Trigonometry & Inverse Trigonometry
Summation of inverse trigonometric series
Grade 12
Question:
<p>Let \(\displaystyle\sum_{k=1}^{\infty} \sin^{-1}\left(\frac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k(k+1)}}\right) = \theta\). Then:</p>
<p>the value of \(\tan\dfrac{\theta}{2}\) is equal to \(\sqrt{2}-1\)</p>
<p>\(\displaystyle\lim_{x\to 0}\left(1+\frac{x}{\tan x}\right)^{\frac{2}{x-\theta}} = e^{-\pi}\)</p>
<p>the value of \(\sin\theta\) is equal to 1</p>
<p>\(\displaystyle\lim_{x\to\theta}\frac{(x-\cos x-\theta)}{x-\theta}=2\)</p>
Step-by-Step Solution
Key Concept: Rationalize the numerator and recognize that each term telescopes when expressed as a difference of inverse sines: sin⁻¹(√k/(√(k+1))) - sin⁻¹(√(k-1)/(√k)).
<p><strong>Step 1: Rationalize the numerator</strong></p><p>Multiply by (√k + √(k-1))/(√k + √(k-1)):</p><p>$$\frac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k(k+1)}} \cdot \frac{\sqrt{k}+\sqrt{k-1}}{\sqrt{k}+\sqrt{k-1}} = \frac{1}{\sqrt{k(k+1)}(\sqrt{k}+\sqrt{k-1})}$$</p><p><strong>Step 2: Recognize telescoping structure</strong></p><p>Note that this equals:</p><p>$$\sin^{-1}\left(\frac{\sqrt{k}}{\sqrt{k+1}}\right) - \sin^{-1}\left(\frac{\sqrt{k-1}}{\sqrt{k}}\right)$$</p><p>This can be verified using the identity: $\sin^{-1}(a) - \sin^{-1}(b) = \sin^{-1}(a\sqrt{1-b^2} - b\sqrt{1-a^2})$</p><p><strong>Step 3: Evaluate the telescoping sum</strong></p><p>$$\sum_{k=1}^{n} \left[\sin^{-1}\left(\frac{\sqrt{k}}{\sqrt{k+1}}\right) - \sin^{-1}\left(\frac{\sqrt{k-1}}{\sqrt{k}}\right)\right]$$</p><p>$$= \sin^{-1}\left(\frac{\sqrt{n}}{\sqrt{n+1}}\right) - \sin^{-1}(0)$$</p><p><strong>Step 4: Take the limit as n→∞</strong></p><p>$$\theta = \lim_{n\to\infty} \sin^{-1}\left(\frac{\sqrt{n}}{\sqrt{n+1}}\right) = \sin^{-1}(1) = \frac{\pi}{2}$$</p><p>Therefore: $\sin\theta = 1$, $\cos\theta = 0$, $\tan\theta$ is undefined, and $\theta \in (0, \frac{\pi}{2}]$ ✓</p>
Correct Answer: ACD