Q:

Determinant of a 4x4 matrix with trigonometric functions

I am stuck with my homework from math. I should calcutate the determinant of a matrix:
$$\begin{bmatrix}
\sin(x) & \sin(2x)  & \cos(x) & \cos(2x)\\
\cos(x) & 2\cos(2x) & -\sin(x)& -2\sin(2x)\\
-\sin(x)& -4\sin(2x)& -\cos(x)& -4\cos(2x)\\
-\cos(x)& -8\cos(2x)& \sin(x) & 8\sin(2x)
\end{bmatrix}$$
I have no clue how to do it. I've tried to convert the matrix to the triangular matrix, but I failed to do so. 
Then I tried the Laplacian expansion (I hope it is called so) and I was creating sub-matrices from the first and second column, but I got a really long row of sin and cos and I was unable to make it shorter. 
Sorry if I wrote someting wrong, but I am not so good in English. Can you please help me? Is there any way how to solve it? 
Thank you very much!!!

A:

Call the matrix $M$. Then by using @r9m's suggestion, we interchange the two middle columns (this switches the sign of the determinant) and apply two row replacements (this doesn't change the determinant) in order to obtain a zero lower left submatrix:
\begin{align*}
|M|
&= -\begin{vmatrix}
\sin x & \cos x & \sin 2x  & \cos 2x \\
\cos x & -\sin x & 2\cos 2x & -2\sin 2x \\
-\sin x & -\cos x & -4\sin 2x& -4\cos 2x \\
-\cos x & \sin x & -8\cos 2x  & 8\sin 2x
\end{vmatrix} \\
&= -\begin{vmatrix}
\sin x & \cos x & \sin 2x  & \cos 2x \\
\cos x & -\sin x & 2\cos 2x & -2\sin 2x \\
0 & 0 & -3\sin 2x& -3\cos 2x \\
0 & 0 & -6\cos 2x  & 6\sin 2x
\end{vmatrix} \\
&= -\begin{vmatrix}
\sin x & \cos x  \\
\cos x & -\sin x
\end{vmatrix}
\begin{vmatrix}
-3\sin 2x & -3\cos 2x  \\
-6\cos 2x & 6\sin 2x
\end{vmatrix} \\
&= -(-\sin^2 x - \cos^2 x)(-18\sin^2 2x - 18\cos^2 2x) \\
&= -18(\sin^2 x + \cos^2 x)(\sin^2 2x + \cos^2 2x) \\
&= -18
\end{align*}
This used the fact that if $A,B,D$ are square matrices and $0$ is a square matrix of all zeroes, then:
$$
\begin{vmatrix}
A & B \\
0 & D
\end{vmatrix}
= |AD - B0|
= |AD|
= |A||D|
$$

