
theorem Thm9:
  sin(PI/3) = sqrt 3 /2
  proof
A1: (sin(PI/3)) ^2 + (cos(PI/3)) ^2 =1 by SIN_COS:29;
A2: (cos(PI/3)) ^2 = cos(PI/3) * cos(PI/3) by SQUARE_1:def 1
    .=1/4 by Thm8;
    sin(PI/3) >=0
    proof
A3: (PI/2-(PI/3)) in ]. 0 , PI/2 .[
      proof
        set n=PI/2-(PI/3);
        PI/6 < PI/2 by COMPTRIG:5,XREAL_1:76;
        hence thesis by COMPTRIG:5,XXREAL_1:4;
      end;
      sin(PI/3)=cos(PI/2-(PI/3)) by SIN_COS:79;
      hence thesis by A3,SIN_COS:81;
    end;
    then sin(PI/3) = sqrt (3/4) by A1,A2,SQUARE_1:22;
    hence thesis by SQUARE_1:20,SQUARE_1:30;
  end;
