How do I propagate relative uncertainty through atan2? I've got $y = \sin(\theta)$ and $x = \cos(\theta)$ with some relative error on both.  If I compute $\theta$ with $\theta = \operatorname{atan2}(y,x)$, how do I propagate the relative error from the inputs to a relative error on the angle?
 A: The mapping from polar to Cartesian is
$$\binom xy = \binom{r\cos \theta}{r\sin \theta}$$
with total differential
$$
\binom{dx}{dy}
=\binom{\cos\theta\,dr-r\sin\theta\,d\theta}{\sin\theta\,dr+r\cos\theta\,d\theta} 
=\underbrace{\begin{pmatrix}
\cos\theta & -r\sin\theta \\
\sin\theta & r\cos\theta
\end{pmatrix}}_{\textstyle=:M} \binom{dr}{d \theta}
$$
The matrix $M$ encodes how small changes in $r$ and $\theta$ translate to small changes in $x$ and $y$.
If you are mapping the other way, then
$$\binom{dr}{d \theta} = M^{-1}\binom{dx}{dy}$$
So all you have to do is to invert $M$ because using $\operatorname{atan2}(y,x)$ just gives you the $\theta$-coordinate for cartesian $(x,y)$.  Notice that this just gives you the ideal error propagation and does not care for gory detail of your atan2 implementation.
More spacifically,
$$M^{-1} = \frac 1{|M|}\begin{pmatrix}
r\cos\theta & r\sin\theta \\
-\sin\theta & \cos\theta
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta & \sin\theta \\
-\dfrac1r\sin\theta & \dfrac1r\cos\theta
\end{pmatrix}
$$
so that
$$d\theta = \frac1r (-\sin\theta ~~ \cos\theta)
\binom{dx}{dy}
$$
Maybe you find it more useful to express the derivative in terms of $x$ and $y$.  In that case:
$$d\theta = \frac1{x^2+y^2} (-ydx +xdy)$$
See here for example. Thus for $\delta\theta$:
$$\delta\theta^2 = \frac1{(x^2+y^2)^2}(y^2\delta x^2 + x^2\delta y^2)$$
In the case where $\delta x_r=\delta x/x$ denotes the relative error:
$$\delta\theta^2 = \frac{x^2y^2}{(x^2+y^2)^2}(\delta x_r^2 + \delta y_r^2)$$
