<p>Let \(z = r(\cos\theta + i\sin\theta)\). The least value of \(\dfrac{\sin 5\theta}{(\sin\theta)^5}\) expressed in terms of \(\cot\theta\) is:</p>
Step-by-Step Solution
Key Concept: Use Chebyshev polynomials or De Moivre's theorem to expand sin(5θ) in terms of sin(θ) and cos(θ), then express the ratio as a polynomial in cot(θ) to find its minimum.
<p><strong>Step 1:</strong> Use De Moivre's theorem: cos(5θ) + i·sin(5θ) = (cos(θ) + i·sin(θ))⁵</p><p><strong>Step 2:</strong> Expand using binomial theorem and extract imaginary part:<br/>sin(5θ) = sin(θ)[16cos⁴(θ) - 12cos²(θ) + 1]</p><p><strong>Step 3:</strong> Therefore: sin(5θ)/sin⁵(θ) = [16cos⁴(θ) - 12cos²(θ) + 1]/sin⁴(θ)</p><p><strong>Step 4:</strong> Divide numerator and denominator by sin⁴(θ):<br/>= 16cot⁴(θ) - 12cot²(θ) + 1</p><p><strong>Step 5:</strong> Let x = cot²(θ) where x ≥ 0:<br/>f(x) = 16x² - 12x + 1</p><p><strong>Step 6:</strong> Find minimum by taking derivative: f'(x) = 32x - 12 = 0<br/>x = 3/8</p><p><strong>Step 7:</strong> Minimum value = 16(9/64) - 12(3/8) + 1 = 9/4 - 9/2 + 1 = 9/4 - 18/4 + 4/4 = <strong>-5/4</strong></p><p>∴ Answer: A</p>
Correct Answer: A