reserve n,m for Element of NAT;
reserve h,k,r,r1,r2,x,x0,x1,x2,x3 for Real;
reserve f,f1,f2 for Function of REAL,REAL;
reserve S for Seq_Sequence;

theorem
  (for x holds f.x=1/(sin(x))^2) & sin(x+h/2)<>0 & sin(x-h/2)<>0 implies
  cD(f,h).x = 16*cos(x)*sin((-h)/2)*cos((-h)/2)*sin(x)/((cos(2*x)-cos(h))^2)
proof
  assume that
A1:for x holds f.x=1/(sin(x))^2 and
A2:sin(x+h/2)<>0 & sin(x-h/2)<>0;
  cD(f,h).x = f.(x+h/2) - f.(x-h/2) by DIFF_1:5
    .= 1/(sin(x+h/2))^2 - f.(x-h/2) by A1
    .= 1/(sin(x+h/2))^2 - 1/(sin(x-h/2))^2 by A1
    .= (1*(sin(x-h/2))^2-1*(sin(x+h/2))^2)/((sin(x+h/2))^2*(sin(x-h/2))^2)
                                              by A2,XCMPLX_1:130
    .= ((sin(x-h/2))^2-(sin(x+h/2))^2)/((sin(x+h/2)*sin(x-h/2))^2)
    .= ((sin(x-h/2))^2-(sin(x+h/2))^2)/((-(1/2)*(cos((x+h/2)+(x-h/2))
       -cos((x+h/2)-(x-h/2))))^2) by SIN_COS4:29
    .= ((sin(x-h/2))^2-(sin(x+h/2))^2)/((1/4)*(cos(2*x)-cos(h))^2)
    .= ((sin(x-h/2))^2-(sin(x+h/2))^2)/(1/4)/((cos(2*x)-cos(h))^2)
                                                            by XCMPLX_1:78
    .= 4*((((sin(x-h/2))-(sin(x+h/2)))*((sin(x-h/2))+(sin(x+h/2))))
       /((cos(2*x)-cos(h))^2))
    .= 4*(((2*(cos(((x-h/2)+(x+h/2))/2)*sin(((x-h/2)-(x+h/2))/2)))
       *((sin(x-h/2))+(sin(x+h/2))))/((cos(2*x)-cos(h))^2)) by SIN_COS4:16
    .= 4*(((2*(cos((2*x)/2)*sin((-h)/2)))*(2*(cos((-h)/2)*sin((2*x)/2))))
       /((cos(2*x)-cos(h))^2)) by SIN_COS4:15
    .= 16*cos(x)*sin((-h)/2)*cos((-h)/2)*sin(x)/((cos(2*x)-cos(h))^2);
  hence thesis;
end;
