Trigonometry & Inverse Trigonometry
Summation of inverse trigonometric series
Grade 12
Question:
<p>If \( S_n = \sum_{r=1}^{n} \cot^{-1}(r^2 + 3r + 3) \), then:</p>
<p>(a) \( S_\infty = \cot^{-1}(2) \)</p>
<p>(b) \( S_5 = \cot^{-1}(3) \)</p>
<p>(c) \( S_6 = \cot^{-1}\left(\dfrac{17}{6}\right) \)</p>
<p>(d) \( S_8 = \cot^{-1}(5) \)</p>
Step-by-Step Solution
Key Concept: Use the telescoping property: cot⁻¹(r² + 3r + 3) = tan⁻¹(1/(r² + 3r + 3)) can be decomposed as tan⁻¹(r+2) - tan⁻¹(r+1) using the identity tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)). This creates a telescoping series where consecutive terms cancel.
<p><strong>Step 1:</strong> Recognize that cot⁻¹(x) = tan⁻¹(1/x). So cot⁻¹(r² + 3r + 3) = tan⁻¹(1/(r² + 3r + 3))</p><p><strong>Step 2:</strong> Use the identity: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)). Set a = r+2 and b = r+1.</p><p><strong>Step 3:</strong> Verify: (a-b)/(1+ab) = ((r+2)-(r+1))/(1+(r+2)(r+1)) = 1/(1+r²+3r+2) = 1/(r²+3r+3) ✓</p><p><strong>Step 4:</strong> Therefore: cot⁻¹(r² + 3r + 3) = tan⁻¹(r+2) - tan⁻¹(r+1)</p><p><strong>Step 5:</strong> Sum telescopes:</p><p>S_n = Σ[tan⁻¹(r+2) - tan⁻¹(r+1)]</p><p>= [tan⁻¹(3) - tan⁻¹(2)] + [tan⁻¹(4) - tan⁻¹(3)] + ... + [tan⁻¹(n+2) - tan⁻¹(n+1)]</p><p>= tan⁻¹(n+2) - tan⁻¹(2)</p><p><strong>Step 6:</strong> As n → ∞: S_∞ = π/2 - tan⁻¹(2)</p><p>∴ Answer: A</p>
Correct Answer: A