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

Step 1: Simplify the given sum expression.

We are given the sum:
$\sum_{r=1}^{13} \frac{1}{\sin \frac{\pi}{6}} \sin \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right) \sin \left( \frac{\pi}{4} + \frac{\pi}{6} \right)$

By using trigonometric identities, this becomes:
$\frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \sin \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right)$

Further simplification leads to:
$\sum_{r=1}^{13} \cot \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right) - \cot \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right)$

Step 2: Identify constants $a$ and $b$

From the result:
$2\sqrt{3} - 2 = a\sqrt{3} + b$

Step 3: Compute $a^2 + b^2$

By comparing terms, we have:
$a^2 + b^2 = 8$