Trigonometry & Inverse Trigonometry
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: Recognize that the argument of sin⁻¹ can be simplified using the telescoping identity sin⁻¹(a) - sin⁻¹(b) = sin⁻¹(a√(1-b²) - b√(1-a²)), combined with the rationalization of the denominator to reveal a difference of inverse sines.
<p><strong>Step 1:</strong> Simplify the denominator by rationalizing. Multiply numerator and denominator by (√(ω²+2ω) - √(ω²-1)):</p><p>Denominator becomes: (ω²+2ω) - (ω²-1) = 2ω+1</p><p><strong>Step 2:</strong> The argument becomes:</p><p>$$\frac{(2\omega+1)(\sqrt{\omega^2+2\omega} - \sqrt{\omega^2-1})}{(2\omega+1)} = \sqrt{\omega^2+2\omega} - \sqrt{\omega^2-1}$$</p><p><strong>Step 3:</strong> Notice that √(ω²+2ω) = √(ω(ω+2)) and √(ω²-1) = √((ω-1)(ω+1)). Recognize this equals sin⁻¹(√(ω+1)) - sin⁻¹(√ω).</p><p><strong>Step 4:</strong> Apply telescoping sum from ω=1 to n:</p><p>$$\sum_{\omega=1}^{n} [\sin^{-1}(\sqrt{\omega+1}) - \sin^{-1}(\sqrt{\omega})] = \sin^{-1}(\sqrt{n+1}) - \sin^{-1}(1)$$</p><p><strong>Step 5:</strong> As n→∞, sin⁻¹(√(n+1)) → π/2 and sin⁻¹(1) = π/2</p><p>∴ Answer: **π/2** or equivalent</p>
Correct Answer: D