Trigonometry & Inverse Trigonometry
Summation of inverse trigonometric series
Grade 12
Question:
<p>The value of \( \displaystyle\sum_{\omega=1}^{\infty} \sin^{-1}\left[\dfrac{2\omega+1}{\omega(\omega+1)(\sqrt{\omega^2+2\omega}+\sqrt{\omega^2-1})}\right] \) is equal to:</p>
<p>(a) \( \dfrac{\pi}{4} \)</p>
<p>(b) \( \dfrac{\pi}{6} \)</p>
<p>(c) \( \dfrac{3\pi}{4} \)</p>
<p>(d) \( \dfrac{\pi}{2} \)</p>
Step-by-Step Solution
Key Concept: Rationalize the denominator by multiplying by the conjugate to simplify the argument of arcsin, then recognize that the resulting expression can be written as a telescoping sum using the identity sin⁻¹(a) - sin⁻¹(b) = sin⁻¹(a√(1-b²) - b√(1-a²)).
<p><strong>Step 1:</strong> Rationalize the denominator by multiplying by the conjugate:</p><p>Multiply numerator and denominator by (√(ω²+2ω) - √(ω²-1)):</p><p>Denominator becomes: (ω²+2ω) - (ω²-1) = 2ω+1</p><p>This gives: <br/><span style='font-family:monospace'>[2ω+1]/[ω(ω+1)(2ω+1)] · [√(ω²+2ω) - √(ω²-1)]</span></p><p><strong>Step 2:</strong> Simplify to get the argument:</p><p>= [√(ω²+2ω) - √(ω²-1)]/[ω(ω+1)]</p><p>= √[ω(ω+2)]/[ω(ω+1)] - √[(ω-1)(ω+1)]/[ω(ω+1)]</p><p>= √[(ω+2)/(ω+1)]/√ω - √[(ω-1)/ω]/√(ω+1)</p><p><strong>Step 3:</strong> Recognize the telescoping pattern:</p><p>Notice that sin⁻¹(√[(ω+1)/(ω+2)]) - sin⁻¹(√[ω/(ω+1)]) produces exactly this argument.</p><p><strong>Step 4:</strong> Sum the telescoping series:</p><p>∑ᵩ₌₁^∞ {sin⁻¹(√[(ω+1)/(ω+2)]) - sin⁻¹(√[ω/(ω+1)])}</p><p>As ω→∞: First term → sin⁻¹(1) = π/2; Last uncancelled term (ω=1) → sin⁻¹(√(1/2)) = π/4</p><p>∴ Answer: <strong>π/2 - π/4 = π/4</strong></p>
Correct Answer: D