I recently learned that if the entries of a matrix are sines of distinct integers, the determinant could equal $0$. For example,
$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 \\ \sin 4 & \sin 5 & \sin 6 \\ \sin 7 & \sin 8 & \sin 9\end{bmatrix}=0$
And I learned that if the entries of a matrix are distinct prime numbers, the determinant could equal $0$. For example,
$\det\begin{bmatrix}3&5&7\\13&17&19\\23&29&31\end{bmatrix}=0$
This leads to my question:
If the entries of a matrix are sines of distinct prime numbers, could the determinant equal $0$ ?
I have tried to apply the ideas contained in the answers and comments in the links above, to prove that the answer to my question is yes, without success.
If $n \geq 3$, then for any constants $m, b_1, \ldots, b_n$, the any $n \times n$ matrix with respective $(i, j)$ entries $$\sin[b_i + (j - 1) m]$$ has determinant $0$; see this answer for a slick proof.
So, to answer our question in the affirmative it suffices to find $n$ nonoverlapping sequences of $n$ primes in arithmetic progression with the same common difference $m$. (It turns out that $m$ must be divisible by the primorial $n\#$ of $n$.)
For $n = 3$ we can take $m = 6$ (since $3\# = 6$, $m = 6$ the smallest possible value), that is, choose any $3$ nonoverlapping sexy prime triplets $(b_i, b_i + 6, b_i + 12)$ of primes. The lexicographically first example is $b_1 = 5, b_2 = 7, b_3 = 31$: $$\det \pmatrix{\sin 5 & \sin 11 & \sin 17 \\ \sin 7 & \sin 13 & \sin 19 \\ \sin 31 & \sin 37 & \sin 43} = 0 .$$
For $n = 4$ we can again take $m = 6$ and so use sexy prime quadruplets $(b_i, b_i + 6, b_i + 12, b_i + 18)$, but for $n \geq 5$, $m$ must be divisible by $5\# = 30$.
The Green-Tao Theorem implies that one can find matrices of sines of primes of arbitrarily large size satisfying this property, and moreover that we can demand that all $n^2$ primes that appear form an arithmetic sequence.