1. **State the problem:** We want to find the value of $$\arcsin\left(\sin\left(\frac{100\pi}{11}\right)\right)$$. 2. **Recall the domain and range:** The function $$\arcsin(x)$$ returns values in the range $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$. The sine function is periodic with period $$2\pi$$. 3. **Simplify the inner sine argument:** Since sine is periodic with period $$2\pi$$, we reduce $$\frac{100\pi}{11}$$ modulo $$2\pi$$. Calculate: $$\frac{100\pi}{11} = 9\times 2\pi + \frac{2\pi}{11}$$ Because: $$9 \times 2\pi = 18\pi$$ and $$100\pi/11 - 18\pi = \frac{100\pi - 198\pi}{11} = -\frac{98\pi}{11}$$ Actually, better to find remainder modulo $$2\pi$$: Divide $$100/11 \approx 9.09$$, so: $$100\pi/11 = 9 \times 2\pi + \frac{2\pi}{11}$$ So: $$\sin\left(\frac{100\pi}{11}\right) = \sin\left(\frac{2\pi}{11}\right)$$ 4. **Evaluate $$\arcsin(\sin(\frac{2\pi}{11}))$$:** Since $$\frac{2\pi}{11} \approx 0.571 < \frac{\pi}{2}$$, it lies within the principal range of $$\arcsin$$. Therefore: $$\arcsin\left(\sin\left(\frac{100\pi}{11}\right)\right) = \frac{2\pi}{11}$$ **Final answer:** $$\boxed{\frac{2\pi}{11}}$$