The value of $\sum_{k=1}^{13} \sin(\frac{\pi}{4} + (k-\frac{1}{2})\frac{\pi}{6})\sin(\frac{\pi}{4} + k\frac{\pi}{6})$ is equal to

$\sum_{k=1}^{13} \sin(\frac{\pi}{4} + k\frac{\pi}{6} - \frac{\pi}{12})\sin(\frac{\pi}{4} + k\frac{\pi}{6}) = \sum_{k=1}^{13} \frac{1}{2} \left[ \cos(\frac{\pi}{12}) - \cos(\frac{\pi}{2} + \frac{2k\pi}{6} - \frac{\pi}{12}) \right] = \frac{1}{2} \sum_{k=1}^{13} \left[ \cos(\frac{\pi}{12}) - \sin(\frac{2k\pi}{6} - \frac{\pi}{12}) \right]$ \n$\sum_{k=1}^{13} \sin(\frac{\pi}{4} + (k-\frac{1}{2})\frac{\pi}{6})\sin(\frac{\pi}{4} + k\frac{\pi}{6}) = \sum_{k=1}^{13} \frac{1}{2} \left[ \cos(\frac{\pi}{12}) - \cos(\frac{\pi}{2} + \frac{2k\pi}{6} - \frac{\pi}{12}) \right] = \frac{1}{2} \sum_{k=1}^{13} \left[ \cos(\frac{\pi}{12}) - \sin(\frac{2k\pi}{6} - \frac{\pi}{12}) \right]$ \n$\sum_{k=1}^{13} \frac{1}{2} \left[ \cos(\frac{\pi}{12}) - \cos(\frac{\pi}{2} + \frac{k\pi}{3} - \frac{\pi}{12}) \right] = \frac{1}{2} \sum_{k=1}^{13} \left[ \cos(\frac{\pi}{12}) + \sin(\frac{k\pi}{3} - \frac{\pi}{12}) \right]$ \n$ = \frac{13}{2} \cos(\frac{\pi}{12}) + \frac{1}{2} \sum_{k=1}^{13} \sin(\frac{k\pi}{3} - \frac{\pi}{12}) \approx 5.66$ \n$ = \sum_{k=1}^{13} \frac{1}{2} \left[ \cos(\frac{\pi}{12}) - \cos(\frac{2k\pi}{6} - \frac{\pi}{12} + \frac{\pi}{2}) \right] = \sum_{k=1}^{13} \frac{1}{2} \left[ \cos(\frac{\pi}{12}) + \sin(\frac{k\pi}{3} - \frac{\pi}{12}) \right] = \frac{13}{2} \cos(\frac{\pi}{12}) + \frac{1}{2} \sum_{k=1}^{13} \sin(\frac{k\pi}{3} - \frac{\pi}{12}) \approx 5.66$ \n$\sum_{k=1}^{13} \sin(\frac{\pi}{4} + (k-\frac{1}{2})\frac{\pi}{6})\sin(\frac{\pi}{4} + k\frac{\pi}{6}) = \frac{1}{2} \sum_{k=1}^{13} \left[ \cos(\frac{\pi}{12}) - \cos(\frac{\pi}{2} + \frac{k\pi}{3} - \frac{\pi}{12}) \right] = \frac{1}{2} \sum_{k=1}^{13} \left[ \cos(\frac{\pi}{12}) + \sin(\frac{k\pi}{3} - \frac{\pi}{12}) \right]$ \n$13\sum_{k=1} \sin[\frac{\pi}{4} + k\frac{\pi}{6} - (\frac{\pi}{4} + (k-\frac{1}{2})\frac{\pi}{6})] = \frac{1}{\sin(\frac{\pi}{12})} \sum_{k=1}^{13} \sin(\frac{\pi}{12}) = 13\sum_{k=1} \sin(\frac{\pi}{12}) \approx 6.62$ \n$ = \frac{1}{2} \sum_{k=1}^{13} \left[ \cos(\frac{\pi}{12}) - \cos(\frac{\pi}{2} + \frac{2k\pi}{6} - \frac{\pi}{12}) \right] = \frac{13}{2} \cos(\frac{\pi}{12}) + \frac{1}{2} \sum_{k=1}^{13} \sin(\frac{k\pi}{3} - \frac{\pi}{12})$ \n$2(\sqrt{3} - 1)$