Trigonometry & Inverse Trigonometry
Trigonometric Sums
Grade 11
Question:
<p>If \(\sum_{m=1}^{6} \csc\left(\alpha + (m-1)\dfrac{\pi}{4}\right)\csc\left(\alpha + \dfrac{m\pi}{4}\right) = 4\sqrt{2}\), where \(\alpha \in (0, \pi)\) then \(\alpha\) can be:</p>
<p>\(\dfrac{\pi}{12}\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
<p>\(\dfrac{5\pi}{12}\)</p>
<p>\(\dfrac{\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the telescoping identity: csc(A)csc(B) = cot(A) - cot(B) when B - A = π/4. This converts the sum into a telescoping series where consecutive terms cancel.
<p><strong>Step 1: Identify the telescoping pattern</strong></p><p>Use the identity: csc(A)csc(B) = [cot(A) - cot(B)]/sin(B-A) when B - A is constant.</p><p>Here, consecutive angles differ by π/4, so: csc(θ)csc(θ + π/4) = [cot(θ) - cot(θ + π/4)]/sin(π/4) = √2[cot(θ) - cot(θ + π/4)]</p><p><strong>Step 2: Apply to each term</strong></p><p>For m = 1 to 6, let θₘ = α + (m-1)π/4:</p><p>∑ csc(θₘ)csc(θₘ + π/4) = √2 ∑[cot(θₘ) - cot(θₘ + π/4)]</p><p><strong>Step 3: Evaluate the telescoping sum</strong></p><p>This telescopes to: √2[cot(α) - cot(α + 6π/4)] = √2[cot(α) - cot(α + 3π/2)]</p><p>Since cot(α + 3π/2) = cot(α + π/2) = -tan(α):</p><p>√2[cot(α) + tan(α)] = √2 · [cos(α)/sin(α) + sin(α)/cos(α)] = √2 · [cos²(α) + sin²(α)]/[sin(α)cos(α)]</p><p>= √2 · 2/sin(2α) = 2√2/sin(2α)</p><p><strong>Step 4: Solve for α</strong></p><p>2√2/sin(2α) = 4√2</p><p>1/sin(2α) = 2</p><p>sin(2α) = 1/2</p><p>2α = π/6 or 2α = 5π/6 (in [0, 2π])</p><p>α = π/12 or α = 5π/12 (both in (0, π)) ✓</p><p>∴ Answer: A</p>
Correct Answer: A