A triangle determinant that is always zero How do we prove, without actually expanding, that
$$\begin{vmatrix} 
\sin {2A}& \sin {C}& \sin {B}\\
\sin{C}& \sin{2B}& \sin {A}\\
\sin{B}& \sin{A}& \sin{2C}
\end{vmatrix}=0$$
where $A,B,C$ are angles of a triangle?
I tried adding and subtracting from the rows and columns and I even tried using the sine rule, but to no avail.

$$\begin{pmatrix}
\sin(2A) & \sin C & \sin B \\
\sin C & \sin (2B) & \sin A \\
\sin B & \sin A & \sin (2C) \\
\end{pmatrix} = 
\begin{pmatrix}
\sin(2A) & \sin (\pi-A-B) & \sin (\pi-A-C) \\
\sin (\pi-A-B) & \sin (2B) & \sin (\pi-B-C) \\
\sin (\pi-A-C) & \sin (\pi-B-C) & \sin (2C) \\
\end{pmatrix} $$
$$=
\begin{pmatrix}
\sin A \cos A + \cos A \sin A & \sin A \cos B+\cos A  \sin B & \sin A \cos C + \sin C \cos A \\
\sin B \cos A + \cos B \sin A & \sin B \cos B + \cos B \sin B & \sin B \cos C + \cos B \sin C \\
\sin A \cos C + \sin C \cos A & \sin B \cos C + \cos B \sin C & \sin C \cos C + \cos C \sin C
\end{pmatrix}
$$
$$=\begin{pmatrix}
\sin A & \cos A \\ \sin B & \cos B \\ \sin C & \cos C 
\end{pmatrix} 
\begin{pmatrix} 
\cos A & \cos B & \cos C \\ \sin A & \sin B & \sin C 
\end{pmatrix}$$
My other post has a moral: exponentials are easier than trig functions. I can't see a moral in this one.
