Commutative property of Chebyshev polynomial. I'm trying to understand a few properties of $T_n$, Chebychev polynomials of the first kind.  That is, those of the form: $\cos(kz)=T_k(\cos(z))$.  For example, $T_{2}(z)=2z^2-1$.
I don't understand why this isn't a trivial fact: for all $m,n$, show that $T_n$ commutes with $T_m$.  Thinking about a concrete example, $T_2$ and $T_3$.  It's clear that $(4x^3-3)(2x^2-1)=(2x^2-1)(4x^3-3)$ since multiplication of polynomial is commutative.  Is there something deeper here where the commutative property is in doubt?

Commutativity here means $T_m(T_n(x))=T_n(T_m(x))$. The reason for this
is that $T_m(T_n(x))=T_{mn}(x)$. The reason for that is that
$$T_m(T_n(\cos y))=T_m(\cos ny)=\cos mny.$$
