Prove $f(r, \theta) = (\cos\theta, \sin\theta)$ is continuous

Question

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.

Answer 1

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

Answer 2

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.