Solution
= Solution
\b[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.