
theorem Thm18:
  for a be Real holds
    sin (3 * a) = 4 * sin a * sin (PI/3 + a) * sin (PI/3 - a)
  proof
    let a be Real;
A1: (sqrt 3 /2)^2 = (sqrt 3 /2) * (sqrt 3 /2) by SQUARE_1:def 1
    .= sqrt 3 * sqrt 3 /2/2
    .= sqrt (3 * 3) /2/2 by SQUARE_1:29
    .= sqrt (3^2) /2/2 by SQUARE_1:def 1
    .= 3 /2/2 by SQUARE_1:22
    .= 3/4;
    sin( 3 * a ) =-4 * (sin(a)) |^3 + 3 * sin(a) by SIN_COS5:16
    .=4*sin(a)*3/4 - 4* (sin(a)) |^ (2+1)
    .=4*sin(a)*3/4 - 4 * ((sin(a)) |^2 * (sin(a))) by NEWTON:6
    .=4*sin(a)*((sqrt 3/2)^2-(sin(a)) |^2) by A1
    .=4*sin(a)*( (sin (PI/3))^2-(sin a )^2) by Thm9,NEWTON:81
    .=4*sin(a)*(((sin (PI/3))-(sin a ))* (sin(PI/3) + sin a)) by SQUARE_1:8
    .=4*sin(a)*( sin (PI/3) + sin a )* (sin (PI/3) - sin a)
    .=4*sin(a)*( 2*(cos(((PI/3)-a)/2)*sin(((PI/3)+a)/2)))*
    (sin (PI/3) - sin a) by SIN_COS4:15
    .=4*sin(a)*( 2*(cos(((PI/3)-a)/2)*sin(((PI/3)+a)/2)))*
    (2*(cos(((PI/3)+a)/2)*sin(((PI/3)-a)/2))) by SIN_COS4:16
    .=4*sin(a)*(2*sin(((PI/3)-a)/2)*cos(((PI/3)-a)/2))*(2*sin(((PI/3)+a)/2)*
    cos(((PI/3)+a)/2))
    .=4*sin(a)*(sin(2*(((PI/3)-a)/2)))*(2*sin(((PI/3)+a)/2)*
    cos(((PI/3)+a)/2)) by SIN_COS5:5
    .=4*sin(a)*(sin(2*(((PI/3)-a)/2)))*(sin(2*(((PI/3)+a)/2))) by SIN_COS5:5
    .=4*sin(a)*sin(((PI/3)-a))*sin(((PI/3)+a));
    hence thesis;
  end;
