Why are $\sin3(\arcsin \, t) \: (-1\leq t\leq 1) $ and $\cos 4(\arccos \, t) \: (-1\leq t\leq 1) $ considered  polynomials? 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?
 A: $$\sin(3x)=-4\sin^3 x+3\sin x$$
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.
A: 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...
Addendum:
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.)
A: 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 $x=\cos t$.
