Polynomial equation: $P(\sin t) = P(\cos t)$ Let $P(X)$ be a polynomial with real coefficients such that $P(\sin t) = P(\cos t), \, \forall t \in \mathbb R$. 
Prove that there exists a unique polynomial $Q(Y)$ with real coefficients, such that $P(X) = Q(X^4-X^2)$. 
(The converse is trivially true.)   
NOTE:  This problem has almost nothing to do with trigonometry, despite appearances to the contrary. It's really about polynomials.
Edit:  It occurred to me I should add this - for full disclosure: I created this problem many years ago, for the math Olympiad in the East European country I grew up in; I am not looking for a solution for myself. I am offering it as a fun challenge to the math fans on this forum.
 A: From $P(x)=P(\sqrt{1-x^2})$ we conclude $P$ is comprised of even powers, so $P(x)=L(x^2)$.
Now the condition reads $L(u)=L(1-u)$. Divide $L(u)$ by $u(1-u)$ to obtain quotient $q(u)$ and remainder $r(u)$. Substitute $L(u)=q(u)\,u(1-u)+r(u)$ into $L(u)=L(1-u)$ then reduce modulo $u(1-u)$ to obtain $au+b\equiv a(1-u)+b$, hence $a=0$. Then the functional equation descends to $q(u)=q(1-u)$ for the quotient $q$. Induct on degree to get $L(u)=Q(u(1-u))$.
Therefore $P(x)=L(x^2)=Q(x^2(1-x^2))$.
A: $P(\sin t) = P(\cos t)$ for $t \in \mathbb{R}$ implies that $P(x) = P(\sqrt{1 - x^2})$ for all $x \in [-1, 1]$, because for any $x \in [-1, 1]$ there is $t \in \mathbb{R}$ such that $x = \sin(t)$ and $\sqrt{1-x^2} = \cos(t)$, and so  $P(x) = P(\sin(t)) = P(\cos(t)) = P(\sqrt{1-x^2})$. 
Does it make it any easier for you? I can give you more hints if you need them.
A: We have $$P(-\sin t)=P(\sin t)$$ because $$P(- \sin t)=P(\sin (-t))=P(\cos (-t))=P(\cos t)=P(\sin t)$$ It follows that the polynomial $P$ is an even function since a polynomial of degree $n$ is completely determined by $n+1$ distinct points. Hence $P(x)$ have even exponents only.
On the other hand, it is easy to verify the following double identity:
$$\sin^4 t-\sin^2 t=\cos^4 t-\cos^2 t=-(\sin^2 t \cos^2 t)$$
Because of $P$ is an even function, one has $P(-(\sin x \cos x)^2)=P((\sin x \cos x)^2)$ so (to work with positive but not necessary) 
$$ P((\sin x \cos x)^2)=P(\sin^4 x-\sin^2 x)$$
The function $g(x)=(\sin x\cos x)^2$ is an even function of domain $\mathbb R$ which has a máximum equal to $\frac 14$ at each point $x=\frac{(1+2k)\pi}{4}$ . 
We have 
$$P(g(X))=Q(X)=P(X^4-X^2)\qquad (*)$$ 
The equality of polynomials  $(*)$ is valid for all value $X=\sin t$ so is valid for all $X$.
FINAL NOTE.- We have proved that there is a polynomial $Q(X)$ such that $Q(X)=P(X^4-X^2)$ but the proposition ask about  $P(X)= Q(X^4-X^2)$ Are these two equalities equivalent or is there a typo?. Anyway,the relation $(*)$  is true.
