1. **State the problem:** We want to find the value of the infinite sum $$\sum_{k=0}^\infty \left( \sin \left( \frac{(k+1)\pi}{12k+11} \right) - \sin \left( \frac{k\pi}{12k-1} \right) \right).$$ 2. **Analyze the terms:** Notice the sum is of the form $$\sum_{k=0}^\infty (a_{k+1} - a_k)$$ where $$a_k = \sin \left( \frac{k\pi}{12k-1} \right).$$ 3. **Check the telescoping pattern:** The sum can be rewritten as $$\sum_{k=0}^\infty \left( a_{k+1} - a_k \right) = \lim_{n \to \infty} (a_{n+1} - a_0).$$ 4. **Evaluate the boundary terms:** - When $$k=0$$, $$a_0 = \sin \left( \frac{0 \cdot \pi}{12 \cdot 0 - 1} \right) = \sin(0) = 0.$$ - When $$k = n+1$$, $$a_{n+1} = \sin \left( \frac{(n+1)\pi}{12(n+1) - 1} \right) = \sin \left( \frac{(n+1)\pi}{12n + 11} \right).$$ 5. **Find the limit of $$a_{n+1}$$ as $$n \to \infty$$:** As $$n \to \infty$$, $$\frac{(n+1)\pi}{12n + 11} \approx \frac{n\pi}{12n} = \frac{\pi}{12}.$$ So, $$\lim_{n \to \infty} a_{n+1} = \sin \left( \frac{\pi}{12} \right).$$ 6. **Calculate the final sum:** $$\sum_{k=0}^\infty \left( \sin \left( \frac{(k+1)\pi}{12k+11} \right) - \sin \left( \frac{k\pi}{12k-1} \right) \right) = \sin \left( \frac{\pi}{12} \right) - 0 = \sin \left( \frac{\pi}{12} \right).$$ 7. **Simplify the sine value:** $$\sin \left( \frac{\pi}{12} \right) = \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}.$$