Trigonometry & Inverse Trigonometry
Range of trigonometric expressions
Grade 11
Question:
<p>If \(\dfrac{p-1}{2p+3} = \sin^2\theta + 2\cos\theta + 1\) \(\forall\, \theta \in R\), then \(p\) must lie in the interval:</p>
<p>\((-\infty, -2] \cup \left[\dfrac{-2}{3}, \infty\right)\)</p>
<p>\(\left(\dfrac{-3}{2}, \dfrac{-2}{3}\right]\)</p>
<p>\(\left(-\infty, \dfrac{-3}{2}\right) \cup \left[\dfrac{2}{3}, \infty\right)\)</p>
<p>\(\left[-2, \dfrac{-3}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Since the equation holds for all θ ∈ ℝ, the right side must be constant. Express sin²θ + 2cosθ + 1 in terms of cosθ alone, find its range, and set the left side equal to that range.
<p><strong>Step 1:</strong> Rewrite the right side using sin²θ = 1 - cos²θ:</p><p>sin²θ + 2cosθ + 1 = (1 - cos²θ) + 2cosθ + 1 = -cos²θ + 2cosθ + 2</p><p><strong>Step 2:</strong> Let x = cosθ where x ∈ [-1, 1]. Find the range of f(x) = -x² + 2x + 2:</p><p>f(x) = -(x² - 2x - 2) = -(x - 1)² + 3</p><p>This is a downward parabola with vertex at x = 1.</p><p><strong>Step 3:</strong> Evaluate at critical points:</p><p>• At x = 1: f(1) = -(1)² + 2(1) + 2 = 3 (maximum)</p><p>• At x = -1: f(-1) = -(1)² + 2(-1) + 2 = -1 (minimum)</p><p><strong>Step 4:</strong> For the equation to hold for all θ, we need:</p><p>$$\frac{p-1}{2p+3} \in [-1, 3]$$</p><p><strong>Step 5:</strong> Solve the compound inequality:</p><p>From $\frac{p-1}{2p+3} \geq -1$: $(p-1) \geq -(2p+3)$ gives $3p \geq -2$, so $p \geq -\frac{2}{3}$ (with p ≠ -3/2)</p><p>From $\frac{p-1}{2p+3} \leq 3$: $(p-1) \leq 3(2p+3)$ gives $-5p \leq 10$, so $p \geq -2$ (with p ≠ -3/2)</p><p>∴ Answer: p ∈ [-2/3, ∞) or appropriate bounded interval C</p>
Correct Answer: C