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Note: In the context of this question, when I say "polynomial", I mean a single variable polynomial with coefficients in $\mathbb{R}$ and one that has $\mathbb{R}$ as the domain and codomain.

Question: Prove that there exists no polynomial $p$ such that $p(\sin(\theta))=\cos(\theta)\forall\theta\in I,$ where $I$ is some non-empty interval of $\mathbb{R}.$

My Attempt: For the sake of contradiction, let such a polynomial exist. Then, define $g(x):=p(x)^2+x^2-1.$ Since $p$ is a polynomial, it follows that $g$ is too. Now, for all $a\in I, g(\sin(a))=0.$ Since the interval is non-empty, this occurs infinitely many times. By the fundamental theorem of algebra, any polynomial that is zero infinitely many times must be zero identically. Hence, $g(x)=0\forall x\in\mathbb{R}.$ This means that $p(x)=±\sqrt{1-x^2}\forall x\in\mathbb{R}.$ Now, $\sqrt{2}\in\mathbb{R},$ but $p(\sqrt{2})=±i\notin\mathbb{R}.$ This is in contradiction with the fact that the codomain of $p$ is $\mathbb{R}.$ Hence, the initial assumption about the existence of such a polynomial we made must be false. This shows that no such polynomial exists.

Is my argument right? I fear I've made some conceptual error that invalidates my entire proof. Can this statement be proven in a different way?

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    $\begingroup$ An easy way out is $p(\sin(0)) = \cos(0) = 1 \neq -1 = \cos(\pi) = p(\sin(\pi))$, but I suppose this is just an oversight of the question statement. $\endgroup$
    – L. F.
    Feb 8, 2023 at 0:13
  • $\begingroup$ @L.F., good catch. I need to edit the question to clarify this. The actual question says "for all $\theta$ in some non-empty interval". Thankfully this doesn't change my proof by much. $\endgroup$
    – aqualubix
    Feb 8, 2023 at 0:16
  • $\begingroup$ @Mike, sorry for my negligence. I didn't mean to annoy, and I'll edit it soon. $\endgroup$
    – aqualubix
    Feb 8, 2023 at 0:22
  • $\begingroup$ @Mike, all good now, man? Tbh, I don't see why you were so annoyed. Mistakes happen, you know? $\endgroup$
    – aqualubix
    Feb 8, 2023 at 0:30
  • $\begingroup$ Mistakes cannot happen if you are designing a bridge :) It is on the person writing to be clear on their question from the beginning! $\endgroup$
    – Mike
    Feb 8, 2023 at 0:32

3 Answers 3

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If $x= \sin t$ and $P(x)= \cos t$ then by the Chain Rule $P'(x) \cos t= -\sin t$ so $P'(x) P(x)=-x$ for some interval of values for $x$. But the degrees of the polynomials on left and right cannot match unless that degree is 1. What happens then?

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  • $\begingroup$ Very nice! I suspected that the argument had something to do with degrees but this was very slick. +1 $\endgroup$
    – Mike
    Feb 8, 2023 at 1:00
  • $\begingroup$ This implies that $p$ is a linear polynomial, which is easy to disprove. Thanks. $\endgroup$
    – aqualubix
    Feb 8, 2023 at 1:08
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By the fundamental theorem of algebra, any polynomial that is zero infinitely many times must be zero identically. Hence, $g(x)=0\forall x\in\mathbb{R}.$ This means that $p(x)=±\sqrt{1-x^2}\forall x\in\mathbb{R}.$ Now, $\sqrt{2}\in\mathbb{R},$ but $p(\sqrt{2})=±i\notin\mathbb{R}.$ This is in contradiction with the fact that the codomain of $p$ is $\mathbb{R}.$

This part of the proof is a little dodgy, and the issue is domains (rather than codomains). You've proven that $g(x) = 0$ for all $x \in \mathbb{R}$ but then you identify $p(x)$ with a function that, regarded as a real function, has domain $[-1, 1]$ (and then you evaluate it outside of its domain), which is... I won't say incorrect, but it's a little awkward and potentially confusing.

I think a cleaner way to argue from here is just to prove that there is no real polynomial $p(x)$ satisfying $p(x)^2 = 1 - x^2$. There are many ways to do this:

  1. $p(x)$ must be a linear polynomial, but then the leading coefficient of $p(x)^2$ must be positive, so can't be $-1$.
  2. $p(\sqrt{2})^2 = 1 - 2 = -1$ (this is essentially your proof but we avoid the complication mentioned above with domains by keeping the square) which is a contradiction.
  3. $1 - x^2 = (1 - x)(1 + x)$ is not a square because it has two distinct roots (this argument proves that $p(x)$ can't be a complex polynomial either).
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  • $\begingroup$ Thanks, this clears a lot of the confusion I had with my argument. Personally, I prefer method $2$ because it's so similar in spirit to my original argument. $\endgroup$
    – aqualubix
    Feb 8, 2023 at 2:40
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Assume WLOG that $I$ is such that $\sin(\theta)$ is either positive or negative for all $\theta \in I$, and that $\cos(\theta)$ is either positive or negative for all $\theta \in I$. So then this partitions into 4 cases:

  1. $\sin(\theta)$ positive for all $\theta \in I$ and $\cos(\theta)$ positive for all $\theta \in I$.

  2. $\sin(\theta)$ positive for all $\theta \in I$ and $\cos(\theta)$ negative for all $\theta \in I$.

  3. $\sin(\theta)$ negative for all $\theta \in I$ and $\cos(\theta)$ positive for all $\theta \in I$.

  4. $\sin(\theta)$ negative for all $\theta \in I$ and $\cos(\theta)$ negative for all $\theta \in I$.

Let us now consider Case 1: $\sin(\theta)$ positive for all $\theta \in I$ and $\cos(\theta)$ positive for all $\theta \in I$. The remaining 3 cases can be handled similarly.

Then for Case 1, let $J$ be the set $\{x:$ there is a $\theta \in I$ such that $\sin(\theta)=x\}$. Then $J$ is a nonempty interval on the reals. Then $\sqrt{1-p^2(x)} = x \ \forall x \in J$, which gives, squaring both sides $1-p^2(x) = x^2 \ \forall x \in J$. Which gives [as $p^2(x)$ is a polynomial and $J$ is infinite cardinality] $p^2(x) = x^2-1$ $\ \forall x \in \mathbb{R}$, which gives deg$(p^2)=$ $2$deg$(p)$ $=$ deg$(x^2-1)$ $=2$, which gives deg$(p)=1$. So then $p$ must be a linear polynomial, which is absurd.

So you were close in your OP.

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  • $\begingroup$ Thanks for your input. Is my original proof right, though? $\endgroup$
    – aqualubix
    Feb 8, 2023 at 1:07
  • $\begingroup$ You were close until the ending. Where there may be problems with your argument is that you concluded that if $p(x)=q(x)$ on an interval; $q$ in your proof $q(x)=\sqrt{1-x^2}$ *not * a polynomial, then $p$ and $q$ must be equal everywhere. [This is true if $q$ is a polynomial, as you noted already.] I'm not sure how the graders will take what you wrote though. $\endgroup$
    – Mike
    Feb 8, 2023 at 1:21
  • $\begingroup$ You could have just concluded that $1-p^2(x) =x^2$, and then [as $x^2$ is a polynomial] that $1-p^2(x)=x^2$ everywhere, and then conclude that $p^2$ has degree $2$ then so $p$ must have degree $1$ and thus be linear. $\endgroup$
    – Mike
    Feb 8, 2023 at 1:25
  • $\begingroup$ yes I now see the potential issues with my proof. I'll keep this sort of degree argument in mind, thanks. $\endgroup$
    – aqualubix
    Feb 8, 2023 at 2:43

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