Question

The value of $\displaystyle\sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$ is equal to

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

Answer 1

Answer: (C)

$\displaystyle\sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$
$=2 \displaystyle\sum_{k=1}^{13} \frac{\sin \left[\left(\frac{\pi}{4}+\frac{k \pi}{6}\right)-\left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right)\right]}{\sin \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$
$=2 \displaystyle\sum_{k=1}^{13}\left[\cot \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)\right]$
$=2\left[\cot \frac{\pi}{4}-\cot \left(\frac{\pi}{4}+\frac{\pi}{6}\right)+\cot \left(\frac{\pi}{4}+\frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{2 \pi}{6}\right)\right.$
$\left.+\ldots+\cot \left(\frac{\pi}{4}+\frac{12 \pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{13 \pi}{6}\right)\right]$
$=2\left[\cot \frac{\pi}{4}-\cot \left(\frac{\pi}{4}+\frac{13 \pi}{6}\right)\right]$
$=2[1-(2-\sqrt{3})]$
$=2(\sqrt{3}-1)$