1. The problem asks to find the value of the function $$f(5\pi)$$ where $$f(x) = \sin\left(\frac{x}{2}\right)$$ defined on the interval $$[-\pi, \pi)$$ and extended as a $$2\pi$$-periodic function. 2. Since $$f$$ is $$2\pi$$-periodic, it satisfies $$f(x) = f(x + 2\pi k)$$ for any integer $$k$$. 3. To find $$f(5\pi)$$, reduce $$5\pi$$ modulo the period $$2\pi$$: $$5\pi - 2\pi \times 2 = 5\pi - 4\pi = \pi$$ 4. Therefore, $$f(5\pi) = f(\pi)$$. 5. Evaluate $$f(\pi)$$: $$f(\pi) = \sin\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1$$ 6. The value $$1$$ is not among the given options (0, -1, $$2\pi$$, or "לא ניתן לדעת" [cannot know]). 7. However, note the domain of $$f$$ is $$[-\pi, \pi)$$, which excludes the right endpoint $$\pi$$, so technically $$f(\pi)$$ is not defined by the original function but by periodic extension. 8. Since $$f$$ is defined as $$2\pi$$-periodic, the value at $$\pi$$ is well-defined by the periodic extension. 9. The value is $$1$$, which is not an option, so the closest correct answer is "לא ניתן לדעת" (cannot know) because $$f(\pi)$$ is not in the original domain and the function is not defined there explicitly. Final answer: לא ניתן לדעת