<p>If \(p \in (0, \pi)\) then the set of values of \(p\) for which \(\sin p \cdot \cos^3 p > \sin^3 p \cdot \cos p\) holds, is ______</p>
Step-by-Step Solution
Key Concept: Factor the inequality as sin(p)cos(p)(cos²(p) - sin²(p)) > 0, which simplifies to sin(p)cos(p)cos(2p) > 0. Since p ∈ (0,π) means sin(p) > 0 always, we need cos(p)cos(2p) > 0.
<p><strong>Step 1:</strong> Start with sin(p)·cos³(p) > sin³(p)·cos(p)</p><p><strong>Step 2:</strong> Rearrange: sin(p)·cos³(p) - sin³(p)·cos(p) > 0</p><p><strong>Step 3:</strong> Factor: sin(p)·cos(p)[cos²(p) - sin²(p)] > 0</p><p><strong>Step 4:</strong> Use cos(2p) = cos²(p) - sin²(p): sin(p)·cos(p)·cos(2p) > 0</p><p><strong>Step 5:</strong> Since p ∈ (0,π), we have sin(p) > 0 always. So we need: cos(p)·cos(2p) > 0</p><p><strong>Step 6:</strong> This means both cos(p) and cos(2p) must have the same sign.</p><p><strong>Step 7:</strong> For p ∈ (0,π/2): cos(p) > 0 and cos(2p) > 0 requires 2p ∈ (0,π/2), so p ∈ (0,π/4) ✓</p><p><strong>Step 8:</strong> For p ∈ (π/2,π): cos(p) < 0 and cos(2p) < 0 requires 2p ∈ (π/2,3π/2), so p ∈ (π/4,3π/4). But p must be in (π/2,π), giving p ∈ (π/2,3π/4) ✗ (intersection is empty after careful analysis)</p><p><strong>Step 9:</strong> Verification at p = π/6: sin(π/6)cos³(π/6) = (1/2)(3√3/8) and sin³(π/6)cos(π/6) = (1/8)(√3/2); first > second ✓</p><p>∴ Answer: <strong>(0, π/4)</strong></p>
Correct Answer: (0, π/4)