Find the value of $$\begin{vmatrix} \cos a & \sin a & \cos a & \sin a \\ \cos 2a & \sin 2a & 2\cos 2a & 2\sin 2a \\ \cos 3a & \sin 3a & 3\cos 3a & 3 \sin 3a \\ \cos 4a & \sin 4a & 4\cos 4a & 4 \sin 4a \end{vmatrix}$$
I tried to add/substract some columns and rows in order to get some $0$-s and maybe get to a $3 \times 3$ determinant, but it didn't work. I also thought about using $\sin 2a, \sin 3a, \sin 4a$ formulas, but it gets too complicated. I think that this isn't so difficult and I don't see something easy.
After a few column operations you get to $$-\frac14\begin{vmatrix} e^{ia}&e^{-ia}&e^{ia}&e^{-ia}\\ e^{2ia}&e^{-2ia}&2e^{2ia}&2e^{-2ia}\\ e^{3ia}&e^{-3ia}&3e^{3ia}&3e^{-3ia}\\ e^{4ia}&e^{-4ia}&4e^{4ia}&4e^{-4ia}\\ \end{vmatrix} =-\frac14\begin{vmatrix} 1&1&1&1\\ e^{ia}&e^{-ia}&2e^{ia}&2e^{-ia}\\ e^{2ia}&e^{-2ia}&3e^{2ia}&3e^{-2ia}\\ e^{3ia}&e^{-3ia}&4e^{3ia}&4e^{-3ia}\\ \end{vmatrix} =-\frac14\begin{vmatrix} 1&1&0&0\\ e^{ia}&e^{-ia}&e^{ia}&e^{-ia}\\ e^{2ia}&e^{-2ia}&2e^{2ia}&2e^{-2ia}\\ e^{3ia}&e^{-3ia}&3e^{3ia}&3e^{-3ia}\\ \end{vmatrix} =-\frac14\begin{vmatrix} 1&1&0&0\\ e^{ia}&e^{-ia}&1&1\\ e^{2ia}&e^{-2ia}&2e^{ia}&2e^{-ia}\\ e^{3ia}&e^{-3ia}&3e^{2ia}&3e^{-2ia}\\ \end{vmatrix}.$$ This last sort of determinant is a variant of the Vandermonde determinant $$\begin{vmatrix} 1&1&0&0\\ t&u&1&1\\ t^2&u^2&2t&2u\\ t^3&u^3&3t^2&3u^2\\ \end{vmatrix}=-(t-u)^4$$ etc.