Evaluate Sine Expression Given Product of Sine Ratios

To find the value of the given expression, we first calculate the value of $K$.

Given: $K = \sin\left(\frac{\pi}{18}\right) \sin\left(\frac{5\pi}{18}\right) \sin\left(\frac{7\pi}{18}\right)$

Using the co-function identity $\sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right)$, we can rewrite each factor: $\sin\left(\frac{7\pi}{18}\right) = \cos\left(\frac{\pi}{2} - \frac{7\pi}{18}\right) = \cos\left(\frac{2\pi}{18}\right) = \cos\left(\frac{\pi}{9}\right)$ $\sin\left(\frac{5\pi}{18}\right) = \cos\left(\frac{\pi}{2} - \frac{5\pi}{18}\right) = \cos\left(\frac{4\pi}{18}\right) = \cos\left(\frac{2\pi}{9}\right)$ $\sin\left(\frac{\pi}{18}\right) = \cos\left(\frac{\pi}{2} - \frac{\pi}{18}\right) = \cos\left(\frac{8\pi}{18}\right) = \cos\left(\frac{4\pi}{9}\right)$

Substituting these back into the expression for $K$: $K = \cos\left(\frac{\pi}{9}\right) \cos\left(\frac{2\pi}{9}\right) \cos\left(\frac{4\pi}{9}\right)$

Multiplying and dividing by $8\sin\left(\frac{\pi}{9}\right)$, we get: $K = \frac{8\sin\left(\frac{\pi}{9}\right) \cos\left(\frac{\pi}{9}\right) \cos\left(\frac{2\pi}{9}\right) \cos\left(\frac{4\pi}{9}\right)}{8\sin\left(\frac{\pi}{9}\right)}$

Applying the double-angle formula $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ repeatedly: $K = \frac{4\sin\left(\frac{2\pi}{9}\right) \cos\left(\frac{2\pi}{9}\right) \cos\left(\frac{4\pi}{9}\right)}{8\sin\left(\frac{\pi}{9}\right)}$ $K = \frac{2\sin\left(\frac{4\pi}{9}\right) \cos\left(\frac{4\pi}{9}\right)}{8\sin\left(\frac{\pi}{9}\right)}$ $K = \frac{\sin\left(\frac{8\pi}{9}\right)}{8\sin\left(\frac{\pi}{9}\right)}$

Since $\sin\left(\frac{8\pi}{9}\right) = \sin\left(\pi - \frac{\pi}{9}\right) = \sin\left(\frac{\pi}{9}\right)$, we obtain: $K = \frac{\sin\left(\frac{\pi}{9}\right)}{8\sin\left(\frac{\pi}{9}\right)} = \frac{1}{8}$

Now, we need to evaluate the required expression: $\sin\left(\frac{10K\pi}{3}\right)$

Substitute $K = \frac{1}{8}$: $\frac{10K\pi}{3} = \frac{10 \times \frac{1}{8} \times \pi}{3} = \frac{10\pi}{24} = \frac{5\pi}{12}$

Thus, we compute: $\sin\left(\frac{5\pi}{12}\right) = \sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right)$

Using the angle sum identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$: $\sin\left(\frac{5\pi}{12}\right) = \sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{6}\right)$ $\sin\left(\frac{5\pi}{12}\right) = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

Therefore, the correct option is A.