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Suppose $-\frac{\pi}{2}+\epsilon\leq\theta\leq\frac{\pi}{2}-\epsilon$ for some $\epsilon>0$ and $x,y\in\mathbb{R}$. I am wondering whether there exists $\delta>0$ such that:

$$\left|\tan^{-1}\left(\frac{\sin(\theta)+x}{\cos(\theta)+y}\right)-\theta\right|\leq \left\{\begin{array}{ll}f(x,y)&~\text{if}~|x|,|y|\leq\delta\\c_2+f(x,y)&\text{otherwise}\end{array}\right.,$$

where $f(x,y)=|x\cos(\theta)-y\sin(\theta)+c(x^2+y^2)|$ and $c_1,c_2\geq0$ are constants.

Basically, I am wondering does there exist a small region around the origin where the error in approximating $|\theta|\leq\frac{\pi}{2}-\epsilon$ from the noisy in-phase and quadrature components $\sin(\theta)+x$ and $\cos(\theta)+y$ such that it 1) is governed by $f(x,y)$ if $(x,y)$ is in that region; and 2) is governed by a constant plus $f(x,y)$ everywhere else.

The two-dimensional Taylor series expansion of $\tan^{-1}\left(\frac{\sin(\theta)+x}{\cos(\theta)+y}\right)$ can be used to show that the first assertion is true, since $f(x,y)$ is essentially the first term (linear) and the remainder (quadratic) in this expansion, and it converges for $|x|,|y|$ small enough.

I am having trouble with the second assertion. It seems true by inspections of the plots, with $c_2\leq\pi$. Unfortunately, I have no idea how to prove this. Can anyone help?

This is related to my previous question.

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    $\begingroup$ Have you tried reformulating the problem in terms of polar coordinates? $f(r,\theta)=\vert r\cos(2\theta)+c r^2\vert$. $\endgroup$ – John Wayland Bales Jun 29 '16 at 4:41
  • $\begingroup$ @JohnWaylandBales Yes, except that I couldn't figure out what to do with the LHS. $\endgroup$ – M.B.M. Jun 29 '16 at 4:46
  • $\begingroup$ Yes, it would be nice if that denominator were $\cos\theta-y$, then the LHS would equal $\vert\tan^{-1}r\vert$. $\endgroup$ – John Wayland Bales Jun 29 '16 at 4:56
  • $\begingroup$ $\tan^{-1}\left(\dfrac{\tan A+\tan B}{1-\tan A\tan B}\right)=A+B$ so$\tan^{-1}\left(\dfrac{\tan\theta+r}{1-r\tan\theta}\right)=\tan^{-1}\left( \dfrac{ \tan \theta+\tan(\tan^{-1}r)}{1-\tan\theta\tan(\tan^{-1} r)}\right)=\theta+\tan^{-1}( r)$, then $\theta$ is subtracted. $\endgroup$ – John Wayland Bales Jun 29 '16 at 5:15
  • $\begingroup$ Had a devil of a time getting that formatted right. Whew! $\endgroup$ – John Wayland Bales Jun 29 '16 at 5:18

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