I'm having difficulty seeing the trick for this problem. There's something obvious I'm missing:
Use the Lagrange interpolation polynomial to derive a quadrature rule of the form
$$\hat{I}_{[0,1]}\,(f) = w_0\,f(1/3) + w_1\,f(2/3)$$
That approximates the integral
$$I_{[0,1]}\,(f) = \int_0^1 f(x) \,dx$$
I tried to write a Lagrange Interpolation formula with points $x_0 = 0, x_1 = 1/3, x_2 = 2/3, x_3 = 1$ but the first and last point does not cancel. My intuition tells me this is used to approximate $f(x) = sin(x)$ but I'm not sure how to get rid of the first and last term. There's something obvious I'm missing.
Let $x_0=\frac{1}{3}$ and $x_1=\frac{2}{3}$, then the interpolation polynomial of $f(x)$ is given by
$$L(x) = f(1 / 3) \frac{x-2 / 3}{1 / 3 - 2 / 3} + f(2 / 3) \frac{x-1 / 3}{2 / 3 - 1 / 3} = f(1/ 3) \cdot (2-3x) + f(2/ 3) \cdot(3x-1)$$
Then, \begin{align}\hat{I}_{[0,1]}(f) &= \int_0^1 L(x) \,dx\\ &= \left[f(1/ 3)\cdot\left(2x-\tfrac{3}{2}x^2\right) + f(2/ 3)\cdot\left(\tfrac{3}{2}x^2-x\right)\right]_0^1\\ &= \tfrac{1}{2}f(1/ 3) + \tfrac{1}{2}f(2/ 3)\end{align}