Calculation giving me complex numbers exen though it shlouldn't

Hi.

I have the following code:

%angle in degrees
alpha = [28.8 41.4 68.4 90 117 126 133];
%angle in radians
alpha1 = (alpha./180).*pi;

%calculation:
I_RMS = 947.5 .* sqrt( (2./pi ).*( pi - cos( pi ).*sin( pi ) - alpha1 + cos( alpha1 ).*sin( alpha1 ) )./2 );

EDIT: the above code doesn't deliver complex numbers after all (I had retyped it to shorten my question): Here is exactly what I am trying to do (I have also changed the model-function so it's easier to read):

   alpha1g = [28.8 41.4 68.4 90 117 126 133];
    alpha1 = (alpha1g./180).*pi;
    Ieff1 = [930 920 820 660 400 350 270];
    %preparation for nlinfit:
    Modelfun1 = @(I01,xx)(I01(1).*sqrt((1/pi).*(pi-xx+cos(xx).*sin(xx))));
    I01 = 947.5;
    I02 = 947.5;
    I03 = 800;

%create fit
  TI1 = nlinfit(alpha1, Ieff1, Modelfun1, I01);
  alpha = linspace(0,pi,50000);
  Trend1 = Modelfun1(TI1, alpha);

    alpha = linspace(0,180,50000);

%plot original scatter and fit over alpha  
  scatter(alpha1g, Ieff1, 'r', 'filled'); hold on
  plot(Trend1, alpha, 'r', 'LineWidth', 1.5);

This returns the message:

Warning: Imaginary parts of complex X and/or Y arguments ignored 
> In LE_TRIAC at 52

The fitted plot does not fit at all.

[Azzi Abdelmalek]

I'am not getting complex numbers

[Roger Stafford]

That expression of yours looks decidedly fishy. Why have you included sin(pi) which is zero? Why have you multiplied by 2 and later divided by 2 in the argument for 'sqrt'? Perhaps you actually gave matlab a slightly different expression.

[Walter Roberson]

Is there a reason for taking cos(pi)*sin(pi) ? You know the result is going to be 0.

[Marc Jakobi]

No, not really. I just integrated sin²(x) from alpha to pi and multiplied it with 2/pi then took the square root without thinking of what would happen to it in the sines and cosines. Same goes for the multiplying by 2/2 that resulted.

[Marc Jakobi]

Azzi, strange, I just copied and pasted it into the command window, now I'm not getting complex numbers either. It must be something in my function then.

[Walter Roberson]

You mean as in cos(alpha)+1 ?

[Image Analyst]

There are sind() and cosd() versions of the functions that work directly in degrees you know.

[Marc Jakobi]

I have updated the question with more detail...

[Walter Roberson]

What value shows up for TI1 ?

[Marc Jakobi]

TI1 shows up as 938.1222, which is pretty close to I01 (which it is supposed to replace) What do you mean by your question before?

Thanks, Image Analyst, I was wondering if there is something like that.

[Alfonso Nieto-Castanon]

that's probably just a rounding error (sin(pi) is not exactly 0), switching to sind & cosd should get rid of the imaginary parts (sind(180) IS exactly 0)... and I believe you want to use plot(alpha, Trend1, ...), not plot(Trend1, alpha, ...)

[Marc Jakobi]

Thanks Alfonso and Image Analyst! I switched it to sind(180) and left alpha1 in degrees, too. Now it works. Yea, I had the plot order wrong, too.

The fit doesn't quite fit yet, but I think that's because it assumes that the resistance of the load is constant. In reality, it changes with the current (because of the changing temperature), so I will have to find some parameters for a variable resistance that depends on the current's RMS.