Determinant of a 3*3 matrix of cosines

Question

I need help in evaluating the following determinant: $$\begin{vmatrix} \cos\frac{2\pi}{63} & \cos\frac{3\pi}{90} & \cos\frac{4\pi}{77} \\ \cos\frac{\pi}{72} & \cos\frac{\pi}{40} & \cos\frac{3\pi}{88} \\ 1 & \cos\frac{\pi}{90} & \cos\frac{2\pi}{99} \end{vmatrix}$$

I noticed that there is a pattern in the right row ($\frac{4\pi}{77}$,$\frac{3\pi}{88}$ & $\frac{2\pi}{99}$) but I don't know how to utilize the pattern. I also don't see any other patterns: subtracting or adding two rows seems pointless and would further complicate an already complicated determinant.

Is there any way of evaluating this determinant?

Answer 1

We assume the second entry to be $\cos(3\pi/70)$.

Multiply the last row by $-\cos(\pi/72)$ and add it the the second row; multiply the last row by $-\cos(2\pi/63)$ and add it to the first row. Then use Laplace expansion to obtain $$\begin{vmatrix} -\cos(\frac{\pi}{90}) \cos(\frac{\pi}{72}) + \cos(\frac{\pi}{40})& -\cos(\frac{\pi}{72}) \cos(\frac{2 \pi}{99}) + \cos(\frac{3 \pi}{88})\\ -\cos(\frac{\pi}{90}) \cos(\frac{2 \pi}{63}) + \cos(\frac{3 \pi}{70}) & -\cos(\frac{2 \pi}{99}) \cos(\frac{2 \pi}{63}) + \cos(\frac{4 \pi}{77}) \end{vmatrix}. $$ Key observation is that each entry is of the form $-\cos(x)\cos(y)+\cos(x+y)$. But the latter sum equals $\sin(x)\sin(y)$. Hence we have $$\begin{vmatrix} \sin(\pi/90) \sin(\pi/72)& \sin(\pi/72) \sin(2 \pi/99)\\ \sin(\pi/90) \sin(2 \pi/63) & \sin(2 \pi/99) \sin(2 \pi/63) \end{vmatrix}, $$ which is trivially zero.

Answer 2

I tried WhatsUp's suggestion and evaluated the following using maple: the answer was: $$-0.0000098824$$

On replacing the middle term with $cos(\frac {3\pi}{70})$ and evaluating, the answer comes out to be zero, which corresponds with the answer mentioned. This is, unfortunately, a printing mistake :P

Thank you to all for their advice.