Trigonometry & Inverse Trigonometry
Trigonometric inequalities
Grade 11
Question:
<p>Let \(x = \sin\theta\cos^3\theta\) and \(y = \sin^3\theta\cos\theta\), then:</p>
<p>(a) if \(0 < \theta < \dfrac{\pi}{4}\), then \(x + y > 0\)</p>
<p>(b) if \(\dfrac{\pi}{4} < \theta < \dfrac{\pi}{2}\), then \(x < y\)</p>
<p>(c) if \(\dfrac{\pi}{2} < \theta < \dfrac{3\pi}{4}\), then \(x + y > 0\)</p>
<p>(d) if \(\dfrac{3\pi}{4} < \theta < \pi\), then \(x < y\)</p>
Step-by-Step Solution
Key Concept: Factor out common terms from x and y to express them in terms of sin(2θ), then analyze the relationship between x+y and x-y using double angle formulas to determine valid inequalities.
<p><strong>Step 1:</strong> Express x and y in terms of double angles.</p><p>x = sin θ cos³θ = (sin θ cos θ)cos²θ = (sin 2θ/2) · (1 + cos 2θ)/2</p><p>y = sin³θ cos θ = (sin θ cos θ)sin²θ = (sin 2θ/2) · (1 - cos 2θ)/2</p><p><strong>Step 2:</strong> Find x + y and x - y.</p><p>x + y = (sin 2θ/2)[(1 + cos 2θ)/2 + (1 - cos 2θ)/2] = (sin 2θ/2) = (1/4)sin 4θ</p><p>x - y = (sin 2θ/2)[(1 + cos 2θ)/2 - (1 - cos 2θ)/2] = (sin 2θ/2) · cos 2θ = (1/4)sin 4θ</p><p><strong>Step 3:</strong> Determine maximum and minimum values.</p><p>Since -1 ≤ sin 4θ ≤ 1, we have |x + y| ≤ 1/4</p><p>Since x·y = (sin²θ cos²θ)(sin²θ cos²θ) = sin⁴θ cos⁴θ = (sin 2θ/2)⁴ ≤ 1/16</p><p><strong>Step 4:</strong> Test boundary conditions (θ = π/4, θ = 0, etc.) to verify which statements hold for all valid θ.</p><p>∴ Answer: ABD</p>
Correct Answer: ABD