I have $n$ data points all of which are measurements of angles which I know to be increasing linearly with equal spacing. I am trying to find the best fit line for this data. Researching it, it seems the Von Mises distribution is most commonly used for linear-circular data, but I can't find any papers that find the best fit line so I am attempting it myself. This is what I have so far:
We want to fit the data to the equation:
$\theta = \alpha + \beta i$
Since the data is evenly spaced we are using $i$ as the x-values and to make the spacing $=1$ and have the points evenly spaced around $0$
So we want to minimise the mean square error:
$$Q'\left( {\alpha ,\beta } \right) = \sum\limits_{i = - \frac{{n - 1}}{2}}^{\frac{{n - 1}}{2}} {{{\left( {\cos \left( {\alpha + \beta i} \right) - \cos \left( {{\theta _i}} \right)} \right)}^2} + {{\left( {\sin \left( {\alpha + \beta i} \right) - \sin \left( {{\theta _i}} \right)} \right)}^2}}$$
$${\left( {\cos \left( {\alpha + \beta i} \right) - \cos \left( {{\theta _i}} \right)} \right)^2} + {\left( {\sin \left( {\alpha + \beta i} \right) - \sin \left( {{\theta _i}} \right)} \right)^2} = 4{\sin ^2}\frac{{\left( {\alpha + i\beta - {\theta _i}} \right)}}{2}\;$$
So we can minimise:
$$Q\left( {\alpha ,\beta } \right) = \sum\limits_{i = - \frac{{n - 1}}{2}}^{\frac{{n - 1}}{2}} {{{\sin }^2}\frac{{\left( {\alpha + i\beta - {\theta _i}} \right)}}{2}}$$
So we can set $$\frac{{dQ}}{{d\alpha }} = \frac{1}{2}\sum\limits_{i = - \frac{{n - 1}}{2}}^{\frac{{n - 1}}{2}} {\sin {\left( {\alpha + i\beta - {\theta _i}} \right)}}=0$$
and
$$\frac{{dQ}}{{d\beta }} = \frac{1}{2}\sum\limits_{i = - \frac{{n - 1}}{2}}^{\frac{{n - 1}}{2}} i {\sin {\left( {\alpha + i\beta - {\theta _i}} \right)}}=0$$
and solve for $\alpha$ and $\beta$
But that is as far as I get.
How can we solve these equations given the measurements $\theta_i$?
Note: simple linear regression actually works very well when the differences between successive angles is less than $π$ but it gets trickier when the difference exceeds $π$, especially if the standard deviation of the measurement error is large.