<p>We have, \(\begin{vmatrix} a^2 & a & 1 \\ \cos nx & \cos(n+1)x & \cos(n+2)x \\ \sin nx & \sin(n+1)x & \sin(n+2)x \end{vmatrix}\) is independent of:</p>
Step-by-Step Solution
Key Concept: Use the property that cos and sin satisfy the recurrence relation: cos(n+2)x = 2cos(n+1)x·cos(x) - cos(nx), and similarly for sin. This means the second and third rows are linearly dependent on each other through a fixed relationship, making the determinant independent of n and x.
<p><strong>Step 1:</strong> Recognize that for trigonometric functions, we have the recurrence relation:</p><p>cos(n+2)x = 2cos(x)·cos(n+1)x - cos(nx)</p><p>sin(n+2)x = 2cos(x)·sin(n+1)x - sin(nx)</p><p><strong>Step 2:</strong> Apply column operation C₃ → C₃ - 2cos(x)·C₂ + C₁ to both trigonometric rows:</p><p>For row 2: cos(n+2)x - 2cos(x)·cos(n+1)x + cos(nx) = 0</p><p>For row 3: sin(n+2)x - 2cos(x)·sin(n+1)x + sin(nx) = 0</p><p><strong>Step 3:</strong> After this column operation, the third column becomes [a², 0, 0]ᵀ</p><p><strong>Step 4:</strong> The determinant now has the form where two rows have zeros in the third column, making it independent of n and x (the determinant equals 0 or depends only on a).</p><p>∴ The determinant is independent of <strong>n</strong> and <strong>x</strong>, so the answer is <strong>ac</strong> (both n and x)</p>
Correct Answer: ac