Are there polynomials $p(x)$ such that $p(\sin x)=\sin(2x)$ for all $x$?

Question

Are there any polynomials $p(x)$ such that $$p(\sin x)= \sin (2x)\;\;\;\;\; \forall x \in \mathbb{R}\,?$$

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This is what i did: Anyway, thanks for your hint, I think I've found the solution. Here's my reasoning: we would have $|p(t)|=2t\sqrt{1−t^2}$, $\forall t\in\mathbb{R}$, so for the identity principle $|p(t)|=2t\sqrt{1−t^2}$ but this is not a polynomial. Is it right?

Answer 1

Suppose there is. Then we have $p(x)$ such that

$$ (p(\sin x))^2 = 4\sin^2 x (1-\sin^2 x)$$

put $t=\sin x$ then we have $$p(t)^2 = \underbrace{4t^2(1-t^2)}_{q(t)};\;\;\;\;\;\;\forall t\in[-1,1]$$

So $p(t)^2$ and $q(t)$ match for infinite values so they match for all values. So $p(t)$ is of second degree, so $p(t)= at^2+bt+c$ and $a\ne 0$. Now we have

$$ a^2t^4+2abt^3+(2ac+b^2)t^2+2bct+c^2 = -4t^4+4t^2$$ so we have $a^2=-4m$, $2ab=0$, $2ac+b^2 =4$ and $c=0$. From second equation we get $b=0$ and from second $b=\pm 2$, so there is no such polynomial.

Answer 2

Here is a simplification of greedoid's answer, assuming $p$ is a real polynomial.

We have $p(t)^2=4t^2(1-t^2)=q(t)$ for all $t\in[-1,1]$ and so for all $t \in \mathbb R$. But then $$ 0 \le p(2)^2 = q(2) = -12 < 0, $$ a contradiction.

Answer 3

The easiest approach I can think of is to note that $$\sin\left(\frac{\pi}{3}\right)=\sin\left(\frac{2\pi}{3}\right)$$ but $$\sin\left(2\cdot\frac{\pi}{3}\right)\neq \sin\left(2\cdot\frac{2\pi}{3}\right)\,.$$ This means: not only there does not exist a polynomial function $p:\mathbb{R}\to\mathbb{R}$ such that $p\big(\sin(x)\big)=\sin(2x)$ for all $x\in\mathbb{R}$, but you also know that there does not exist any function $p:[-1,+1]\to\mathbb{R}$ at all, polynomial or not, that satisfies $p\big(\sin(x)\big)=\sin(2x)$ for every $x\in\mathbb{R}$.

Answer 4

Let's prove a more general result which states the only polynomial satisfying $P(\sin x)=P(\sin 2x)$ is the constant polynomial.

First we observe the sequence $\sin(\pi/2^n),\sin(\pi/2^{n-1}),\sin(\pi/2^{n-2}),...,\sin(\pi/2)$ are $n$ distinct numbers in an increasing order as $\sin x$ is monotonic on $[0, pi/2]$. If there exists such non-constant polynomial, by the assumption, we have $P(\sin(\pi/2^n))=P(\sin(\pi/2^{n-1})=P(\sin(\pi/2^{n-2})=...=P(\sin(\pi/2))$,by applying the MVT, we can construct $n-1$ distinct numbers(let's call them $a_i$) s.t. $P'(a_i)=0$, where $\sin(\pi/2^i)<a_i<\sin(\pi/2^{i+1}), 1\le i<n$. Which means the degree of polynomial $P'(x)$ is greater than or equal to $n-1$, which gives the degree of $P(x)$ is greater than or equal to $n$. However, the choice of $n$ is arbitrary, hence we have a contradiction.

Answer 5

As an alternative, similar to Batominovski’s answer, we have that

$$p\left(\frac \pi 4\right)= p\left(\frac{\sqrt 2}2\right)=\sin \left(\frac \pi 2\right)=1$$

but

$$p\left(\frac {3\pi}4\right)= p\left(\frac{\sqrt 2}2\right)=\sin \left(\frac{3 \pi }2\right)=-1$$

therefore no polynomial nor function defined on $\mathbb R$ exists with such property.