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A trigonometric polynomial ( INMO 2020 Problem 2)

Indian National Math Olympiad (INMO 2020) Solution and sequential hints to problem 2

Suppose $P(x)$ is a polynomial with real coefficients, satisfying the condition $P(\cos \theta+\sin \theta)=P(\cos \theta-\sin \theta)$, for every real $\theta$. Prove that $P(x)$ can be expressed in the form$$P(x)=a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}$$for some real numbers $a_0, a_1, \dots, a_n$ and non-negative integer $n$.

Using a very standard trigometric identity , we can easily convert the following ,
\begin{align*} P(\cos\theta + \sin\theta) &= P(\cos\theta - \sin\theta) \\ \implies P\left(\sqrt{2}\sin\left(\frac{\pi}{4} + \theta\right)\right) &= P\left(\sqrt{2}\cos\left(\frac{\pi}{4} + \theta\right)\right) \\ \implies P(\sqrt{2}\sin x) &= P(\sqrt{2}\cos x) \\ \end{align*} Assuming ,  $ (\frac{\pi}{4}+\theta) = x$ for all reals \(x\). So,
\[P(-\sqrt{2}\sin(x)) = P(\sqrt{2}\sin(-x)) = P(\sqrt{2}\cos(-x)) = P(\sqrt{2}\cos(x)) = P(\sqrt{2}\sin(x))\]for all \(x\in\mathbb{R}\). Since \(P(x) = P(-x)\) holds for infinitely many \(x\), it must hold for all \(x\) (since \(P(x)\) is a polynomial). so we get that ,  $P(x)$ is a even polynomial .

Also
\[P(\sqrt{2}\cos(x)) = P(\sqrt{2}\sin(x))\] implies that
\[P(t) = P(\sqrt{2}\sin(\cos^{-1}(t/\sqrt{2})))\]putting , $x=\cos^{-1}(t/\sqrt{2})$
for infinitely many \(t\) $\in [-\sqrt2 ,\sqrt2]$.
\[\sqrt{2}\sin(\cos^{-1}(t/\sqrt{2})) = \sqrt{2 - t^2}\]so we get , \(P(x) = P(\sqrt{2-t^2})\)
Again as it is a polynomial function we can extend it all $\mathbb{R} $. And we get , \(P(x) = P(\sqrt{2-x^2})\) for all reals \(x\)
Since \(P(x)\) is even , we can choose a even polynomial $Q(x)$ such that ,\(Q(x) = P(\sqrt{x+1})\). \[P(\sqrt{1+x}) = Q(x) = a_0 + a_1x^2 + a_2x^4 + \cdots + a_nx^{2n}\]now take , $\sqrt{1+x} = y$ and you get the polynomial of required form .

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