Solution
= Solution
\b[True.] Define the <Chebyshev polynomials> by $T_0(x)=1$, $T_1(x)=x$, and $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$. The <cosine> addition identities give the same recurrence for $\cos(nt)$, so <mathematical induction> proves $T_n(\cos t)=\cos(nt)$. For $n\ge1$, the leading coefficient is $2^{n-1}$, proving that its <polynomial degree> is exactly $n$.