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January 20, 2020

A trigonometric polynomial ( INMO 2020 Problem 2)

[et_pb_section fb_built="1" admin_label="Blog Hero" _builder_version="3.22" use_background_color_gradient="on" background_color_gradient_start="rgba(114,114,255,0.24)" background_color_gradient_end="#ffffff" background_blend="multiply" custom_padding="0|0px|0|0px|false|false" animation_style="slide" animation_direction="top" animation_intensity_slide="2%" locked="off"][et_pb_row _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" custom_margin="|||" custom_padding="27px|0px|27px|0px" custom_width_px="1280px"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_text_color="#474ab6" text_line_height="1.9em" background_size="initial" background_position="top_left" background_repeat="repeat" text_orientation="center" max_width="540px" module_alignment="center" locked="off"]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 $n$.

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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 .[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]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 \(x\)[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]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 .[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" admin_label="Footer" _builder_version="3.22" background_color="#f7f8fc" custom_padding="0px|0px|2px|0px|false|false" animation_style="zoom" animation_direction="bottom" animation_intensity_zoom="6%" animation_starting_opacity="100%" saved_tabs="all"][et_pb_row column_structure="1_2,1_4,1_4" use_custom_gutter="on" gutter_width="2" _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" custom_padding="24px|0px|145px|0px|false|false"][et_pb_column type="1_2" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_text_color="#7272ff" header_font="|on|||" header_text_color="#7272ff" header_font_size="36px" header_line_height="1.5em" background_size="initial" background_position="top_left" background_repeat="repeat" custom_margin="||20px|" animation_style="slide" animation_direction="bottom" animation_intensity_slide="10%"]

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