We are asked to use Lagrange interpolation to approximate $\sin(\pi/3)$, given the function $y = \sin(x)$ with three data points:

• $x_0 = 0$, $y_0 = \sin(0) = 0$
• $x_1 = \frac{\pi}{4}$, $y_1 = \sin\left(\frac{\pi}{4}\right) = 0.7071$
• $x_2 = \frac{\pi}{2}$, $y_2 = \sin\left(\frac{\pi}{2}\right) = 1$

The Lagrange interpolation formula is:

$L(x) = \sum_{i=0}^{n} y_i \prod_{j=0, j\neq i}^{n} \frac{x - x_j}{x_i - x_j}$

For $n = 2$ (since we have 3 points), the formula becomes:

$L(x) = y_0 \frac{(x - x_1)(x - x_2)}{(x_0 - x_1)(x_0 - x_2)} + y_1 \frac{(x - x_0)(x - x_2)}{(x_1 - x_0)(x_1 - x_2)} + y_2 \frac{(x - x_0)(x - x_1)}{(x_2 - x_0)(x_2 - x_1)}$

Substitute the values of $x_0 = 0$, $x_1 = \frac{\pi}{4}$, and $x_2 = \frac{\pi}{2}$, along with their corresponding $y$-values:

+ 0.7071 \cdot \frac{(x - 0)(x - \frac{\pi}{2})}{(\frac{\pi}{4} - 0)(\frac{\pi}{4} - \frac{\pi}{2})} + 1 \cdot \frac{(x - 0)(x - \frac{\pi}{4})}{(\frac{\pi}{2} - 0)(\frac{\pi}{2} - \frac{\pi}{4})}$$ Now, let's evaluate this at $$x = \frac{\pi}{3}$$. I'll calculate this step by step for you.Using Lagrange interpolation, the value of $$\sin\left(\frac{\pi}{3}\right)$$ is approximately $$0.8508$$. Would you like further details on the process or have any questions? Here are five related questions to explore: 1. How does Lagrange interpolation differ from Newton interpolation? 2. Can Lagrange interpolation be used for non-equally spaced data points? 3. What are the limitations of Lagrange interpolation in approximating functions? 4. How does increasing the number of data points affect the accuracy of interpolation? 5. Can Lagrange interpolation be applied to multidimensional data? **Tip:** Lagrange interpolation works best for a small number of points. For larger datasets, spline interpolation or other methods might be more efficient and accurate.