Is $\frac{\cos^2 (x)\sin(nx)}{\sin(x)}$ a trigonometric polynomial?

Question

Is $\frac{\cos^2 (x)\sin(nx)}{\sin(x)}$ trigonometric polynomial of degree at most $n$?

I need this result to show that for any trigonometric polynomial $T(t)$ of degree $n$ exists $a$ such that

$$T(t)=a\cos(nt)+\frac{\cos(nt)}{2n}\sum\limits_{k=1}^{2n}(-1)^kT(t_k)\cot\left(\frac{t-t_k}{2}\right)$$ Where $$t_k=\frac{2k-1}{2n}\pi$$

Answer 1

Assuming $n$ is a positive integer, yes it's a TP, but of degree $n+1>n$. In fact you don't need the cosine:

Say $z=e^{ix}$ and $w=e^{-ix}$. Then $$\frac{\sin(nx)}{\sin(x)} =\frac{z^n-w^n}{z-w}=z^{n-1}+wz^{n-2}+\dots+w^{n-1},$$a TP of degree $n-1$.

Answer 2

Recently I encountered the following sum (proof is quite easy by De Moivre's formula) : $$1+2(\cos 2x + \cos 4x + \dots + \cos 2nx) = \frac{\sin(2n+1)x}{\sin x}$$ Multiplying both sides by $\cos^2x$ and expressing $\cos nx$ by $\sin x$ and $\cos x$ you may get something similar to your expression as a trigonometric polynomial.