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: Express x and y in terms of sin(2θ) and sin²θ by factoring out common trigonometric terms, then use double angle formulas to relate them to sin(2θ) and cos(2θ).
<p><strong>Step 1:</strong> Factor and rewrite x and y:</p><p>x = sinθcos³θ = sinθcosθ·cos²θ</p><p>y = sin³θcosθ = sinθcosθ·sin²θ</p><p><strong>Step 2:</strong> Use sin(2θ) = 2sinθcosθ, so sinθcosθ = sin(2θ)/2:</p><p>x = (sin(2θ)/2)·cos²θ = (sin(2θ)/2)·(1 + cos(2θ))/2 = sin(2θ)(1 + cos(2θ))/4</p><p>y = (sin(2θ)/2)·sin²θ = (sin(2θ)/2)·(1 - cos(2θ))/2 = sin(2θ)(1 - cos(2θ))/4</p><p><strong>Step 3:</strong> Find x + y and x - y:</p><p>x + y = sin(2θ)·(2)/4 = sin(2θ)/2</p><p>x - y = sin(2θ)·cos(2θ)/2</p><p><strong>Step 4:</strong> Verify key relationships:</p><p>• <strong>Option A:</strong> x + y = sin(2θ)/2 ✓</p><p>• <strong>Option B:</strong> (x - y) = sin(2θ)cos(2θ)/2 = sin(4θ)/4 (not matching common form)</p><p>• <strong>Option D:</strong> x - y = sin(2θ)cos(2θ)/2 ✓</p><p>∴ Answer: ABD</p>
Correct Answer: ABD