Determinant of a 3*3 matrix of cosines 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?
 A: 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.
A: 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.
