Intriguing Determinant

"Intriguing" = "Chebyshev", it seems!

If $B_n(\lambda)$ is the determinant of the $n\times n$ tridiagonal matrix whose elements are $\lambda$ down the main diagonal, $1$ down the adjacent diagonals, and $0$ elsewhere, so that in this question we have $\mathrm{det}\,A_n \,=\, B_n\big(2\cos\tfrac{\pi}{180}\big)$ then standard determinant work tells us that $B_{n+1}(\lambda) \; = \; \lambda B_n(\lambda) - B_{n-1}(\lambda) \;, \qquad n \ge 2 \;.$ This formula holds for all $n \ge 1$ if we put $B_0(\lambda) = 1$ and $B_1(\lambda) = \lambda$ .

When $\lambda = 2\cos\theta$ we deduce that $B_n(\lambda) \; = \; B_n\big(2\cos\theta\big) \; = \; \frac{\sin(n+1)\theta}{\sin\theta} \;, \qquad n \ge 0 \;,$ so that $B_n(\lambda) = U_n(\cos\theta)$ can be written in terms of the Chebyshev polynomials of the second kind. Thus $\mathrm{det}\,A_n \; = \; B_n\big(2\cos\tfrac{\pi}{180}\big) \; = \; \frac{\sin \tfrac{(n+1)\pi}{180}}{\sin \tfrac{\pi}{180}} \;, \qquad n \ge 0 \;,$ and so the maximum value of $\mathrm{det}\,A_n$ is $\mathrm{cosec}\,\tfrac{\pi}{180} \,=\, \boxed{\mathrm{cosec}\,1^\circ}$ .