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a) Are there polynomials $ p (x) $ satisfying $ p(\sin x) = \sin (2x) \quad\quad \forall x \in \mathbb{R} $ ?

b) An extension of this problem is:

1) If $n$ is even, then there does not exist a polynomial $P$ satisfying $P\left(\sin x\right)=\sin\left(nx\right)$ for all $x\in\mathbb{R}$.

2) If $n$ is odd, then there exists a polynomial $P$ satisfying $P\left(\sin x\right)=\sin\left(nx\right)$ for all $x\in\mathbb{R}$.

3) There exists a polynomial $P$ satisfying $P\left(\cos x\right)=\cos\left(nx\right)$ for all $x\in\mathbb{R}$

Where $n \in \mathbb{N}$

For the non-existence in 1) (and also for a) point) we can simply note that $\sin x=\sin (\pi-x)$ and hence $P(\sin x)=P(\sin (\pi-x))$ but $\sin(n(\pi-x))=-\sin nx$ for even $n$. Contradiction.

But I can’t solve point 2) and 3) can somebody help me?

EditPiAf
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Mat15
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2 Answers2

1

You can easily construct an example for 3).

Take $P(x)=2x^2-1$. Then, $P(\cos x)=2\cos^2x-1=\cos2x$

For 2), You can take $P(x)=3x-4x^3$. So, $P(\sin x)=3\sin x-4\sin^3x=\sin3x$

smanoos
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1

You're looking for Chebyshev polynomials.

flawr
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