Compute a determinant I want to compute this determinant:
$$
\begin{vmatrix}
\sin(2x)&\sin(3x)&\sin(4x)\\
\sin(3x)&\sin(4x)&\sin(5x)\\
\sin(4x)&\sin(5x)&\sin(6x)
\end{vmatrix}
$$
 A: Using Euler's formula, your determinant is :
$$ D = \begin{vmatrix} \frac{e^{2ix}-e^{-2ix}}{2i} & \frac{e^{3ix}-e^{-3ix}}{2i} & \frac{e^{4ix}-e^{-4ix}}{2i} \\ \frac{e^{3ix}-e^{-3ix}}{2i} & \frac{e^{4ix}-e^{-4ix}}{2i} & \frac{e^{5ix}-e^{-5ix}}{2i} \\ \frac{e^{4ix}-e^{-4ix}}{2i} & \frac{e^{5ix}-e^{-5ix}}{2i} & \frac{e^{6ix}-e^{-6ix}}{2i} \end{vmatrix} $$
Factor out the $2i$ factors, and use multilinearity : you obtain
$$ D = \frac1{(2i)^3}\begin{vmatrix} e^{2ix}-e^{-2ix} & e^{3ix}-e^{-3ix} & e^{4ix}-e^{-4ix} \\ e^{3ix}-e^{-3ix} & e^{4ix}-e^{-4ix} & e^{5ix}-e^{-5ix} \\ e^{4ix}-e^{-4ix} & e^{5ix}-e^{-5ix} & e^{6ix}-e^{-6ix} \end{vmatrix}$$
$$= \frac1{(2i)^3}\sum_{(\epsilon_1, \epsilon_2, \epsilon_3)\in\{\pm 1\}^3} \epsilon_1 \epsilon_2 \epsilon_3\begin{vmatrix} e^{2i\epsilon_1 x} & e^{3i\epsilon_2 x} & e^{4i\epsilon_3 x} \\ e^{3i\epsilon_1 x} & e^{4i\epsilon_2 x} & e^{5i\epsilon_3 x} \\ e^{4i\epsilon_1 x} & e^{5i\epsilon_2 x} & e^{6i\epsilon_3 x} \end{vmatrix}$$
Factor out the elements from the first line. Each of the determinant in the sum over $(\epsilon_1, \epsilon_2, \epsilon_3)$ is
$$ = e^{(2\epsilon_1 + 3\epsilon_2 + 4\epsilon_3)ix}\ \times\begin{vmatrix} 1 & 1 & 1 \\ e^{i\epsilon_1 x} & e^{i\epsilon_2 x} & e^{i\epsilon_3 x} \\ e^{2i\epsilon_1 x} & e^{2i\epsilon_2 x} & e^{2i\epsilon_3 x} \end{vmatrix}. $$
Now for any $(\epsilon_1, \epsilon_2, \epsilon_3)\in\{\pm 1\}^3$, at least two of the columns will be the same (there's only two choices for each $\epsilon_j$ !) : so the last determinant is always zero and $D=0$.
