 Solution

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