The value of the determinant left|begin{array | vedclass

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(A) Given the determinant $\Delta = \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \ \sin \beta & \cos \beta & \sin (\beta+

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(A) Given the determinant $\Delta = \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \ \sin \beta & \cos \beta & \sin (\beta+\delta) \ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}ight|$.Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can expand the third column:$\Delta = \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \cos \delta + \cos \alpha \sin \delta \ \sin \beta & \cos \beta & \sin \beta \cos \delta + \cos \beta \sin \delta \ \sin \gamma & \cos \gamma & \sin \gamma \cos \delta + \cos \gamma \sin \delta\end{array}ight|$.By the property of determinants,we can split this into the sum of two determinants:$\Delta = \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \cos \delta \ \sin \beta & \cos \beta & \sin \beta \cos \delta \ \sin \gamma & \cos \gamma & \sin \gamma \cos \delta\end{array}ight| + \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \cos \alpha \sin \delta \ \sin \beta & \cos \beta & \cos \beta \sin \delta \ \sin \gamma & \cos \gamma & \cos \gamma \sin \delta\end{array}ight|$.Factoring out $\cos \delta$ from the third column of the first determinant and $\sin \delta$ from the third column of the second determinant:$\Delta = \cos \delta \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \ \sin \beta & \cos \beta & \sin \beta \ \sin \gamma & \cos \gamma & \sin \gamma\end{array}ight| + \sin \delta \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \cos \alpha \ \sin \beta & \cos \beta & \cos \beta \ \sin \gamma & \cos \gamma & \cos \gamma\end{array}ight|$.In the first determinant,column $1$ and column $3$ are identical,so its value is $0$.In the second determinant,column $2$ and column $3$ are identical,so its value is $0$.Therefore,$\Delta = \cos \delta (0) + \sin \delta (0) = 0$.

The value of the determinant $\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|$ is equal to

Similar Questions

Let $ \Delta = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix} $ and $ \Delta_1 = \begin{vmatrix} A & B & C \\ x & y & z \\ zy & zx & xy \end{vmatrix} $,then:

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