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](https://doi.org/10.1090/S0002-9947-04-03687-6)", Transactions of the American Mathematical Society, 357, pp. 965-981 (2004), [MR2110427](https://www.ams.org/mathscinet-getitem?mr=2110427), [Zbl 1067.12001](https://zbmath.org/?q=an%3A1067.12001).


[2] David P. Jacobs and Mohamed O. Rayes and Vilmar Trevisan, "
[The resultant of Chebyshev polynomials](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/D3FA1C0EBDAB83534C86375A0D711BF7/S0008439500017677a.pdf/the-resultant-of-chebyshev-polynomials.pdf)",
Canadian Mathematical Bulletin 54, No. 2, 288-296 (2011), [MR2884245](https://mathscinet.ams.org/mathscinet-getitem?mr=MR2884245), [Zbl 1272.12006](https://zbmath.org/?q=an%3A1272.12006).



Resultant of linear combinations of Chebyshev polynomials of the second kind
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.


