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 $P(x)=2^{1-n}T_n(x)$.

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**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][1] contains a survey of this and many other extremal properties of Chebychev polynomials.


  [1]: http://books.google.co.uk/books?id=8FHf0P3to0UC&printsec=frontcover&dq=chebyshev+polynomials+mason&source=bl&ots=1fUJhUKr8o&sig=T6g3dzffCkP1Ip3IDe_jmU2uGx8&hl=en&ei=o5P3S72EBqCy0gTy78npBw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBcQ6AEwAA#v=onepage&q&f=false