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I am trying to prove that $f(r, \theta)= (\cos\theta, \sin\theta)$ is continuous using an $\varepsilon - \delta$ definition. Here is my work so far.

Let $f(r, \theta) = (\cos\theta, \sin\theta)$ and $(r_0, \theta_0) \in \mathbb{R}^2$. We will show that $f_r$ is continuous at any point $(r_0, \theta_0) \in \mathbb{R}^2$. Given $\varepsilon > 0$, let $\delta = ??$. Then we have that \begin{align*} ||f_r(r, \theta) - f_r(r_0, \theta_0)|| &= ||(\cos\theta, \sin\theta) - (\cos\theta_0, \sin\theta_0)||\\ &= \sqrt{(\cos\theta_0 - \cos\theta)^2 + (\sin \theta_0 - \sin \theta)^2}\\ &= \sqrt {(\cos\theta - \cos\theta_0)^2 + (\sin\theta - \sin\theta_0)^2}\\ &= \sqrt{\cos^2\theta + \cos^2\theta_0 - 2\cos\theta\cos\theta_0 + \sin^2\theta + \sin^2\theta_0 - 2\sin\theta\sin\theta_0}\\ &=\sqrt {2 - 2\cos(\theta - \theta_0)}\\ &= 2\sin\frac {\theta - \theta_0}{2} \text{ (Thanks to Doug's answer)} \end{align*} whenever $||(r, \theta) - (r_0, \theta_0)|| < \delta$. However, the problem is that the $\delta$-inequality is dependent on $r$ and $\theta$, while the $\varepsilon$-inequality is only dependent on $\theta$. I can't see how to manipulate this so we can use the $\delta$-inequality to complete the proof.

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    $\begingroup$ @OolongMilktea A small correction: not all metrics on $\mathbb R^2$ induce the same topology, but all metrics induced by norms do. $\endgroup$ Commented Nov 8, 2022 at 1:48
  • $\begingroup$ @EthanMartin You're right. I'll Delete mine since there are answers. $\endgroup$ Commented Nov 8, 2022 at 2:00

2 Answers 2

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$\sqrt {(\cos\theta - \cos\theta_0)^2 + (\sin\theta - \sin\theta_0)^2}\\ \sqrt{\cos^2\theta + \cos^2\theta_0 - 2\cos\theta\cos\theta_0 + \sin^2\theta + \sin^2\theta_0 - 2\sin\theta\sin\theta_0}\\ \sqrt {2 - 2\cos(\theta - \theta_0)}\\ 2\sin\frac {\theta - \theta_0}{2}$

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  • $\begingroup$ How did you go from the 3rd to the fouth line? $\endgroup$ Commented Nov 8, 2022 at 1:51
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    $\begingroup$ @ClydeKertzer he has applied the identity $2\sin^{2}(x) = 1 - \cos(2x)$. $\endgroup$ Commented Nov 8, 2022 at 1:52
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    $\begingroup$ Half-angle identity. $\endgroup$
    – user317176
    Commented Nov 8, 2022 at 1:53
  • $\begingroup$ Okay, so we chose $\delta = \varepsilon/2$, correct? $\endgroup$ Commented Nov 8, 2022 at 1:55
  • $\begingroup$ $2\sin \frac {\theta - \theta_0}{2} < \theta - \theta_0 < \epsilon$ What metric have you chosen for $|d(r,\theta) - d(r_0,\theta_0)|$? If you want to be simple-minded and take $|d(r,\theta) - d(r_0,\theta_0)| = |r-r_0| + |\theta - \theta_0|$ then when $\delta < \epsilon$ then $|f(r,\theta) - f(r_0,\theta_0)| < \epsilon$ $\endgroup$
    – user317176
    Commented Nov 8, 2022 at 2:00
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First, are you sure the map is not $f(r,\theta) = (r\cos \theta, r\sin \theta)$?

Anyway my suggestion is to use the following inequality:

$$\|f(r,\theta) - f(r_0,\theta_0)\| = \|(\cos \theta, \sin \theta) - (\cos\theta_0, \sin \theta_0)\| \leq \|(\cos \theta,\sin \theta) - (\cos\theta_0,\sin \theta)\| + \|(\cos \theta_0 , \sin \theta) - (\cos \theta_0 , \sin \theta_0)\|$$

Now since $\|(x,y)\| = \sqrt{x^2 + y^2} \leq \sqrt{2} \max\{|x|,|y|\}$ it suffices to show that $|\cos (\theta) - \cos(\theta_0)|$ and $|\sin(\theta)-\sin(\theta_0)|$ are small.

Here you can either use the $\varepsilon-\delta$ proof continuity of $\cos,\sin$ which I am sure you already know.

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  • $\begingroup$ Doug, why did you delete your answer? I think it was the most helpful. $\endgroup$ Commented Nov 8, 2022 at 1:46
  • $\begingroup$ Sorry, I thought I was commenting on the original post, and not an answer to that post. I will delete my comments. $\endgroup$
    – user317176
    Commented Nov 8, 2022 at 1:50

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