Method: Telescoping Series

The given summation is of the form $\sum \frac{1}{\sin(A) \sin(B)}$, which often suggests a telescoping series approach.

1. Analyze the General Term

   Let the general term of the series be $T_k$. $T_k = {1 \over {\sin \left( {{\pi \over 4} + {{\left( {k - 1} \right)\pi } \over 6}} \right)\sin \left( {{\pi \over 4} + {{k\pi } \over 6}} \right)}}$ Let's define the angles in the sine functions: Let $A_k = {\pi \over 4} + {{k\pi } \over 6}$. Then the previous angle is $A_{k-1} = {\pi \over 4} + {{\left( {k - 1} \right)\pi } \over 6}$. So, the general term can be written as: $T_k = \frac{1}{\sin(A_{k-1}) \sin(A_k)}$

2. Express the General Term as a Difference

   Calculate the difference between the angles: A_k - A_{k-1} = \left( {\pi \over 4} + {{k\pi } \over 6}} \right) - \left( {\pi \over 4} + {{\left( {k - 1} \right)\pi } \over 6}} \right) = {{k\pi } \over 6} - {{(k-1)\pi } \over 6} = {\pi \over 6} This difference is a constant. We can use the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$.

   Let's multiply the numerator and denominator of $T_k$ by $\sin(A_k - A_{k-1}) = \sin(\pi/6)$. $T_k = \frac{1}{\sin(\pi/6)} \cdot \frac{\sin(A_k - A_{k-1})}{\sin(A_{k-1}) \sin(A_k)}$ Now, expand the numerator: $T_k = \frac{1}{\sin(\pi/6)} \cdot \frac{\sin(A_k)\cos(A_{k-1}) - \cos(A_k)\sin(A_{k-1})}{\sin(A_{k-1}) \sin(A_k)}$ Split the fraction: $T_k = \frac{1}{\sin(\pi/6)} \left( \frac{\sin(A_k)\cos(A_{k-1})}{\sin(A_{k-1})\sin(A_k)} - \frac{\cos(A_k)\sin(A_{k-1})}{\sin(A_{k-1})\sin(A_k)} \right)$ $T_k = \frac{1}{\sin(\pi/6)} \left( \cot(A_{k-1}) - \cot(A_k) \right)$ We know $\sin(\pi/6) = 1/2$, so $1/\sin(\pi/6) = 2$. $T_k = 2 \left( \cot(A_{k-1}) - \cot(A_k) \right)$

3. Evaluate the Summation

   The sum is $S = \sum\limits_{k = 1}^{13} T_k = \sum\limits_{k = 1}^{13} 2 \left( \cot(A_{k-1}) - \cot(A_k) \right)$. $S = 2 \sum\limits_{k = 1}^{13} \left( \cot(A_{k-1}) - \cot(A_k) \right)$ This is a telescoping series. Let's write out the first few and the last terms: For $k=1$: $2(\cot(A_0) - \cot(A_1))$ For $k=2$: $2(\cot(A_1) - \cot(A_2))$ For $k=3$: $2(\cot(A_2) - \cot(A_3))$ ... For $k=13$: $2(\cot(A_{12}) - \cot(A_{13}))$

   Summing these terms, all intermediate terms cancel out: $S = 2 \left[ (\cot(A_0) - \cot(A_1)) + (\cot(A_1) - \cot(A_2)) + \dots + (\cot(A_{12}) - \cot(A_{13})) \right]$ $S = 2 \left( \cot(A_0) - \cot(A_{13}) \right)$

4. Calculate the Required Values

   We need to find the values of $\cot(A_0)$ and $\cot(A_{13})$.

   • For $k=1$, $A_0 = {\pi \over 4} + {{(1-1)\pi } \over 6} = {\pi \over 4}$. $\cot(A_0) = \cot({\pi \over 4}) = 1$
   • For $k=13$, we need $A_{13}$, which is given by $A_k$ with $k=13$. $A_{13} = {\pi \over 4} + {{13\pi } \over 6} = {{3\pi + 26\pi} \over 12} = {{29\pi} \over 12}$ We can simplify the angle using the periodicity of the cotangent function (period $\pi$): $A_{13} = {{24\pi + 5\pi} \over 12} = 2\pi + {{5\pi} \over 12}$ $\cot(A_{13}) = \cot\left(2\pi + {{5\pi} \over 12}\right) = \cot\left({{5\pi} \over 12}\right)$ To evaluate $\cot(5\pi/12)$, we can write $5\pi/12 = 75^\circ = 45^\circ + 30^\circ$. $\cot(75^\circ) = \cot(45^\circ + 30^\circ) = \frac{\cot(45^\circ)\cot(30^\circ) - 1}{\cot(45^\circ) + \cot(30^\circ)}$ Since $\cot(45^\circ) = 1$ and $\cot(30^\circ) = \sqrt{3}$: $\cot(75^\circ) = \frac{1 \cdot \sqrt{3} - 1}{1 + \sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ Rationalizing the denominator: $\cot(75^\circ) = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$ So, $\cot(A_{13}) = 2 - \sqrt{3}$.

5. Final Calculation

   Substitute the values back into the expression for $S$: $S = 2 \left( \cot(A_0) - \cot(A_{13}) \right) = 2(1 - (2 - \sqrt{3}))$ $S = 2(1 - 2 + \sqrt{3})$ $S = 2(\sqrt{3} - 1)$

This matches option C. The value is $2\sqrt{3} - 2 \approx 2(1.732) - 2 = 3.464 - 2 = 1.464$.