Trigonometry & Inverse Trigonometry
Summation of trigonometric series
Grade 11
Question:
<p>The value of \(\displaystyle\sum_{r=0}^{10} \cos^3\dfrac{r\pi}{3}\) is equal to \(\dfrac{-a}{b}\), then the value of \(b\) is (where g.c.d of \((a,b)\) is 1).</p>
Step-by-Step Solution
Key Concept: Use the triple angle formula cos(3θ) = 4cos³(θ) - 3cos(θ) to express cos³(θ) = [cos(3θ) + 3cos(θ)]/4, then exploit the periodicity of cosine with period 2π to evaluate the sum over r = 0 to 10.
<p><strong>Step 1:</strong> Use the triple angle reduction formula: cos³(θ) = [cos(3θ) + 3cos(θ)]/4</p><p><strong>Step 2:</strong> Apply to our sum: Σ(r=0 to 10) cos³(rπ/3) = (1/4)Σ(r=0 to 10)[cos(rπ) + 3cos(rπ/3)]</p><p><strong>Step 3:</strong> Evaluate Σ(r=0 to 10) cos(rπ): The values alternate ±1 with period 2. We get 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 = 1</p><p><strong>Step 4:</strong> Evaluate Σ(r=0 to 10) cos(rπ/3): Since cos has period 6, the sequence repeats. One full cycle (r=0 to 5) gives: cos(0) + cos(π/3) + cos(2π/3) + cos(π) + cos(4π/3) + cos(5π/3) = 1 + 1/2 - 1/2 - 1 - 1/2 + 1/2 = 0. For r=0 to 10, we have one full cycle (0-5) plus one partial cycle (6-10): partial = 1 + 1/2 - 1/2 - 1 - 1/2 = -1/2</p><p><strong>Step 5:</strong> Total: Σ cos(rπ/3) = 0 + (-1/2) = -1/2</p><p><strong>Step 6:</strong> Therefore: Σ cos³(rπ/3) = (1/4)[1 + 3(-1/2)] = (1/4)[1 - 3/2] = (1/4)(-1/2) = -1/8</p><p>∴ Answer: -1/8 = -a/b, so a = 1, b = 8. Therefore <strong>b = 8</strong></p>
Correct Answer: -1