The monic Chebyshev polynomial appears perhaps a bit more naturally as a unique minimizer in the following $L^2$-problem.
Problem: find a monic polynomial $P(x)$ of degree $n$ which minimizes the weighted norm $$\|P\|^2=\int_{-1}^{1}P^2(x)\frac{dx}{\sqrt{(1-x^2)}}.$$
The proof is straightforward. First, we check the orthogonality property $$\int_{-1}^{1}T_i(x)T_k(x)\frac{dx}{\sqrt{(1-x^2)}}=\delta_{ik}\frac{\pi}{2},\quad i,k=0,1,2,...,$$ which is equivalent to the orthogonality property of the sequence $\cos kx$ in $L^2(0,\pi)$. Next, we have for an arbitrary monic polynomial of degree $n$ $$P(x)=\sum\limits_{k=1}^n a_kT_k(x),\quad a_n=2^{1-n}.$$ Therefore $$\|P(x)-2^{1-n}T_n(x)\|^2=\|P\|^2+\|2^{1-n}T_n(x)\|^2-2^{2-n}\int_{-1}^{1}\sum\limits_{k=1}^n a_kT_k(x)T_n(x)\frac{dx}{\sqrt{(1-x^2)}}= $$ $$=\|P\|^2-\|2^{1-n}T_n(x)\|^2.$$ So $\|P\|\geq \|2^{1-n}T_n(x)\|,$ and the equality is possible if and only if $P(x)=2^{1-n}T_n(x)$.
Edit added. By the way, $2^{1-n}T_n(x)$ minimizes all weighted $L^p$-norms $$\left[\int_{-1}^{1}|P_n(x)|^p\frac{dx}{\sqrt{(1-x^2)}}\right]^{\frac{1}{p}},\quad 1\leq p\leq\infty,$$ over monic polynomials $P_n(x)$ of degree $n$. This book contains a survey of this and many other extremal properties of Chebyshev polynomials.
