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 $$\sum_{k=1}^n(-1)^kk^{2m}B_{2m+1}\left(\frac{n-k}2\right)=0,$$ where $B_j(x)$ denotes the Bernoulli polynomial of degree $j$.