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Question
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
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Solution
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
\[ = \begin{vmatrix}\sin\alpha\sin\delta & \cos\alpha\cos\delta & \cos(\alpha + \delta) \\ \sin\beta\sin\delta & \cos\beta\cos\delta & \cos(\beta + \delta) \\ \sin\gamma\sin\delta & \cos\gamma\cos\delta & \cos(\gamma + \delta)\end{vmatrix} \left[\text{ Applying }C_1 \to \sin\delta C_1 \text{ and }C_2 \to \cos\delta C_2 \right]\]
\[ = \begin{vmatrix}\sin\alpha\sin\delta & \cos(\alpha + \delta) & \cos(\alpha + \delta) \\ \sin\beta\sin\delta & \cos(\beta + \delta) & \cos(\beta + \delta) \\ \sin\gamma\sin\delta & \cos(\gamma + \delta) & \cos(\gamma + \delta)\end{vmatrix} \left[ \text{ Applying }C_2 \to C_2 - C_1 \right]\]
\[ = 0\]
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