In class, the professor asked us to find the determinant of the matrix $$\left(\cos(i+j-1)\right)_{1\leq i,j\leq n}=\begin{pmatrix}\cos(1) & \cos(2) & \cdots & \cos(n)\\ \cos(2) & \cos(3) & \cdots & \cos(n+1)\\ \vdots & \vdots & \ddots & \vdots \\ \cos(n) & \cos(n+1) & \cdots & \cos(2n-1)\end{pmatrix},$$ and I have found that all the matrices return a determinant of zero, if the matrix is larger than a $2 \times 2$.
I'm trying to understand why the determinant is always zero, is it because they cancel each other out in reduced form causing it all to be zeroes?
You can show that the columns are not linearly independent if the matrix is larger than $2\times 2$, using angle addition formulas.
Let the matrix be $M$ with elements $M_{i,j}=\cos(i+j-1)$. Then $$M_{i,2}=\cos((i+1-1)+1)=\cos(i+1-1)\cos(1)-\sin(i+1-1)\sin(1)\\ M_{i,2}=\cos(1)M_{i,1}-\sin(i+1-1)\sin(1)$$ or $$\frac{M_{i,2}-\cos(1)M_{i,1}}{-\sin(1)}=\sin(i+1-1)$$
So $\sin(i+1-1)$ is a linear combination of the first two columns. We can write any other column as a linear combination of $\sin(i+1-1)$ and $\cos(i+1-1)$: $$\cos(i+j-1)=\cos(i+1-1)\cos(j-1)-\sin(i+1-1)\sin(j-1)$$
So the columns are not linearly independent, so the determinant is 0.