$T_n(\cos \theta )=\cos (n\theta )$ and $U_n(\cos \theta )\sin \theta =\sin ((n+1)\theta )$ for every $n\in \mathbb{N}$

The two identities follow directly from the initial polynomial values at $n=0$ and $n=1$.

Assume the identities at $n$ and $n-1$.

The recurrences and the addition formulas give the usual second-order recurrences for $\cos ((n+1)\theta )$ and $\sin ((n+2)\theta )$.

Hence the identities hold at $n+1$, so induction proves them for every natural $n$. Separately, $\cos 0=1=(-1{)}^0$ by [F1]; if $\cos (j\pi )=(-1{)}^j$, the shift in [F1] gives $\cos ((j+1)\pi )=-\cos (j\pi )=(-1{)}^{j+1}$. Natural induction therefore proves $\cos (j\pi )=(-1{)}^j$ for every natural $j$. For $n\ge 1$ and $0\le j\le n$, substituting $\theta =j\pi \mathrm{/}n$ in the first proved identity gives $T_n(\cos (j\pi \mathrm{/}n))=\cos (j\pi )=(-1{)}^j$, as claimed.