# A surprising identity: $\det[\cos\pi\frac{jk}n]_{1\le j,k\le n}=(-1)^{\lfloor\frac{n+1}2\rfloor}(n/2)^{(n-1)/2}$

On the basis of my computation, here I pose my following conjecture involving the cosine function.

Conjecture. For any positive integer $$n$$, we have the identity $$\frac1{2n}\det\left[\cos\pi\frac{jk}n\right]_{0\le j,k\le n}=\det\left[\cos\pi\frac{jk}n\right]_{1\le j,k\le n}=(-1)^{\lfloor\frac{n+1}2\rfloor}(n/2)^{(n-1)/2}.$$

This is a part of Conjecture 5.7 in my preprint arXiv:1901.04837. The paper contains more similar conjectures.

Any ideas towards a solution of the conjecture?

First of all, we use the formula $$D:=\det [x_j^k+x_j^{-k}]_{j,k=0,\dots,m-1}=\prod_{l This follows from the observation that $$x^k+x^{-k}=p_k(x+x^{-1})$$ for a polynomial $$p$$ of degree $$k$$ with leading coefficient 1, so our matrix is the Vandermonde matrix for $$x_j+x_j^{-1}$$ times some unitriangular matrix.
For the determinant when $$j$$ and $$k$$ vary from 0 to $$n$$, this is a special case when $$x_j=e^{\pi i j/n}$$ for $$j=0,1,\dots,n$$ (with $$m=n+1$$). The sign of $$D$$ clearly equals $$(-1)^{\binom{n+1}2}$$, since all differences $$x_j+x_j^{-1}-x_l-x_l^{-1}$$ are negative reals, and we have to find the absolute value of $$D$$. $$|D|=4\left|\prod_{0 It is well known that the absolute value of the product over $$a,b$$ (i.e. of the discriminant of the polynomial $$f_{2n}(z)=z^{2n}-1$$) equals $$(2n)^n$$ (for example, because $$|f_{2n}'(\omega)|=2n$$ for any root $$\omega$$ of $$f_0$$), and $$\prod_{j=1}^{n-1}(1-x_j^2)=n$$ for the same reason (i.e. $$f_n'(1)=n$$, $$x_j^2$$ are roots of $$f_n$$).
Now about the determinant $$\det [x_j^k+x_j^{-k}]_{j,k=1,\dots,n}$$ for $$x_j=e^{i\pi j/n}$$ (divided by $$2^n$$). In order to reduce it to the determinant $$\tilde{D}$$ when $$j$$ varies from 1 to $$n$$ but $$k$$ from 0 to $$n-1$$, we introduce the coefficients $$a_0,a_1,\dots,a_{n-1}$$ such that $$x^n+x^{-n}-\sum_{j=0}^{n-1} a_j(x^j+x^{-j})=\prod_{j=1}^n(x+x^{-1}-x_j-x_j^{-1})$$. Then subtracting from the last column the linear combination of first $$n-1$$ columns we get the column of the form $$a_0(x_j^0+x_j^{-0})$$, thus our determinant equals $$(-1)^{n-1}a_0\tilde{D}$$.
It remains to calculate $$\tilde{D}$$ and $$a_0$$. We may compare $$\tilde{D}$$ with above $$D$$. We have $$D=(-1)^n \cdot 2\tilde{D}\cdot \prod_{a^{2n}=1,a\ne 1} (1-a)=4(-1)^nn\tilde{D}.$$
Now about $$a_0$$. $$-a_0$$ is the constant term of the Laurent polynomial $$\prod_{j=1}^n(x+x^{-1}-x_j-x_j^{-1})=x^{-n}\prod_{j=1}^n(x-x_j)(x-x_j^{-1})=x^{-n}(x^{2n}-1)\cdot \frac{x+1}{x-1}=x^{-n}(x+1)(1+x+\dots+x^{2n-1}),$$ thus $$a_0=-2$$.