theorem Th13:
  cos (PI/4) = sqrt 2/2
proof
A1: sqrt 2 > 0 by SQUARE_1:25;
  cos ((PI/2)/2) = sqrt((1+cos(PI/2))/2) by Lm3,Th12
    .= 1 /sqrt 2 by SIN_COS:77,SQUARE_1:18,30
    .= (sqrt(2) * 1)/(sqrt(2)*sqrt(2)) by A1,XCMPLX_1:91
    .= (sqrt(2) * 1)/(sqrt(2))^2;
  hence thesis by SQUARE_1:def 2;
end;
