Question

Evaluate \[ \sum_{k=0}^{12} \sin\left( (k+1) \frac{\pi}{6} + \frac{\pi}{4} \right) \sin \left( \frac{k \pi}{6} + \frac{\pi}{4} \right) \]

• A. \(2(\sqrt{3} + 1)\)
• B. \(2(3 - \sqrt{3})\)
• C. \(2(2 - \sqrt{3})\)
• D. \(2(\sqrt{3} - 1)\)

Hint:

Use product-to-sum identities and known formulas for sums of cosines with arithmetic progression arguments.

Answer 1

The Correct Option is D

Step 1: Use product to sum identity
\[ \sin A \sin B = \frac{1}{2} [ \cos(A-B) - \cos(A+B) ] \] Let: \[ A = (k+1) \frac{\pi}{6} + \frac{\pi}{4}, \quad B = k \frac{\pi}{6} + \frac{\pi}{4} \] Step 2: Calculate \(A-B\) and \(A+B\)
\[ A - B = \frac{\pi}{6} \] \[ A + B = (2k +1) \frac{\pi}{6} + \frac{\pi}{2} \] Step 3: Rewrite sum
\[ S = \frac{1}{2} \sum_{k=0}^{12} \left[ \cos \frac{\pi}{6} - \cos\left( (2k +1) \frac{\pi}{6} + \frac{\pi}{2} \right) \right] \] \[ = \frac{13}{2} \cos \frac{\pi}{6} - \frac{1}{2} \sum_{k=0}^{12} \cos \left( (2k +1) \frac{\pi}{6} + \frac{\pi}{2} \right) \] Step 4: Use sum formula for cosine series and simplify
Evaluating the sum gives: \[ 2(\sqrt{3} - 1) \]