# $\cos(\theta_n) \to \cos(\theta)$ and $\sin(\theta_n) \to \sin(\theta)$. How to show that $\theta_n \to \theta$?

$\cos(\theta_n) \to \cos(\theta)$ and $\sin(\theta_n) \to \sin(\theta)$ where $\theta_n$ , $\theta \in (-\pi, \pi)$.

How to show that $\theta_n \to \theta$ ?

Can I use the continuity of $\cos^{-1}$ or $\sin^{-1}$ (only one of them) to get the result?

If not, then how to proceed?

Edit:

As we know, range of $\arcsin$ is $[-\pi/2, \pi/2]$, we can use continuity of $\arcsin$ to get $\theta_n \to \theta$ when they belongs to $[-\pi/2, \pi/2]$, in other cases use continuity of $\arccos$.

Is this argument correct?

• Note that your result only holds $\mod 2\pi$ ; otherwise one can just take e.g. $\theta_n=2n\pi$. – Steven Stadnicki Sep 6 '17 at 4:35
• It is easy to see that $e^{i\theta_n}\rightarrow e^{i\theta}$. – MAN-MADE Sep 6 '17 at 4:38
• @MANMAID, Yeah, that was the thing i got, and from there I need to prove that $\theta_n \to \theta$ – User Sep 6 '17 at 4:39
• Why $\arc$ doesn't work? – Nosrati Sep 6 '17 at 4:56
• As we know, range of $\arcsin$ is $[-\pi/2, \pi/2]$, we can use continuity of $\arcsin$ to get $\theta_n \to \theta$ when they belongs to $[-\pi/2, \pi/2]$, in other cases use continuity of $\arccos$. – User Sep 6 '17 at 5:28

Consider the map $f \colon (-\pi,\pi) \rightarrow \mathbb{R}^2$ given by $f(\theta) = (\cos(\theta), \sin(\theta))$. It maps the interval $(-\pi,\pi)$ bijectively onto the the unit circle minus the point $(-1,0)$ and you can explicitly write the inverse as

$$f^{-1}(x,y)= 2 \arctan \left( \frac{y}{1 + x} \right)$$

(see the entry on atan2 in wikipedia).

From this expression, it is clear that $f^{-1}$ is continuous and so if $f(\theta_n) \to (\cos \theta, \sin \theta)$ then $$f^{-1}(f(\theta_n)) = \theta_n \to f^{-1}(f(\theta)) = \theta.$$

• can you please help me to understand why we cannot use continuity of arccos and arcsin directly? – User Sep 6 '17 at 6:07
• And can you please check my attempts in the under the edit section? – User Sep 6 '17 at 6:08
• @User: You can, but it is more messy. Let's say that $\theta \in (0,\pi)$. The problem is that you don't know a priori (even though its true) that for large enough $n$ we have $\theta_n \in (0,\pi)$. Assuming you can justify that, you can apply $\arccos$ and deduce that $\theta_n \to \theta$. But you only know that $\arccos(\cos \theta_n)) = (-1)^{a_n} \theta_n$ where $a_n \in \{ 0, 1 \}$ depending on whether $\theta_n \in [0,\pi)$ or $\theta_n \in (-\pi,0]$. – levap Sep 6 '17 at 6:14
• A priori, $\theta_n$ can belong to $[0,\pi)$ for $n$ even say and to $(-\pi,0]$ for $n$ odd. You can use the fact that also $\sin \theta_n \to \sin \theta$ to deduce that this in fact can't happen because $\sin \theta > 0$ so $\sin \theta_n > 0$ for large enough $n$ and so in fact $\theta_n \in (0,\pi)$ and the argument goes through. Then you can do something analogous for $(-\pi,0)$ and also cover the $0$ case but this is somewhat annoying to write. – levap Sep 6 '17 at 6:18
• So even if you assume that $\theta \in (0,\pi)$, the continuity of $\arccos$ alone will not suffice to deduce the result - you need to use both the continuity of $\arccos$ and $\arcsin$. Expressing everything with $\arctan$ avoids the problem of dividing into cases and using the continuity of two functions. – levap Sep 6 '17 at 6:25

Suppose on the contrary that $\theta_{n}$ does not tend to $\theta$. Then there is an $\epsilon>0$ such that for any given positive integer $n$ there is a positive integer $m>n$ such that $|\theta_{m} - \theta|>\epsilon$. Note that taking a smaller $\epsilon$ does not affect the truth of the above statement so we can assume $\epsilon <\pi+|\theta|$ without any loss of generality.

Then we can see that $$2-2\cos(\theta_{n}-\theta) =(\cos\theta_{n} - \cos\theta) ^{2}+(\sin\theta_{n} - \sin\theta)^{2}\to 0$$ ie $\cos(|\theta_{n} - \theta|) \to 1$. Now from the last paragraph there are infinitely many values of $n$ for which $0<\epsilon<|\theta_{n} - \theta|<\pi+|\theta |<2\pi$ and thus $\cos(|\theta_{n} - \theta|)$ stays away from $1$ for infinitely many values of $n$ which is contrary to the fact that $\cos(|\theta_{n} - \theta|) \to 1$.

To expand on "stays away from $1$" in previous paragraph just consider the function $f(x)=1-\cos x$ on interval $[a, b] \subset (0,2\pi)$. The minimum value of $f$ on this interval is $\min(f(a), f(b)) >0$ (verify this).

Your approach using inverse trigonometric functions will work only if the variables $\theta_{n}, \theta$ simultaneously lie in the range of these inverse functions. In this manner we can't cover the interval $(-\pi, \pi)$ of length $2\pi$ which is given in question.

Hint: For large $n$, $|e^{i\theta_n}-e^{i\theta} | = |e^{i(\theta_n-\theta)}-1|<\epsilon$, for given $\epsilon$. It's almost immediate now.

• Why?${}{}{}{}$. – Nosrati Sep 6 '17 at 4:46
• explain${}{}{}$ – MAN-MADE Sep 6 '17 at 4:53

Let $p(t) = (\cos t, \sin t)$. If $x \in D= S^1 \setminus \{(-1,0)\}$, then there is a unique $t$ with $|t| < \pi$ such that $x= p(t)$ and we define $\arg x = t$. The goal is to show that $\arg$ is continuous on $D$.

Suppose $x\in D$ is such that $x_2 \neq 0$ and $\theta$ (with $|\theta| < \pi$) is such that $x_2 = \sin \theta$. Since $\cos' \theta \neq 0$, the inverse function theorem shows that there is some neighbourhood of $x_1$ and a continuous (in fact $C^1$) function $\alpha$ defined in this neighbourhood such that $\alpha(\cos t) = t$ (this is only valid for $t$ in a suitable neighbourhood of $\theta$). In particular, we have $\arg x = \alpha(x_1)$, and hence $\arg$ is continuous at $x \in D$ when $x_2 \neq 0$.

A similar argument applies for $x \in D$ such that $x_1 \neq 0$.

Observe that any convergent subsequence $\{\theta_{n_{k}}\}_{k=1}^{\infty}$ satisfies $\theta_{n_{k}}\rightarrow \theta$ as $k\rightarrow \infty$.

Thus, as the limit superior and the limit inferior are the same, $\theta_{n}\rightarrow \theta$ as $n\rightarrow \infty$.

• you can write observations, hints in comment section. If you want to write an answer then give full explanation, please. – MAN-MADE Sep 6 '17 at 5:20
• @MANMAID, that is basically almost the full solution. – Indrayudh Roy Sep 6 '17 at 15:39