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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. – Qiaochu Yuan Sep 8 '12 at 18:25
What you mean with nobody has any choice in the matter? – Voyska Sep 8 '12 at 18:39
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. – Ben Millwood Sep 8 '12 at 18:46
@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. – Qiaochu Yuan Sep 8 '12 at 18:47
@GustavoBandeira : functions (1) and (2) have domain $[-1,1]$, so one could dispute whether they are polynomials (which have domain at least $\mathbb{R}$). – Stefan Smith Nov 21 '13 at 0:25

$$\sin(3x)=-4\sin^3 x+3\sin x$$ and $$\cos(4x)=8\cos^4 x-8\cos^2x+1$$ so, for example, $$\sin(3\arcsin x)=-4x^3+3x.$$

On the other hand, $\sin(2\arcsin x)=2\sin(\arcsin x)\cos(\arcsin x)=2x\sqrt{1-x^2}$.

Also, note that what you were inputting to Mathematica was wrong. It should be Sin[3 ArcSin[x]], etc.

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Where is the $8\cos^4 x-8\cos^2x+1$ coming from? – Voyska Sep 8 '12 at 18:41
About the Mathematica input: Bizarre, is the input you suggested implicit in the polynomials I suggested? – Voyska Sep 8 '12 at 18:43
Oh, it's $\sin (3*\arcsin x)$, It wasn't clear to me. – Voyska Sep 8 '12 at 18:44
Yeah, that's correct. Also, recall $\cos(2x)=2\cos^2(x)-1$ so $\cos(4x)=\cos(2\cdot 2x)=2(2\cos^2 x-1)^2-1$. – Jonathan Sep 8 '12 at 18:45
You say: "recall $\cos(2x)=2\cos^2(x)-1$. I've never seen it before, where did that come from? It seems to be a function factorization. (If this concept exists, I don't know what it is, but the process you employed reminded me of a factorization o the cosine function. – Voyska Sep 8 '12 at 21:11

Another way of looking at it (for suitably restricted $t$):

$$e^{i \arcsin t} = \sqrt{1-t^2} + i t$$ $$e^{i \arccos t} = t + i \sqrt{1-t^2}$$

This gives:

$$\sin (3\arcsin \, t) =\mathrm{Im} (\sqrt{1-t^2} + i t)^3 = 3t-4t^3$$ $$\cos ( 4 \arccos \, t) = \mathrm{Re} (t + i \sqrt{1-t^2})^4 = 8t^4 - 8 t^2 +1 $$ But $$\sin ( 2 \arcsin \, t) =\mathrm{Im} (\sqrt{1-t^2} + i t)^2 = t \sqrt{1-t^2}$$

Sometimes the $\sqrt{1-t^2}$ gets 'squared out', sometimes it doesn't...


Since $e^{i n \arcsin t} = \sum_{k=0}^n \binom{n}{k}(\sqrt{1-t^2})^{n-k}(it)^k = \sum_{k=0, k \, \mathrm{even}}^n (\cdots)+\sum_{k=1, k \, \mathrm{odd}}^n (\cdots)$, and $i^k$ is real iff $k$ is even, we see that $t \mapsto \sin ( n \arcsin t )$ is a polynomial iff $n$ is odd, and $t \mapsto \cos ( n \arcsin t )$ is a polynomial iff $n$ is odd.

Similar considerations show that $t \mapsto \cos ( n \arccos t )$ is always a polynomial and $t \mapsto \sin ( n \arccos t )$ is never a polynomial.

(Proving the 'is not a polynomial' part involves showing that $t \mapsto \sqrt{1-t^2} p(t)$ is not a polynomial if $p$ is a polynomial. This can be done in a manner similar to showing that $\sqrt{2}$ is not rational.)

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The procedure is a little bit cumbersome. You can read about the Chebyshev polynomials of the first kind, $p_n(t)=\cos(n\arccos t)$. Consider $$p(x)=\sin(2\arcsin x)$$

Let $\sin t=x$. Then $\arcsin x =t$. We're interested in $\sin 2t$, then, given that $\sin t=x$. But $$\sin 2t=2\sin t \cos t$$ so that $$\sin2\arcsin x =2x\sqrt{1-x^2}$$

The general procedure is then based in a closed form of $\sin (nt)$ in terms of $\sin t=x$. For the cosine case, things work out nicer, because

$$\cos(2t)=\cos^2 t-\sin ^2 t$$

Thus, assuming $x=\cos t$, we get so that $$\cos(2\arccos x)=x^2-\left(\sqrt{1-x^2}\right)^2=2x^2-1$$

The general case is then treated by finding $\cos(nt)$ in terms of $x=\cos t$.

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