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Are there any polynomials $P\in\mathbb R[x]$, other than $P(x)=x^2$, such that $$ P(\sin x) + P(\cos x) = 1,\quad x\in\mathbb R? $$

Notice that there are pairs of polynomials $P,Q$ with $P(\sin x)+Q(\cos x)=1$: say, from $\sin 3x=3\sin x-4\sin^3 x$ and $\cos 3x=4\cos^3 x-3\cos x$ it follows that $$ (3\sin x-4\sin^3 x)^2+(4\cos^3 x-3\cos x)^2=1. $$

My motivation for this question is a pure curiosity; hope it counts.

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    $\begingroup$ Certainly, $P(x) = \frac12$ will do $\endgroup$
    – gt6989b
    Dec 18, 2019 at 15:54
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    $\begingroup$ There's a whole family of those solutions @MatthewTowers: $ax^2 + \frac{1-a}{2}$ will work. $\endgroup$ Dec 18, 2019 at 15:58
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    $\begingroup$ Apparently solved here: artofproblemsolving.com/community/c6h129406p733854 : The general solution is $(x^2-1/2) R((x^2-1/2)^2) + 1/2$ for an arbitrary polynomial $R$. $\endgroup$
    – Martin R
    Dec 18, 2019 at 16:08
  • $\begingroup$ @MartinR: amazing. How did you figure it out? Is there any search engine to check if a problem appears somewhere in the AoPS? $\endgroup$
    – W-t-P
    Dec 18, 2019 at 16:19
  • $\begingroup$ @W-t-P: Found with Approach0 $\endgroup$
    – Martin R
    Dec 18, 2019 at 16:36

2 Answers 2

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This has been solved on Art of Problem Solving: Polynomial - TUYMAADA-2000.

Assume that $P$ is a polynomial satisfying $$ \tag{*} \forall x \in \Bbb R: P(\sin x) + P(\cos x) = 1 \, . $$

  • First show that $P(-y) = P(y)$ for all $y= \sin(x) \in [-1, 1]$ and conclude that $P$ is even, i.e. $P(x) = Q(x^2)$ for some polynomial $Q$.

  • Then show that $Q(\frac 12 +y) + Q(\frac 12 - y) = 1$ for all $y = \sin^2(x) - \frac 12 \in [-1/2, 1/2]$, and conclude that $Q(\frac 12 +y) - \frac 12$ is odd, i.e. $Q(\frac 12 +y) - \frac 12 = y R(y^2)$ for some polynomial $R$.

It follows that $$ P(x) = (x^2- \frac 12) R((x^2 - \frac 12)^2) + \frac 12 $$ for a polynomial $R$. Conversely, every such polynomial satisfies $(*)$.

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I'll answer for $P(\sin^2x)+P(\cos^2x)=1.$

With $t:=\sin^2x-\frac12$, we have the functional equation $$P\left(\frac12+t\right)+P\left(\frac12-t\right)=1,$$

so that $Q(t):=P(t+\frac12)+\frac12$ is an odd function of $t$, such as a polynomial with odd terms.

With $Q(t)=t$, $P(t)=t-\frac12+\frac12=t$.

With $Q(t)=t^3$, $P(t)=(t-\frac12)^3+\frac12=t^3-\frac32t^2+\frac34t+\frac38.$

And so on.

Also, with $Q(t)=16t^3-3t$, $P(\sin^2x)=\sin^2x(3-4\sin^2x)^2$.

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