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Question
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]
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Solution
\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]
\[ = \frac{1}{\sin y\cos y}\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x\sin y & \cos x\sin y & \sin^2 y \\ - \cos x\cos y & \sin x\cos y & - \cos^2 y\end{vmatrix} \left[ \text{ Applying }R_2 \to \sin y R_2 \text{ and }R_3 \to \cos y R_3 \right]\]
\[ = \frac{1}{\sin y\cos y}\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x\sin y - \cos x\cos y & \cos x\sin y + \sin x\cos y & \sin^2 y - \cos^2 y \\ - \cos x\cos y & \sin x\cos y & - \cos^2 y\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 + R_3 \right]\]
\[ = \frac{- 1}{\sin y\cos y}\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ - \cos x\cos y & \sin x\cos y & - \cos^2 y\end{vmatrix}\]
\[ = 0\]
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