Find all polynomial $p(x)$ such that $p(\cos\theta) = p(\sin\theta)$ for all $\theta\in [0, 2\pi)$ Find all real polynomial $p(x)$ such that $p(\cos\theta) = p(\sin\theta)$ for all $\theta\in [0, 2\pi)$.
This problem comes from a past USA TST, but I couldn't find any solutions online. After spending a lot of time on it, I got nowhere, so if anyone can help, please do so.
I suspect that the only polynomials that are solutions are $p(x) = c$ for some constant $c$.
 A: $f(\theta)=p(\cos\theta)$ is an even function, hence we must have $p(\sin\theta)=p(-\sin\theta)$. That implies that $p(x)$ is a polynomial in $x^2$, $p(x)=q(x^2)$, and
$$ q(\sin^2\theta) = q(\cos^2\theta) = q(1-\sin^2\theta), $$
so $q(z)$ is a polynomial symmetric with respect to $z=\frac{1}{2}$:
$$ q(z) = c_0+c_2\left(z-\frac{1}{2}\right)^2 + \ldots + c_{2m}\left(z-\frac{1}{2}\right)^{2m} $$
and
$$ p(x) = c_0+c_2\left(x^2-\frac{1}{2}\right)^2 + \ldots + c_{2m}\left(x^2-\frac{1}{2}\right)^{2m}.$$
Since $1,z(1-z),z^2(1-z)^2,z^3(1-z)^3,\ldots$ is a base of the space of polynomials symmetric with respect to $z=\frac{1}{2}$, we may re-write the last line in the more compact way:
$$\boxed{\, p(x) = r\!\left(x^2(1-x^2)\right)\quad\text{for some polynomial }r.}$$
It is straightforward to check that this set of polynomials fulfill the given constraints.
A: Partial solution.
Consider polinomial $p(x)=a+cx^2-cx^4$, where $a,c$ - some arbitrary constants.
Then,
$$
p(x)=a+cx^2(1-x^2)\\
p(\sin(\theta))=a+c\sin^2(\theta)\cos^2(\theta)\\
p(\cos(\theta))=a+c\cos^2(\theta)\sin^2(\theta)
$$
Edit
Actually, it easy to see natural generalization:
$$
p(x)=a+cx^{2n}(1-x^2)^n
$$
where $n$ - nonnegative integer.
A: For all $\theta$ we ha
$$
P(\sin(\theta))=P(\cos(\theta))=P(\cos(-\theta))=P(\sin(-\theta))=P(-\sin(\theta))
$$
This shows that $\sin(\theta)$ is a solution to $P(-X)=P(X)$, hence this equation has infinitely many solutions.
This implies that $P$ is an even polynomial. Therefore, there exists a polynomial $Q(X)$ such that
$$P(X)=Q(X^2)$$
The given equation then becomes 
$$Q(\sin^2(\theta))=Q(1-\sin^2(\theta))$$
This shows that $Q(X)=Q(1-X)$ has infinitely many solutions, thus we must have equality 
$$Q(X)=Q(1-X)$$
From the above it follows that 
$$P(X)=Q(X^2) \mbox{ where } Q(X)=Q(1-X)$$
and it is trivial to check that any polynomial satisfying this is a soluton.
We next solve $Q(X)=Q(1-X)$. Let $Y=\frac{1}{2}-X$. Then, the equation becomes 
$$Q(Y)=Q(-Y)$$
which is equivalent to $Q$ being an even polynomial in $Y$.
Therefore, all solutions are of the form $Q(Y)=\sum_{k=0}^n a_{2k}Y^{2k}$.
Going back we get
$$Q(X)=\sum_{k=0}^n a_{2k}(\frac{1}{2}-X)^{2k} \\
P(X)=\sum_{k=0}^n a_{2k}(\frac{1}{2}-X^2)^{2k} $$
Therefore, all the solutions are the polynomials of the form
$$P(X)=\sum_{k=0}^n a_{2k}(\frac{1}{2}-X^2)^{2k} $$
Note that for such a polynomial we indeed have
$$P(\sin(\theta))=\sum_{k=0}^n a_{2k}(\frac{1}{2}-\sin^2(\theta))^{2k} =\sum_{k=0}^n a_{2k}(\cos^2(\theta)-\frac{1}{2})^{2k}=P(\cos(\theta))$$
