Let $a_k^{(n)}$ be the $n$-vector whose components are the first $n$ non-null coefficients of the Taylor expansion of $\sin(k x)$ around $0$. Define the matrix $A^{(n)}$ as the matrix whose rows are the vectors $a_1^{(n)};a_2^{(n)}\dots a_n^{(n)}$, i.e. the $k$-th row of $A^{(n)}$ is the vector $a_k^{(n)}$. Then $A^{(n)}$ has determinant given by $$\det A^{(n)} = (-1)^{\lfloor n/2\rfloor}\,,$$ i.e. $\det A^{(1)} = 1$, $\det A^{(2)} = -1$, $\det A^{(3)} = -1$, $\det A^{(4)} = 1$, $\det A^{(5)} = 1$ and so on.
To clarify the first matrices are $$A^{(1)}=\begin{pmatrix}1\end{pmatrix}\,;$$ $$A^{(2)}=\begin{pmatrix}1 & -\frac{1}{6}\\2 & -\frac{4}{3}\end{pmatrix}\,;$$ $$A^{(3)}=\begin{pmatrix}1 & -\frac{1}{6} & \frac{1}{120}\\2 & -\frac{4}{3} & \frac{4}{15}\\ 3&-\frac{9}{2} & \frac{81}{40}\end{pmatrix}\,\dots$$
I guess there must be a simple explanation for such an easy determinant and a smart way to prove it. Any idea?
I came across this problem when thinking about some possible answer to a question asking for some intuitive way behind the fact that the trigonometric functions forms a basis. My idea was that one may show that the monomials $x^{2k+1}$ can be founded by eliminating term by term the unnecessary terms from the Taylor expansion of $\sin$. So I've started by looking at the truncated Taylor series and I found this matrices, whose inverse are the coefficient to find monomials from linear combination of the truncated expansions of $\sin(kx)$. My idea is that actually the fact that the determinant is $1$ may be somehow linked to the fact that the Fourier transform is an isometry, and to the extension of this fact on subspaces of $L^2$ with of polynomials of finite degree $<n$. But maybe I'm hallucinating...
Note that $$A^{(3)}=\pmatrix{1&1&1\\2&2^3&2^5\\3&3^3&3^5}\pmatrix{1&&\\&-1/6&\\&&1/120} =\pmatrix{1&&\\&2&\\&&3}\pmatrix{1&1&1\\1&2^2&2^4\\1&3^2&3^4}\pmatrix{1&&\\&-1/6&\\&&1/120}$$ a product of a Vandermonde matrix and two diagonal matrices. There's a well-known formula for the Vandermonde determinant.
This pattern persists.