Question

\(\begin{vmatrix} \sin\alpha & \cos(\alpha+\theta) & \cos\alpha \\ \sin\beta & \cos(\beta+\theta) & \cos\beta \\ \sin\gamma & \cos(\gamma+\theta) & \cos\gamma\end{vmatrix}=\)

• A. -1
• B. 1
• C. 2
• D. 4
• E. 0

Answer 1

Answer: (E)

We are tasked with evaluating the determinant of the following matrix: \[ \left| \begin{matrix} \sin \alpha & \cos(\alpha + \theta) & \cos \alpha \\ \sin \beta & \cos(\beta + \theta) & \cos \beta \\ \sin \gamma & \cos(\gamma + \theta) & \cos \gamma \end{matrix} \right| \] We will use the property of determinants: the determinant of a matrix with two proportional rows or columns is zero. To check if the rows or columns are proportional, consider the following:
The first column has the values \( \sin \alpha, \sin \beta, \sin \gamma \).
The second column has \( \cos(\alpha + \theta), \cos(\beta + \theta), \cos(\gamma + \theta) \).
The third column has \( \cos \alpha, \cos \beta, \cos \gamma \). Notice that the second and third columns have a form of similarity due to the presence of trigonometric identities.
In particular, the expressions \( \cos(\alpha + \theta) \) and \( \cos \alpha \) have a linear relation due to the angle addition formula for cosine.
This similarity suggests that the rows are linearly dependent, leading to a determinant of zero.

The correct option is (E) : \(0\)

Answer 2

We are asked to find the value of the determinant of the following matrix: \[ \begin{vmatrix} \sin\alpha & \cos(\alpha+\theta) & \cos\alpha \\ \sin\beta & \cos(\beta+\theta) & \cos\beta \\ \sin\gamma & \cos(\gamma+\theta) & \cos\gamma \end{vmatrix} \] To evaluate this determinant, notice that the columns of the matrix are in the form of trigonometric functions with a common shift by \(\theta\) in the second column. The structure of the matrix suggests that there is a linear dependence between the rows or columns, as the trigonometric identities cause certain relationships between the terms. Given this linear dependence, the determinant of the matrix will be zero. Therefore, the value of the determinant is: \[ \boxed{0} \]