Trigonometry & Inverse Trigonometry
Trigonometric expressions and summation
Grade 11
Question:
<p><strong>989.</strong> Let \(T(n) = \cos^2(30° - n°) - \cos(30° - n°)\cos(30° + n°) + \cos^2(30° + n°)\). Find the value of \(4\displaystyle\sum_{n=1}^{30} nT(n)\).</p>
Step-by-Step Solution
Key Concept: Simplify T(n) using the identity (a-b)² + (a+b)² = 2(a²+b²) and product-to-sum formulas, recognizing that T(n) is actually constant and equals 3/4 for all n.
<p><strong>Step 1: Simplify T(n)</strong></p><p>Let A = cos²(30° - n°), B = cos(30° - n°)cos(30° + n°), C = cos²(30° + n°)</p><p>Then T(n) = A - B + C</p><p><strong>Step 2: Use product formula</strong></p><p>cos(30° - n°)cos(30° + n°) = ½[cos(60°) + cos(2n°)] = ½[1/2 + cos(2n°)] = 1/4 + ½cos(2n°)</p><p><strong>Step 3: Use sum of squares</strong></p><p>cos²(30° - n°) + cos²(30° + n°) = ½[2 + cos(60° - 2n°) + cos(60° + 2n°)]</p><p>= 1 + ½[cos(60° - 2n°) + cos(60° + 2n°)] = 1 + cos(60°)cos(2n°) = 1 + ½cos(2n°)</p><p><strong>Step 4: Calculate T(n)</strong></p><p>T(n) = [1 + ½cos(2n°)] - [1/4 + ½cos(2n°)] = 3/4</p><p><strong>Step 5: Evaluate the sum</strong></p><p>4∑(n=1 to 30) nT(n) = 4 · (3/4) · ∑(n=1 to 30) n = 3 · [30·31/2] = 3 · 465</p><p><strong>∴ Answer: 1395</strong></p>
Correct Answer: 1395