reserve th, th1, th2, th3 for Real;

theorem Th16:
  sin(th1)-sin(th2)= 2*(cos((th1+th2)/2)*sin((th1-th2)/2))
proof
  sin(th1)-sin(th2)= sin(th1/2+th1/2)-sin(th2/2+th2/2)
    .= sin(th1/2)*cos(th1/2)+cos(th1/2)*sin(th1/2)-sin(th2/2+th2/2) by
SIN_COS:75
    .= 2*(sin(th1/2)*cos(th1/2)) -(sin(th2/2)*cos(th2/2)+sin(th2/2)*cos(th2/
  2)) by SIN_COS:75
    .= 2*(sin(th1/2)*cos(th1/2)*1-sin(th2/2)*cos(th2/2))
    .= 2*(sin(th1/2)*cos(th1/2)*(cos(th2/2)*cos(th2/2)+sin(th2/2)*sin(th2/2)
  ) -sin(th2/2)*cos(th2/2)*1) by SIN_COS:29
    .= 2*(sin(th1/2)*cos(th1/2)*(cos(th2/2)*cos(th2/2)) +sin(th1/2)*cos(th1/
2)*(sin(th2/2)*sin(th2/2)) -(sin(th2/2)*cos(th2/2)*(cos(th1/2)*cos(th1/2)+sin(
  th1/2)*sin(th1/2)))) by SIN_COS:29
    .= 2*((sin(th1/2)*cos(th2/2)-cos(th1/2)*sin(th2/2)) *(cos(th1/2)*cos(th2
  /2)+-sin(th1/2)*sin(th2/2)))
    .= 2*(sin(th1/2-th2/2)*(cos(th1/2)*cos(th2/2)-sin(th1/2)*sin(th2/2))) by
SIN_COS:82
    .= 2*(sin((th1-th2)/2)*(cos(th1/2+th2/2))) by SIN_COS:75
    .= 2*(cos((th1+th2)/2)*sin((th1-th2)/2));
  hence thesis;
end;
