# 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.

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

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

• How did you go from the 3rd to the fouth line? Commented Nov 8, 2022 at 1:51
• @ClydeKertzer he has applied the identity $2\sin^{2}(x) = 1 - \cos(2x)$. Commented Nov 8, 2022 at 1:52
• Half-angle identity. Commented Nov 8, 2022 at 1:53
• Okay, so we chose $\delta = \varepsilon/2$, correct? Commented Nov 8, 2022 at 1:55
• $2\sin \frac {\theta - \theta_0}{2} < \theta - \theta_0 < \epsilon$ 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 $|f(r,\theta) - f(r_0,\theta_0)| < \epsilon$ Commented Nov 8, 2022 at 2:00

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.

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