0) by SIN_COS:31;
    hence contradiction by A2,A3;
  end;
