Sequences & Series
Telescoping/Trigonometric Series
Grade 11
Question:
<p>The value of \(\displaystyle\sum_{k=1}^{13} \dfrac{1}{\sin\!\left(\dfrac{\pi}{4} + \dfrac{(k-1)\pi}{6}\right)\sin\!\left(\dfrac{\pi}{4} + \dfrac{k\pi}{6}\right)}\) is equal to</p>
<p>\(3 - \sqrt{3}\)</p>
<p>\(2(3 - \sqrt{3})\)</p>
<p>\(2(\sqrt{3} - 1)\)</p>
<p>\(2(2 + \sqrt{3})\)</p>
Step-by-Step Solution
Key Concept: Use the telescoping identity: 1/[sin(A)sin(B)] = (1/sin(B-A))[cot(A) - cot(B)] where B - A = π/6 is constant. This converts the sum into a telescoping series where consecutive terms cancel.
<p><strong>Step 1:</strong> Identify the telescoping pattern. Note that consecutive angles differ by π/6.</p><p>Let A_k = π/4 + (k-1)π/6 and B_k = π/4 + kπ/6, so B_k - A_k = π/6.</p><p><strong>Step 2:</strong> Apply the cotangent difference formula:</p><p>$$\frac{1}{\sin(A_k)\sin(B_k)} = \frac{1}{\sin(\pi/6)}[\cot(A_k) - \cot(B_k)] = 2[\cot(A_k) - \cot(B_k)]$$</p><p><strong>Step 3:</strong> Write the telescoping sum:</p><p>$$\sum_{k=1}^{13} 2[\cot(A_k) - \cot(B_k)] = 2\left[\cot\left(\frac{\pi}{4}\right) - \cot\left(\frac{\pi}{4} + \frac{13\pi}{6}\right)\right]$$</p><p><strong>Step 4:</strong> Simplify the boundary terms. Note that 13π/6 = 2π + π/6, so:</p><p>$$\cot\left(\frac{\pi}{4} + \frac{13\pi}{6}\right) = \cot\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \cot\left(\frac{5\pi}{12}\right)$$</p><p><strong>Step 5:</strong> Calculate: cot(π/4) = 1 and cot(5π/12) = 2 - √3</p><p>$$2[1 - (2 - \sqrt{3})] = 2(\sqrt{3} - 1) = 2\sqrt{3} - 2$$</p><p>∴ Answer: C</p>
Correct Answer: C