AP DSC School Assistant Math Model Test 3

Concept:The determinant of a 3x3 matrix equals zero gives an equation. We expand the determinant using the first row, apply trigonometric identities for sum/difference angles, and simplify to solve for $B$.Explanation:Let $\Delta = \begin{vmatrix} \cos(A+B) & -\sin(A+B) & \cos 2B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B \end{vmatrix} = 0$.Expand along the first row:$\Delta = \cos(A+B) \cdot (\cos A \cos B - \sin A \sin B) \;-\; (-\sin(A+B)) \cdot (\sin A \cos B - (-\cos A)(\sin B)) \;+\; \cos 2B \cdot (\sin A \cdot \sin A - \cos A \cdot (-\cos A))$.Second term simplifies: $+\sin(A+B) \cdot (\sin A \cos B + \cos A \sin B)$. The third term: $\cos 2B \cdot (\sin^2 A + \cos^2 A) = \cos 2B \cdot 1 = \cos 2B$.Now $\cos A \cos B - \sin A \sin B = \cos(A+B)$, and $\sin A \cos B + \cos A \sin B = \sin(A+B)$.So the determinant becomes:$\cos(A+B) \cdot \cos(A+B) + \sin(A+B) \cdot \sin(A+B) + \cos 2B = \cos^2(A+B) + \sin^2(A+B) + \cos 2B$.Using $\cos^2\theta + \sin^2\theta = 1$, we get $1 + \cos 2B = 0 \implies \cos 2B = -1$.Since $\cos 2B = -1$ gives $2B = (2n+1)\pi$, so $B = (2n+1)\frac{\pi}{2}$, where $n \in \mathbb{Z}$.Answer:B. $(2n+1)\frac{\pi}{2}, \; n \in \mathbb{Z}$