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Find all polynomials in $f \in \mathbb{R}[x]$ such that $f(\sin x) = f(\cos x)$ for all $x\in \mathbb{R}$.

The best idea I tried was comparing coefficients of $f(t)$ and $f(\sqrt{1-t^2})$ but it's still quite messy.

Update: Finding all $f$ with $f(t^2) = f(1-t^2)$ would be enough, by Sufficient and essential condition for polynomials $P$ and $Q$ to satisfy $P(\sin x)= Q(\cos x)$

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Yup, I just arrived at that.

Make a substitution $t^2=u$, then we have $f(u)$ can be any polynomial that is symmetric around $u=\frac{1}{2}$.

One example could be $a(u)=(u-\frac{1}{2})^2$ which corresponds to $f(t)=(t^2-\frac{1}{2})^2$.

Another example of a non-polynomial function would be $f(t)=|t^2-\frac{1}{2}|$.

Any polynomial that is a linear combination of even powers of $(t^2-\frac{1}{2})$ will work.

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    $\begingroup$ $f(t) = |t^2 - \frac{1}{2}|$ wouldn't be a polynomial, though. $\endgroup$ – Daniel Schepler Jun 7 at 0:10
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    $\begingroup$ Ah, my bad. @DanielSchepler I think I'll keep it and mention it's not a polynomial $\endgroup$ – Saketh Malyala Jun 7 at 0:17

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