What is the determinant of the Wronskian of the functions $\{\cos\ x, \sin\ x, \cos\ 2x, \sin\ 2x,\ldots, \cos\ nx, \sin\ nx\}$? This determinant seems to be an integer, and the sequence starts with 1, 18, 86400, 548674560000... It is not in the Encyclopedia. 


 **Question**  What is this sequence? I guess it is enough to prove that it consists of integers (constants), because then to compute it, one can simply put $x=0$. 


 **Update 1.**  The fact that the determinant of the Wronskian matrix is a constant is obvious. Take the derivative of the determinant. It is a sum of determinants of matrices each of which has two proportional columns. 


 **Update 2.**  The determinant is equal to the square of the Vandermonde determinant of $1, 2^2,\ldots, n^2$ times $n!$ (alternatively see Felipe Voloch's answer below). It is interesting that for $n=1$ we get just the equality $\cos^2 x+\sin^2 x=1$ (the equation of the circle). So the equality for $n > 1$ can be considered as the generalization of this equation. What is the geometry behind this identity? Of course the parametrization $(\cos x,\sin x,\ldots, \cos nx, \sin nx)$ defines some curve in $\mathbb{R}^{2n}$. What is known about that curve?


 **Update 3.**  Here is an easier formula for the determinant. It is equal to 
$$(1! 3!\ldots (2n-1)!)^2/n!$$


 **Update 4**  I found a related paper: 
Larsen, Mogens Esrom, Wronskian harmony. 
Math. Mag. 63 (1990), no. 1, 33–37. He considers the Wronskian of $\sin x, \ldots, \sin nx$.


 **Update 5.**  The sequence is in the [Encyclopedia](https://oeis.org/A192081)  now. 



The Wronskian of sin(kx) and cos(kx), k=1...n
If you only need that it doesn't depend on $x$ consider the following argument. For any $\pi \in S\_{2n}$ written as $\pi(1)\pi(2)\cdots \pi(n)$, let $[\pi]$ be the set of all permutations $\sigma \in S\_{2n}$ which satisfy $\lbrace\sigma(2k-1),\sigma(2k)\rbrace =\lbrace\pi(2k-1),\pi(2k)\rbrace$ as sets for all $k$. Then $S\_{2n}$ can be written as a disjoint union of such classes $[\pi]$ and so the terms in the determinant can be grouped accordingly. One can see that for each $[\pi]$ the correpsonding terms in the determinant add up to zero if $\pi(2j-1)+\pi(2j)$ is even for some $j$, and to
$$ (-1)^{\text{something}}\prod\_{j=1}^{n}\left( j^{\pi(2j-1)+\pi(2j)}(\sin^2(jx)+\cos^2(jx))\right)$$
otherwise, which makes it clear that the determinant doesn't depend on $x$.


