Matlab curve fitting toolbox acting weird

I am entering v as x variable and G as y variable in matlab toolbox and fitting it using sum of sines, no. of terms =2

v=-90:90;

B=65;

G=(v./B).^2;

I am getting output shown below. This shows a great fit. But when I write the equation using coefficients given by matlab myself, the graph I get is not the same!!!

fitresult(x) =  a1*sin(b1*x+c1) + a2*sin(b2*x+c2)
     Coefficients (with 95% confidence bounds):
       a1 =       504.9  (-6.058e+08, 6.058e+08)
       b1 =     0.01277  (-26.03, 26.06)
       c1 =       1.571  (1.525, 1.617)
       a2 =         505  (-6.058e+08, 6.058e+08)
       b2 =     0.01281  (-25.95, 25.98)
       c2 =      -1.571  (-1.617, -1.525)

[Walter Roberson]

Any time you see fitting coefficient bounds that cross 0 and are roughly the same positive and negative, then you should understand that the curve fitting toolbox was unable to decide on the appropriate sign of the coefficient.

Your two sub-expressions are symmetric to each other, so MATLAB cannot decide between A*sin(B*x+C) + D*sin(E*x+F) compared to D*sin(E*x+F)+A*sin(B*x+C) and therefore the 95% confidence bounds will be of little value, reflecting the full range for both coefficients. Likewise, because sin(-x) = -sin(x), then A*sin(B*x+C) = (-A)*sin((-B)*x+(-C)) and so curve fitting cannot decide between the signs of the terms.

You should be adding in constraints to make the coefficient order unique. For example you can pin down a1 and a2 to be positive while letting b1, b2 be positive or negative, and you can pin down c1 and c2 to be in the range +/- pi . You would also prefer to make a1 <= a2

[Walter Roberson]

But when I write the equation using coefficients given by matlab myself

How are you doing that? Remember that you are only seeing about 4 decimal places. You would want to use coeffvalues(fitresult) to get the full values.

[Haneya Qureshi]

Problem got resolved! Thank you so much! It was the number of decimal places!!! I was taking rounded off values by copy pasting directly from the curve fitting toolbox. Thanks a lot!

[Haneya Qureshi]

Very valuable insights given by you! Much appreciated!