Trigonometry & Inverse Trigonometry
Trigonometric identities and summation
Grade 11
Question:
<p>If <em>T</em>(<em>n</em>) = cos²(30° − <em>n</em>°) − cos(30° − <em>n</em>°)cos(30° + <em>n</em>°) + cos²(30° + <em>n</em>°), find the value of \(4\sum_{n=1}^{30} nT(n)\).</p>
Step-by-Step Solution
Key Concept: Recognize that T(n) is independent of n by using the algebraic identity (a-b)²+(a+b)²-2ab = 2a² and product-to-sum formulas to simplify the expression to a constant.
<p><strong>Step 1: Simplify T(n) using algebraic identity</strong></p><p>T(n) = cos²(30° − n°) − cos(30° − n°)cos(30° + n°) + cos²(30° + n°)</p><p>Let A = cos(30° − n°) and B = cos(30° + n°)</p><p>T(n) = A² − AB + B²</p><p><strong>Step 2: Use the product-to-sum formula</strong></p><p>cos(30° − n°)cos(30° + n°) = ½[cos(60°) + cos(2n°)] = ¼ + ½cos(2n°)</p><p><strong>Step 3: Simplify A² + B²</strong></p><p>A² + B² = cos²(30° − n°) + cos²(30° + n°)</p><p>Using cos²θ = (1 + cos2θ)/2:</p><p>A² + B² = 1 + ½[cos(60° − 2n°) + cos(60° + 2n°)]</p><p>= 1 + ½·2cos(60°)cos(2n°) = 1 + ½cos(2n°)</p><p><strong>Step 4: Calculate T(n)</strong></p><p>T(n) = [1 + ½cos(2n°)] − [¼ + ½cos(2n°)] = ¾</p><p><strong>Step 5: Compute the sum</strong></p><p>∑(n=1 to 30) nT(n) = ¾∑(n=1 to 30) n = ¾ · 30·31/2 = ¾ · 465 = 348.75</p><p><strong>Step 6: Final answer</strong></p><p>4∑nT(n) = 4 × 348.75 = <strong>1395</strong></p>
Correct Answer: 1395