Examples of non-constant functions $f(x)$ such that $f(x)=f(\pi+x)=f(\pi-x)$

Question

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!

Answer 1

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.

Answer 2

$\cos2x$ 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.

Answer 3

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.

Answer 4

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.