Question

If \[ \sum_{r=1}^{13} \frac{1}{\sin \frac{\pi}{4} + (r-1) \frac{\pi}{6}} \sin \frac{\pi}{4} + \frac{\pi}{6} = a \sqrt{3} + b, \quad a, b \in \mathbb{Z}, { then } a^2 + b^2 { is equal to:} \]

• A. 10
• B. 4
• C. 8
• D. 2

Hint:

In problems involving sums of trigonometric expressions, look for trigonometric identities to simplify the sum and identify patterns.

Answer 1

The Correct Option is C

We are given the sum \[ \sum_{r=1}^{13} \frac{1}{\sin \frac{\pi}{6}} \sin \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right) \sin \left( \frac{\pi}{4} + \frac{\pi}{6} \right) \] Step 1: Simplify the sum expression. By using trigonometric identities, we have: \[ \frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \sin \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right) \] which can be rewritten as \[ \sum_{r=1}^{13} \cot \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right) - \cot \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right) \] leading to: \[ 2\sqrt{3} - 2 = \alpha \sqrt{3} + b \] Step 3: so \( a^2 + b^2 \). \[ a^2 + b^2 = 8 \]