<s>[INST] Prove $f(r, \theta) = (\cos\theta, \sin\theta)$ is continuous 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.
 [/INST] $\sqrt {(\cos\theta - \cos\theta_0)^2 + (\sin\theta - \sin\theta_0)^2}\\
\sqrt{\cos^2\theta + \cos^2\theta_0 - 2\cos\theta\cos\theta_0 + \sin^2\theta + \sin^2\theta_0 - 2\sin\theta\sin\theta_0}\\
\sqrt {2 - 2\cos(\theta - \theta_0)}\\
2\sin\frac {\theta - \theta_0}{2}$
</s>