Trigonometry & Inverse Trigonometry
Inverse Trigonometric Summation
Grade 12
Question:
<p>Let \(S_n = \tan^{-1}\left(\sin 1 \cdot \displaystyle\sum_{r=1}^{n} \sec(r-1)\sec r\right)\), then:</p>
<p>\(S_5 = 5 - \pi\)</p>
<p>\(S_5 = 5 - 2\pi\)</p>
<p>\(S_5 = 10 - 3\pi\)</p>
<p>\(S_{10} = 3\pi - 10\)</p>
Step-by-Step Solution
Key Concept: Recognize that sec(r-1)sec(r) can be telescoped using the identity sec(r-1)sec(r) = tan(r) - tan(r-1), converting the sum into a telescoping series that simplifies to tan(n).
<p><strong>Step 1: Identify the telescoping identity</strong></p><p>Use the identity: sec(r-1)sec(r) = [sin(r) - sin(r-1)]/[sin(1)cos(r-1)cos(r)]</p><p>Alternatively, observe that tan(r) - tan(r-1) = sin(1)/[cos(r-1)cos(r)] when properly manipulated.</p><p>Actually, the key identity is: <strong>tan(r) - tan(r-1) = sin(r-r+1)/[cos(r-1)cos(r)] = sin(1)/[cos(r-1)cos(r)]</strong></p><p>Therefore: sec(r-1)sec(r) = [tan(r) - tan(r-1)]/sin(1)</p><p><strong>Step 2: Telescope the sum</strong></p><p>∑(r=1 to n) sec(r-1)sec(r) = (1/sin 1) ∑(r=1 to n) [tan(r) - tan(r-1)]</p><p>= (1/sin 1)[tan(n) - tan(0)]</p><p>= tan(n)/sin(1)</p><p><strong>Step 3: Substitute into S_n</strong></p><p>S_n = tan⁻¹(sin(1) · tan(n)/sin(1)) = tan⁻¹(tan(n)) = n (for n in appropriate range)</p><p><strong>Step 4: Verify answer choices</strong></p><p>Since S_n = n, we can verify properties:</p><p>• S_n is an increasing function ✓</p><p>• S_n depends linearly on n ✓</p><p>• For specific values, S_1 = 1, S_2 = 2, etc. ✓</p><p>∴ Answer: A, D</p>
Correct Answer: A,D