Trigonometry & Inverse Trigonometry
Summation of inverse trigonometric series
Grade 12
Question:
<p>Let \(\sum_{k=1}^{\infty} \sin^{-1}\left(\dfrac{\sqrt{k} - \sqrt{k-1}}{\sqrt{k(k+1)}}\right) = \theta\). Then:</p>
<p>the value of \(\tan\dfrac{\theta}{2}\) is equal to \(\sqrt{2} - 1\)</p>
<p>\(\lim_{x \to 0}\left(1 + \dfrac{x}{\tan x}\right)^{\frac{2}{x-\theta}} = e^{-\pi}\)</p>
<p>the value of \(\sin\theta\) is equal to 1</p>
<p>\(\lim_{x \to \theta} \dfrac{(x - \cos x - \theta)}{x - \theta} = 2\)</p>
Step-by-Step Solution
Key Concept: Recognize that the general term can be telescoped by expressing it as a difference of inverse sine functions: sin⁻¹(√k/√(k+1)) - sin⁻¹(√(k-1)/√k). This converts an infinite series into a telescoping sum.
<p><strong>Step 1:</strong> Rationalize and simplify the general term. Note that:</p><p>√k - √(k-1) = (k - (k-1))/(√k + √(k-1)) = 1/(√k + √(k-1))</p><p><strong>Step 2:</strong> The denominator √(k(k+1)) suggests we rewrite the argument. Observe that:</p><p>sin⁻¹(√k/√(k+1)) - sin⁻¹(√(k-1)/√k) when expanded using sin(A-B) formula yields exactly (√k - √(k-1))/√(k(k+1))</p><p><strong>Step 3:</strong> Therefore, the series becomes telescoping:</p><p>∑(k=1 to ∞) [sin⁻¹(√k/√(k+1)) - sin⁻¹(√(k-1)/√k)]</p><p><strong>Step 4:</strong> Writing out terms:</p><p>[sin⁻¹(√1/√2) - sin⁻¹(0)] + [sin⁻¹(√2/√3) - sin⁻¹(√1/√2)] + [sin⁻¹(√3/√4) - sin⁻¹(√2/√3)] + ...</p><p><strong>Step 5:</strong> All intermediate terms cancel. As k→∞, √k/√(k+1) → 1, so sin⁻¹(√k/√(k+1)) → π/2</p><p>∴ θ = π/2 - sin⁻¹(0) = π/2 - 0 = <strong>π/2</strong></p>
Correct Answer: A