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?