Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 2F (Topics in Analysis)

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 $n$.

True, with $R_n=T_n$. The hyperbolic cosine identity $\cosh ((n+1)t)=2\cosh t\cosh (nt)-\cosh ((n-1)t)$ and the initial values $1,\cosh t$ give $T_n(\cosh t)=\cosh (nt)$ by mathematical induction, using the same Chebyshev polynomial recurrence.

False. A function of $\cos t$ is even in $t$, whereas $\sin (nt)$ is odd. If the identity held, replacing $t$ by $-t$ would imply $\sin (nt)=0$ for every $t$, contradicted by $t=\pi /(2n)$. Thus no polynomial of any degree has the required property.