# Determinant of Matrix $n \times n$ defined using cosine function

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?

• How did you find that they are all zero? If you proved it, there is your explanation. – MathematicianByMistake Mar 24 '16 at 16:17
• using mathematica, I calculated it through and received a zero, though I can only do it up to a 4x4 matrices, I'm trying to find a general solution to why it is always zero. – J Nguyen Mar 25 '16 at 14:31

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 $\cos(i+1-1)$: $$\cos(i+j-1)=\cos(i+1-1)\cos(j-1)-\sin(i+1-1)\sin(j-1)$$