Animated Solution for Mathematics - Trigonometry: The value of ${\sum }_{k=1}^{13}\frac{1}{\sin (\frac{\pi }{4}+\frac{(k-1)\pi }{6})\sin (\frac{\pi }{4}+\frac{k\pi }{6})}$ is equal to

When you encounter a summation like it is natural to feel intimidated. However, the structure of the denominator—a product of two sine functions—is a classic signature of a telescoping series.

Let us define the angles as $A_k=\frac{\pi }{4}+\frac{(k-1)\pi }{6}$ and $A_{k+1}=\frac{\pi }{4}+\frac{k\pi }{6}$.

The difference between these angles is constant:

This constant difference is the heartbeat of the problem. We utilize the trigonometric identity:

To apply this, we multiply and divide the general term $T_k$ by $\sin (\frac{\pi }{6})$. Since $\sin (\frac{\pi }{6})=\frac{1}{2}$, dividing by it is equivalent to multiplying by $2$:

Applying the identity, the expression simplifies beautifully:

Now, we evaluate the summation $S={\sum }_{k=1}^{13}{T}_k$. Writing out the terms reveals the rhythmic cancellation:

Every intermediate term vanishes, leaving only the boundary values:

First, we calculate $A_1$:

Next, we calculate $A_{14}$: Since $\cot (\frac{5\pi }{12})=\cot (75^{\circ })=2-\sqrt{3}$, we substitute these values back into our expression for $S$:

The final result is $2(\sqrt{3}-1)$.