I hope this question is allowed, I am interested how you think someone could come up with the Chebychev polynomials, where I refer to them in the sense that someone would be interested in the polynomial of degree $n$ $p(x)$ with leading coefficient as large as possible so that $|p(x)|\leq 1$ (at the end I present a short argument why it attains the maximal leading coefficient).
The moment one thinks of $cos(x)$ this becomes not impossible, via recalling that $cos(na)$ for natural $n$ is a polynomial in $cos(a)$, and thus by plugging $a=arcos(x)$ for $a$ in the possible range; $[-1,1]$ we get what we want.
However, apriori why should it be reasonable to think of $cos(x)$? I am not satisfied with "well it's a function that is bounded like you want", because there are many other ones.
Some questions that I would be happy to know the answer to (although there may be a good explanation that doesn't answer them) are:
Would there be a way to brute force small cases to find the best polynomial? The cases of degree $0,1$ are trivial, what about $2$? A short argument shows that any polynomial that is at most $1$ in absolute value on $[-1,1]$ of degree $n$ has all coefficients $c_i$ satisfy $c_i\leq m(n)$ where $m(n)$ is a constant depending on $n$, for instance by writing the polynomial as a linear combination of legendere polynomials (with respect to the interval $[-1,1]$) and noting that their coefficients are bounded by the inner product evaluation, and thus also the coefficients of the original polynomial. This means that as mentioned above, there is a polynomial attaining the maximal leading coefficient (because the sequence of polynomials with leading term tending the sup has a converging subsequence).
If we are able to brute-force small cases, I think it's reasonable to see we would be able to see the recurrence relation between the polynomials, is there a way to deduce from it the properties of our polynomial?