0

  1. It is possible to set values for $\theta$ such that $|\cot(\theta)| \leq C$ where $C$ is a positive constant?

  2. I think it would be possible to find values for $\theta$ such that $|\sin(\theta)| \geq \delta $? For some $\delta > 0$.

We know that $$ |\cot(\theta)| = \bigg|\dfrac{\cos(\theta)}{\sin(\theta)}\bigg| $$ is an unbounded function and explodes to values $k\pi$ for $k \in \mathbb{N}$. If $|\sin(\theta)| \geq \delta >0$, then

$$\dfrac{1}{|\sin(\theta)|} \leq \dfrac{1}{\delta} \Rightarrow \bigg|\dfrac{\cos(\theta)}{\sin(\theta)}\bigg| \leq \dfrac{|\cos(\theta)|}{\delta} \leq \dfrac{1}{\delta}.$$

Thus, the answer to the first question would be summarized in answering the second question. Using the continuity of the $\sin$ function would be possible, correct? Then I think:

To demonstrate this, we can consider an arbitrary value $\delta > 0.$ If we take the open interval $(\theta- \delta_{1}, \theta+ \delta_{1})$, where $\delta_{1} < \delta$. Then all values of $\sin(\theta)$ within that range will have absolute values greater than or equal to $\delta$. This happens because the function $\sin$ is continuous and has no discontinuity points in the interval $(\theta- \delta_{1}, \theta+ \delta_{1})$.

user253963
  • 659
  • 1
    It is well-known that $|\sin(x)| < |x|$ holds for all real $x \ne 0$, see for example https://math.stackexchange.com/q/71078/42969 . – Martin R May 11 '23 at 13:11
  • @MartinR Yes, but that negates the fact that we don't get delta values so $|\sin(\theta)| \geq \delta$ for some $\delta > 0$? – user253963 May 11 '23 at 13:16
  • Then I don't understand the question. If $|\sin(\theta)|$ is positive then one can of course find a positive $\delta $ such that $|\sin(\theta)| \geq \delta$. – Martin R May 11 '23 at 13:19
  • @MartinR My question is: I know that $|\sin(\theta)| \leq 1$ and that the function $\cot$ is not bounded. But I could choose $\theta$ (any range that $\theta$ is in) so that $|\cot(\theta)| \leq C$ for some constant $C>0$. But I realized (as I did in the question) that I just need to look for $\theta$ so that there is a $\delta$ such that $|\sin(\theta)| \geq \delta$ – user253963 May 11 '23 at 13:25
  • 1
    I don't get your question. The maximal value that $|\sin(x)|$ achieves is $1$, for $x=\pi/2$ for example. And so: no, there is no $\theta$ such that $|\sin(\theta)|\geq\delta$, whenever $\delta>1$. And if $\delta\leq 1$, then yes, there is such $\theta$, for example $\theta=\pi/2$. – freakish May 11 '23 at 13:34
  • 1
    Consider the values of $\cot(\theta)$ and $\sin(\theta)$ when $\theta = \frac\pi2 + 2k\pi$ for integer $k$. That's a lot of values of $\theta.$ But you seem to want something more; it is unclear what or why. Is there some other problem that you were trying to solve with the answer to this question? If so, it would help if you edited that problem into the question and explained how you think the answer to this question would help to solve that problem. – David K May 11 '23 at 13:38
  • Thank you guys. It's pretty clear now. Yes, it is possible, just take $0 < \delta < 1$ and $\theta = \pi/2 + 2k\pi$. – user253963 May 11 '23 at 13:42

0 Answers0