Resultant of linear combinations of Chebyshev polynomials of the second kind

Question

The Chebyshev polynomial $U_n(x)$ of the second kind is characterized by $$ U_n(\cos\theta)=\frac{\sin(n+1)\theta}{\sin(\theta)}. $$ It seems that $$\operatorname*{Res}_x \left( U_n(x)+tU_{n-1}(x),\sum_{k=0}^{n-1}U_k(x) \right) =(-1)^{\frac{n(n-1)}{2}} t^{\left\lfloor\frac{k}{2} \right\rfloor}2^{n(n-1)},$$ where Res denotes the resultant of two polynomials.

I do not know how to prove this "equality". Is it a known result? Similar results appeared in Dilcher and Stolarsky [1] theorem 2 and in Jacobs, Rayes and Trevisan [2].

References

[1] Karl Dilcher and Kenneth Stolarsky, "Resultants and discriminants of Chebyshev and related polynomials", Transactions of the American Mathematical Society, 357, pp. 965-981 (2004), MR2110427, Zbl 1067.12001.

[2] David P. Jacobs and Mohamed O. Rayes and Vilmar Trevisan, " The resultant of Chebyshev polynomials", Canadian Mathematical Bulletin 54, No. 2, 288-296 (2011), MR2884245, Zbl 1272.12006.

Answer 1

Since $U_n + t U_{n-1}$ is of degree $n$ and $\sum_{k=0}^{n-1} U_k$ is of degree $n-1$ with leading coefficient $2^{n-1}$, the resultant factors as $$ 2^{n(n-1)} (-1)^{n(n-1)} \prod_{j=1}^{n-1} (U_n(x_j) + t U_{n-1}(x_j))$$ where $x_1,\dots,x_{n-1}$ are the zeroes of $\sum_{k=0}^{n-1} U_k$.

Fortunately, these zeroes can be located explicitly using the usual trigonometric addition and subtraction identities. Telescoping the trig identity $\sin k \theta = -\frac{\cos\left(k+\frac{1}{2}\right) \theta - \cos\left(k-\frac{1}{2}\right) \theta}{2 \sin \frac{\theta}{2} }$ we conclude that $$ \sum_{k=0}^{n-1} U_k(\cos \theta) = -\frac{\cos\left(\left(n+\frac{1}{2}\right) \theta\right) - \cos\left(\frac{\theta}{2}\right)}{2 \sin \theta \sin \frac{\theta}{2}} = \frac{\sin\left(\frac{n}{2} \theta\right) \sin\left(\frac{n+1}{2} \theta\right)}{2 \cos \frac{\theta}{2} \sin^2 \frac{\theta}{2}}$$ and so the $n-1 = \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n-1}{2} \rfloor$ zeroes of $\sum_{k=0}^{n-1} U_k$ take the form $\cos( \frac{2\pi j}{n+1} )$ for $1 \leq j < (n+1)/2$ and $\cos( \frac{2\pi j}{n} )$ for $1 \leq j < n/2$.

Since the first class $\cos( \frac{2\pi j}{n+1} )$ of zeroes are also zeroes of $U_n$, and the second class $\cos( \frac{2\pi j}{n} )$ are zeroes of $U_{n-1}$, the resultant therefore simplifies to $$ 2^{n(n-1)} (-1)^{n(n-1)} t^{\lfloor \frac{n}{2} \rfloor} \prod_{1 \leq j < \frac{n+1}{2}} U_{n-1}\left( \cos\left( \frac{2\pi j}{n+1} \right) \right) \prod_{1 \leq j < \frac{n}{2}} U_{n}\left( \cos\left( \frac{2\pi j}{n} \right) \right).$$ But $$ U_{n-1}\left( \cos\left( \frac{2\pi j}{n+1} \right) \right) = \left. \sin\left(n\frac{2\pi j}{n+1}\right) \right/ \sin\left(\frac{2\pi j}{n+1}\right) = -1$$ and similarly $$ U_{n}\left( \cos\left( \frac{2\pi j}{n} \right) \right) = \left. \sin\left((n+1)\frac{2\pi j}{n}\right) \right/ \sin\left(\frac{2\pi j}{n}\right) = +1$$ and the claim then follows after counting up the signs.