Matrices & Determinants
Determinants involving trigonometric functions
Grade 12
Question:
<p>If \(f(\theta) = \begin{vmatrix} \sin\theta & \cos\theta & \sin\theta \\ \cos\theta & \sin\theta & \cos\theta \\ \cos\theta & \sin\theta & \sin\theta \end{vmatrix}\), then</p>
<p>\(f(\theta) = 0\) has exactly 2 real solutions in \([0, \pi]\)</p>
<p>\(f(\theta) = 0\) has exactly 3 real solutions in \([0, \pi]\)</p>
<p>range of function \(\dfrac{f(\theta)}{1 - \sin 2\theta}\) is \(\left[-\sqrt{2}, \sqrt{2}\right]\)</p>
<p>range of function \(\dfrac{f(\theta)}{\sin 2\theta - 1}\) is \([-3, 3]\) is \([-3, 3]\)</p>
Step-by-Step Solution
Key Concept: Expand the determinant using cofactor expansion and use the algebraic identity (a³ + b³ - 3ab²) = (a + b)(a² - ab + b²) after factoring out common terms. Recognize that sin²θ + cos²θ = 1 simplifies the final expression dramatically.
<p><strong>Step 1:</strong> Expand the determinant along the first row:</p><p>f(θ) = sin θ · |sin θ cos θ; sin θ sin θ| - cos θ · |cos θ cos θ; cos θ sin θ| + sin θ · |cos θ sin θ; cos θ sin θ|</p><p><strong>Step 2:</strong> Calculate each 2×2 determinant:</p><p>= sin θ(sin²θ - sin θ cos θ) - cos θ(cos θ sin θ - cos²θ) + sin θ(cos θ sin θ - sin θ cos θ)</p><p>= sin³θ - sin θ cos θ sin θ - cos²θ sin θ + cos³θ + 0</p><p>= sin³θ + cos³θ - sin²θ cos θ - sin θ cos²θ</p><p><strong>Step 3:</strong> Factor:</p><p>= sin³θ + cos³θ - sin θ cos θ(sin θ + cos θ)</p><p><strong>Step 4:</strong> Use sin³θ + cos³θ = (sin θ + cos θ)(sin²θ - sin θ cos θ + cos²θ) = (sin θ + cos θ)(1 - sin θ cos θ):</p><p>= (sin θ + cos θ)(1 - sin θ cos θ) - sin θ cos θ(sin θ + cos θ)</p><p>= (sin θ + cos θ)[1 - sin θ cos θ - sin θ cos θ]</p><p>= (sin θ + cos θ)(1 - 2sin θ cos θ)</p><p><strong>Step 5:</strong> Recognize that 1 - 2sin θ cos θ = sin²θ + cos²θ - 2sin θ cos θ = (sin θ - cos θ)²</p><p>∴ f(θ) = (sin θ + cos θ)(sin θ - cos θ)²</p><p><strong>Conclusion:</strong> The factored form shows f(θ) = 0 when sin θ = -cos θ or sin θ = cos θ, matching options A and C.</p>
Correct Answer: A,C