<s>[INST] Prove that the following determinant is equal to $\sin(2(x+y))$ 
Prove that
  $$\left|\begin {array}   c 
\cos(x+y) & \sin(x+y) & -\cos(x+y)\\ 
\sin(x-y) & \cos(x-y) & \sin(x-y) \\ 
\sin{2x} & 0 & \sin (2y)
\end {array}\right|
=\sin(2(x+y))$$

It is my question. I have tried several types of operations such as $C_1=C_1 + C_3$ and many others but I failed. Somebody please help me.
 [/INST] Posing $X=x+y$ and $Y=x-y$, the determinant becomes:
$$\Delta=\left|\begin {array}   c 
\cos(X) & \sin(X) & -\cos(X)\\ 
\sin(Y) & \cos(Y) & \sin(Y) \\ 
\sin(X+Y) & 0 & \sin (X-Y)
\end {array}\right|$$
Because of the zero at $(3,2)$, Sarrus's rule yields an expression of the determinant as a sum of four terms (instead of six):
$$\Delta=\cos(X)\cos(Y)\sin(X-Y) + \sin(X)\sin(Y)\sin(X+Y) -\sin(X)\sin(Y)\sin(X-Y)\\+\cos(X)\cos(Y)\sin(X+Y)$$
which, after simplifications, turns out to be $\sin(2X)=\sin(2x+2y)$ because:
\begin{align}\Delta &= \sin(X+Y)\big[\cos(X)\cos(Y)+\sin(X)\sin(Y)\big] \\&\qquad\qquad + \sin(X-Y)\big[\cos(X)\cos(Y) -\sin(X)\sin(Y)\big]\\ &= \sin(X+Y)\cos(X-Y) +\sin(X-Y)\cos(X+Y)\\ &= \sin(2X)
\end{align}
</s>