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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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