Calculate Summation of Trigonometric Product Series

To evaluate the summation $S = \sum_{k=1}^{13} \frac{1}{\sin\left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right) \sin\left(\frac{\pi}{4} + \frac{k\pi}{6}\right)}$

Let us define $\theta_k = \frac{\pi}{4} + \frac{k\pi}{6}$.

The difference between two consecutive terms in the argument is constant: $\theta_k - \theta_{k-1} = \left(\frac{\pi}{4} + \frac{k\pi}{6}\right) - \left(\frac{\pi}{4} + \frac{(k-1)\pi}{6}\right) = \frac{\pi}{6}$

Taking the sine on both sides, we get: $\sin(\theta_k - \theta_{k-1}) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

We can rewrite the general term $T_k$ of the series by multiplying and dividing by $\sin\left(\frac{\pi}{6}\right)$: $T_k = \frac{1}{\sin \theta_{k-1} \sin \theta_k} = \frac{1}{\sin\left(\frac{\pi}{6}\right)} \cdot \frac{\sin(\theta_k - \theta_{k-1})}{\sin \theta_{k-1} \sin \theta_k}$

Using the sine subtraction formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we obtain: $\frac{\sin(\theta_k - \theta_{k-1})}{\sin \theta_{k-1} \sin \theta_k} = \frac{\sin \theta_k \cos \theta_{k-1} - \cos \theta_k \sin \theta_{k-1}}{\sin \theta_{k-1} \sin \theta_k} = \cot \theta_{k-1} - \cot \theta_k$

Since $\frac{1}{\sin\left(\frac{\pi}{6}\right)} = 2$, the term $T_k$ simplifies to: $T_k = 2(\cot \theta_{k-1} - \cot \theta_k)$

Now, substituting $T_k$ into the summation yields a telescoping series: $S = \sum_{k=1}^{13} T_k = 2 \sum_{k=1}^{13} (\cot \theta_{k-1} - \cot \theta_k) = 2 (\cot \theta_0 - \cot \theta_{13})$

Next, we calculate the required angles $\theta_0$ and $\theta_{13}$:

1. For $k = 0$: $\theta_0 = \frac{\pi}{4} \implies \cot \theta_0 = \cot\left(\frac{\pi}{4}\right) = 1$

2. For $k = 13$: $\theta_{13} = \frac{\pi}{4} + \frac{13\pi}{6} = 2\pi + \left(\frac{\pi}{4} + \frac{\pi}{6}\right) = 2\pi + \frac{5\pi}{12}$ Using the periodicity of the cotangent function: $\cot \theta_{13} = \cot\left(2\pi + \frac{5\pi}{12}\right) = \cot\left(\frac{5\pi}{12}\right) = \tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}$

Substituting these values back into the telescoping sum: $S = 2 \left[ 1 - (2 - \sqrt{3}) \right] = 2 (1 - 2 + \sqrt{3}) = 2 (\sqrt{3} - 1)$

Thus, the correct option is C.