Abstract
In this paper we obtain a new curious identity involving trigonometric functions. Namely, for any positive odd integer $n$, we prove that $$\sum_{k=1}^n(-1)^k(\cot kx)\sin k(n-k)x=\frac{1-n}2,$$ which is equivalent to the identity $$\sum_{k=1}^n(-1)^kU_{n-k}(\cos kx)=-\frac{n+1}{2},$$ where $U_m(z)$ stands for the $m$th Chebyshev polynomial of the second kind. As a consequence, for any positive odd integer $n$ and positive integer $m$, we obtain the identity $$\sum_{k=1}^n(-1)^kk^{2m}B_{2m+1}\left(\frac{n-k}2\right)=0,$$ where $B_j(x)$ denotes the Bernoulli polynomial of degree $j$.
Published Version
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