How significant is the difference between averaging angles and averaging unit vectors? I understand that numerically averaging angles (which I will call the $shortcut$ average for convenience) is in general going to produce a very different result than converting to cartesian unit coordinates, averaging those values, converting to polar coordinates, and then getting the argument (which I will call the $correct$ average). I will call the difference between these results the $error$.
However, my intuition tells me that the closer the angles are numerically, the closer the $shortcut$ average of those angles is to the $correct$ average. The extreme case being points on a line, where the 'angle' between them is proportional to their straight-line distance, and the $error$ is $0$.
I would like to know how significant the $error$ is between the $correct$ and $shortcut$ averages, depending on how numerically close the angles are, and, specifically, how significant the $error$ is when the angles are within a half turn of each other (but without crossing $0$).

Motivation
As suggested by the way I've named the methods above, the $shortcut$ method is simpler to perform and seems (by inspection) to produce intuitive results in some cases (i.e. when the angles are close); however I do not have the intuition to find examples to determine if the values produced by this method are exactly correct, nearly correct, or actually very incorrect.
Additionally, if it is correct for values within a half turn from each other, I wonder how this can be explained rigorously.

Appendix A : Shortcut method
Directly averaging the angles:
$\bar\phi={1 \over n}\sum_{i=1}^{n}\phi_i$

Appendix B : Correct method
Converting to the corresponding point on the unit circle, finding the average point, then finding the angle of this point:
$\bar\phi=arg({1 \over n} \sum_{k=1}^{n}e^{i\phi_k})$
 A: I assume that by "converting to Cartesian coordinates" you mean constructing the vector $(\cos\theta,\sin\theta)$, or equivalently the complex number $\exp i\theta,$ for an angle $\theta$. Then your question is equivalent to comparing the two angles $\theta_s = \langle\theta\rangle$ and $\theta_c = \arg\langle\exp i\theta\rangle$, where the angle brackets denotes the average over all angles $\theta$.
To start with, let's get rid of the $\arg$ by converting to Cartesian coordinates once again, yielding $z_s=\exp i\langle\theta\rangle$ versus $z_c=\langle\exp i\theta\rangle$: How does the direction of the mean angle compare to the mean direction of the angles? Of course, we should remember to ignore the magnitudes of the vectors in this case, since it is only the direction that matters in the end.
Without loss of generality, we can shift all the angles so that their mean is zero. Then $z_s=\exp0=1$. One simple observation is that if the distribution of angles is symmetrical about the mean, i.e. for every $\theta$ there is also a $-\theta$, then $z_c$ is also purely real, and so their directions agree. Thus for a symmetrical distribution there is no difference between the two quantities.
For a general distribution, we can use the Taylor series of $\exp$, yielding
$$\begin{align}
z_c &= \left\langle 1+i\theta + \frac{i^2}{2!}\theta^2 + \frac{i^3}{3!}\theta^3 + \cdots\right\rangle \\
&= 1 + i\langle\theta\rangle - \frac{1}{2!}\langle\theta^2\rangle - \frac{i}{3!}\langle\theta^3\rangle + \cdots \\
&= \left(1 - \frac12\langle\theta^2\rangle + \cdots\right) + i\left(-\frac16\langle\theta^3\rangle + \cdots\right).
\end{align}$$
If the angles $\theta$ are small, the higher-order terms may be neglected, and we see that the directions of $z_c$ and $z_s$ differ by approximately an angle $\frac{\langle\theta^3\rangle/6}{1 - \langle\theta^2\rangle/2}$. This is related to the skewness of the distribution, $\gamma_1 = \langle\theta^3\rangle/\langle\theta^2\rangle^{3/2}$, which indeed measures the deviation from symmetry.
