reserve x,x0, r, s, h for Real,

  n for Element of NAT,
  rr, y for set,
  Z for open Subset of REAL,

  f, f1, f2 for PartFunc of REAL,REAL;

theorem
  -1 <= r & r <= 1 implies arctan r = -arctan(-r)
proof
A1: [.-PI/4,PI/4.] c= ].-PI/2,PI/2.[ by Lm7,Lm8,XXREAL_2:def 12;
  assume that
A2: -1 <= r and
A3: r <= 1;
A4: -r >= -1 by A3,XREAL_1:24;
A5: --1 >= -r by A2,XREAL_1:24;
  then
A6: arctan(-r) <= PI/4 by A4,Th63;
  -PI/4 <= arctan(-r) by A5,A4,Th63;
  then
A7: -arctan(-r) <= --PI/4 by XREAL_1:24;
  arctan(-r) <= PI/4 by A5,A4,Th63;
  then -PI/4 <= -arctan(-r) by XREAL_1:24;
  then
A8: -arctan(-r) in [.-PI/4,PI/4.] by A7,XXREAL_1:1;
  -PI/4 <= arctan(-r) by A5,A4,Th63;
  then arctan(-r) in [.-PI/4,PI/4.] by A6,XXREAL_1:1;
  then
A9: cos arctan(-r) <> 0 by A1,COMPTRIG:11;
  tan arctan(-r) = -r by A5,A4,Th51;
  then
A10: r = (tan 0 - tan arctan(-r))/(1 + (tan 0) * tan arctan(-r)) by Th41
    .= tan(0 - arctan(-r)) by A9,SIN_COS:31,SIN_COS4:8;
A11: [.-PI/4,PI/4.] c= ].-PI/2,PI/2.[ by Lm7,Lm8,XXREAL_2:def 12;
  then
A12: -arctan(-r) < PI/2 by A8,XXREAL_1:4;
  -PI/2 < -arctan(-r) by A8,A11,XXREAL_1:4;
  hence thesis by A10,A12,Th35;
end;
