Prove that the following determinant is equal to $\sin(2(x+y))$

Question

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.

Answer 1

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}