The $i$-th Chebyshev polynomial of the 2nd kind is defined to be the polynomial $U_i(x)$ that satisfies \begin{equation} U_i(\cos \theta) \sin\theta = \sin(i+1)\theta. \end{equation}
There exists an identity for the product of Chebyshev polynomials of the 2nd kind, which is as follows: \begin{equation} U_i(x)U_j(x) = \sum_{k=0}^j U_{i-j+2k}(x) \qquad (i \geq j) \end{equation} Thus every product of Chebyshev polynomials of the 2nd kind may be written as a sum of Chebyshev polynomials of the 2nd kind.
Now consider some fixed $n$ and consider the extended ring of integers \begin{equation} R=\mathbb{Z}\left[U_1\left(\cos\frac{\pi}{n}\right),\ldots,U_{m}\left(\cos\frac{\pi}{n}\right)\right], \end{equation} where \begin{equation} m= \begin{cases} \frac{n-3}{2} & n \text{ odd,} \\ \frac{n-2}{2} & n \text{ even}. \end{cases} \end{equation}
(The reasoning for this choice of $m$ is that the values of $U_i\left(\cos\frac{\pi}{n}\right)$ are symmetric either side of $U_m\left(\cos\frac{\pi}{n}\right)$)
My question is this: Is this 'decomposition' of a product of Chebyshev polynomials of the second kind into summands 'unique in the ring $R$'? To be more precise, given a product \begin{equation} p=\prod_{k=1}^l U_{i_k}\left(\cos\frac{\pi}{n}\right) \in R \end{equation} with each $i_k \in \{0,\ldots,m\}$, and given two elements \begin{align} r &= \sum_{i=0}^{m} a_i U_i\left(\cos\frac{\pi}{n}\right)\in R \\ s &= \sum_{i=0}^{m} b_i U_i\left(\cos\frac{\pi}{n}\right)\in R \end{align} such that $r=p=s$, do we necessarily have $a_i=b_i$ for all $0 \leq i \leq m$?
(Note that here we have $U_0(\cos\theta)=1$.)
The answer, it seems, is no for general $n$. Here is a simple counterexample.
Consider the case where $n = 9$. With this choice of $n$, our ring $R$ is defined to be \begin{equation*} R=\mathbb{Z}\left[U_1\left(\cos\frac{\pi}{9}\right), U_2\left(\cos\frac{\pi}{9}\right), U_3\left(\cos\frac{\pi}{9}\right)\right] \end{equation*}
Now one can use the identity \begin{equation*} 2\cos\frac{\pi}{9}-2\cos\frac{2\pi}{9}+2\cos\frac{3\pi}{9}-2\cos\frac{4\pi}{9}=1 \end{equation*} along with the fact that $2\cos\frac{3\pi}{9}=1$ to construct a polynomial \begin{equation*} x^3 - 3x - 1 \end{equation*} whose roots are $2\cos\frac{\pi}{9}$, $-2\cos\frac{2\pi}{9}$ and $-2\cos\frac{4\pi}{9}$.
The first few Chebyshev polynomials of the second kind with respect to the variable $\frac{x}{2}$ are \begin{equation*} U_0\left(\frac{x}{2}\right) = 1, \end{equation*} \begin{equation*} U_1\left(\frac{x}{2}\right) = x, \end{equation*} \begin{equation*} U_2\left(\frac{x}{2}\right) = x^2-1, \end{equation*} \begin{equation*} U_3\left(\frac{x}{2}\right) = x^3-2x. \end{equation*} Considering the quotient ring $\mathbb{Z}[x]/(x^3-3x-1)$ we obtain \begin{equation*} \left[U_3\left(\frac{x}{2}\right)\right] = [x+1] \end{equation*}
Thus, we have \begin{equation*} U_3\left(\cos\frac{\pi}{9}\right) = U_0\left(\cos\frac{\pi}{9}\right) + U_1\left(\cos\frac{\pi}{9}\right) \end{equation*} and hence $U_3\left(\cos\frac{\pi}{9}\right)$ is linearly dependent in $R$.
I suspect that the original claim/question holds only if the degree of minimal polynomial of $\cos\frac{\pi}{n}$ is strictly greater than $m$.