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I am trying to find a non-constant functions $f(x)$ satisfying $f(x)=f(\pi+x)=f(\pi-x)$.

I tried to use $f(x)=\sin(x)$, $f(x)=\cos(x)$,$f(x)=\sin(x/2)$, $f(x)=\cos(x/2)$,$f(x)=\sin(2x)$, $f(x)=\cos(2x)$, and some other functions involving other trigonometric ratios, also tried to combine some of them together, but I failed.


Your help would be appreciated. THANKS!

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    $\begingroup$ Any (non-constant) periodic even function with period $\pi$ would suffice. (And only these functions work). So the simplest choice is probably $\cos(2x)$. I suspect when you tried $\cos(2x)$ you made a mistake when substituting $\pi-x$ or $\pi+x$. $\endgroup$
    – Fishbane
    Oct 2, 2022 at 9:08
  • $\begingroup$ Given that you say you tried $\cos(2x)$ (and that is an answer) it would be helpful to share what you tried. $\endgroup$ Oct 2, 2022 at 9:14

4 Answers 4

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Let's start by figuring out some properties such a function would have. So firstly, for any $x$ we have that

$$f(-x)=f(\pi+x),$$

from which it follows that

$$f(x)=f(-x),$$

i.e. $f$ is even. Furthermore, as $f(x)=f(x+\pi)$, we must have that $f$ is $\pi$-periodic. So let us try a $\pi$-periodic even function and see if it works. An easy one to check is

$$f(x)=\cos 2x.$$

And indeed then

$$f(x+\pi)=\cos(2x+2\pi)=\cos2x=f(x),$$

and

$$f(\pi-x)=\cos(2\pi-2x)=\cos(-2x)=\cos 2x=f(x).$$

Thus $f(x)=\cos 2x$ works.

EDIT:

Actually, we can use the same check to show that any even $\pi$-periodic function works. Indeed suppose $f$ is even and $\pi$-periodic. Then

$$f(x+\pi)=f(x)$$

and

$$f(\pi-x)=f(-x)=f(x).$$

It follows then that $f$ satisfies $f(x)=f(\pi+x)=f(\pi-x)$ if and only if $f$ is even and $\pi$-periodic.

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$\cos2x$ does work.

$f(x)=f(\pi+x)$ states that the function has period $\pi/n$ for some natural number $n$. $f(x)=f(\pi-x)$ states that the function is even about $\pi/2$, and the two easily imply $f(x)=f(-x)$. $\cos2x$ satisfies both these conditions.

In general, specifying what the function is on $[0,\pi/2]$ will determine the rest of the function by repeated reflection.

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We are looking for functions that are $\pi$-periodic, so later conditions can be simplified to just be even functions. Example could be $$ \cos(2^n x) $$ where $n = 1, 2, 3, \ldots$.

You must have made some mistake when you checked that $\cos(2x)$ does not work.

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The functions you tried, $\sin x$, $\cos x$ have period $2\pi$. When you try adding half the period to $x$, you get $f(x+\pi)=-f(x)$ for these two functions, so it is not them. The functions you are looking for must have period dividing $\pi$. To change a period from $P$ to $P/n$ consider $f(nx)$. So, $\sin (nx)$ and $\cos (nx)$ have period $2\pi/n$ which divides $\pi$ for $n=2k$ even. Now, $\cos(2kx)$ is an even function, giving $f(\pi-x)=f(x-\pi)=f(x)$. So, some examples are $\cos(2kx)$ where $k$ is a positive integer.

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