I just faced this problem where i am asked to show this matrix has determinant = 0 and I got stuck and can't find a way out of this...would really appreciate if someone could help $$ \begin{pmatrix} \cos \alpha & \sin \alpha & \sin (\alpha + \theta) \\ \cos \beta & \sin \beta & \sin (\beta + \theta) \\ \cos \gamma & \sin \gamma & \sin (\gamma + \theta) \\ \end{pmatrix} $$
My attempt:
The third column is equal to $\sin\theta$ times the first column plus $\cos\theta$ times the second column, by the well-known formula $$ \sin(u)\cos(v) + \cos(u)\sin(v) = \sin(u + v) $$ This makes the columns linearly dependent and therefore the matrix is singular.