<p>If \(\begin{vmatrix} \sin x & \sin y & \sin z \\ \cos x & \cos y & \cos z \\ \cos^3 x & \cos^3 y & \cos^3 z \end{vmatrix} = 0\), then which of the following is/are possible?</p>
Step-by-Step Solution
Key Concept: Factor out column differences using the determinant property that if two columns are proportional or satisfy a linear dependence, the determinant becomes zero. Here, the third row (cosines cubed) can be expressed as a linear combination of rows 1 and 2, or the columns themselves must satisfy a special relationship.
<p><strong>Step 1:</strong> Observe that row 3 contains cos³x, cos³y, cos³z. Rewrite using cos²θ = 1 - sin²θ:</p><p>cos³θ = cosθ · cos²θ = cosθ(1 - sin²θ) = cosθ - cosθ·sin²θ</p><p><strong>Step 2:</strong> This means Row 3 = (Row 2) - (Row 2 ⊙ Row 1²), where ⊙ denotes element-wise multiplication. This creates linear dependence among rows.</p><p><strong>Step 3:</strong> Alternatively, factor the determinant by extracting (sin y - sin x)(sin z - sin x)(sin z - sin y) from column operations, revealing that the determinant = 0 when specific trigonometric relations hold.</p><p><strong>Step 4:</strong> The determinant equals zero when: sinx = siny = sinz, OR cosx = cosy = cosz, OR when (sin x - sin y)(sin y - sin z)(sin z - sin x) = 0 combined with row 3 constraint.</p><p>∴ Answer: ABC (All given conditions are possible)</p>
Correct Answer: ABC