I know that $\arctan$ is not periodic. How can this be proved?
If you have a periodic function, within a non periodic function, does the overal function become non periodic, say for $\arctan(\cos(x))$, if we have $x+t$ where $t = 2\pi$ then $\arctan(\cos(x+t)) = \arctan(cos(x))$ which suggests it is periodic?
Please could someone provide a comprehensive method for solving this?
Using an online calculator has given me a result that contradicts my intuition above. Symbolab is flawed so I apologise, but I just wanted clarification. I have been racking my brain for ages. I dont think I’ll trust an online calculator again.
Yes as in the comments @eyeballfrog mentioned that you have already proven that it is periodic
Since for any periodic functions f(x), $\exists$ some $T$ such that $f(x+T)=f(x)$
As you said in your description, $f(x)=tan^{-1}[cos(x)]$ and $T=2\pi$, so therefore $tan^{-1}[cos(x)]$ is periodic
If you aren't convinced, here's a graph of $tan^{-1}[cos(x)]$