Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 2F / iii / Solution

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.