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I'm reading Barbeau's Polynomials, and there's an exercise where he considers:

$\sin3(\arcsin \, t) \: (-1\leq t\leq 1)\tag{1} $

$\cos 4(\arccos \, t) \: (-1\leq t\leq 1) \tag{2}$

As polynomials, but he doesn't consider:

$\sin 2(\arcsin \, t) \: (-1\leq t\leq 1) \tag{3}$

As a polynomial. I've evaluated them on Mathematica but it gave me:

$\sin 3 (\arcsin\,t)\tag{1}$

$\cos 4 (\arccos\,t)\tag{2}$

$\sin 2 (\arcsin\,t)\tag{1}$

Then I tried to evaluate with a Table function:

Table[Sin[3] ArcSin[t], {t, -1, 1}] Table[Cos[4] ArcCos[t], {t, -1, 1}] Table[Sin[2] ArcSin[t], {t, -1, 1}] 

And then I got three values for each:

$\{-\frac{1}{2}\pi \sin(3) ,0\, ,\frac{1}{2}\sin(3)\}\tag{1}$

$\{\pi \cos (4),\frac{1}{2} \pi \cos (4),0\}\tag{2}$

$\{-\frac{1}{2} \pi \sin (2),\, 0,\, \frac{1}{2} \pi \sin (2)\}\tag{3}$

I tried to do this trying to reveal some possible polynomialicity on it, but I'm as lost as I was in the beginning. For what reason the first and second are polynomials while the third isn't?

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    There is no issue of consideration involved. Once you've fixed the definition of a polynomial, there is a fact of the matter whether or not a given function is a polynomial; nobody has any choice in the matter. The first two functions you describe _are_ polynomials, and the third function is _not._2012-09-08
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    What you mean with *nobody has any choice in the matter*?2012-09-08
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    I think the key is that your expression is $\sin (3\arcsin t)$ and not $(\sin 3)\arcsin t$ as you seem to have interpreted it. Some unhelpful notation on the part of your textbook, perhaps.2012-09-08
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    @Gustavo: I mean that once we've fixed the definition of a polynomial, we don't get to _consider_ a function as being a polynomial or not. It either is or it isn't.2012-09-08
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    @BenMillwood Yep. I've been warned about it by Jonathan.2012-09-08
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    @GustavoBandeira: If you examine the Euler formula for $ e^{i \arcsin t}$ (see below), I think it is straightforward to see why or why not the expression is or is not a polynomial.2012-09-08
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    @GustavoBandeira : functions (1) and (2) have domain $[-1,1]$, so one could dispute whether they are polynomials (which have domain at least $\mathbb{R}$).2013-11-21

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