The interpolation error $R(x) = f(x) - L(x)$ of the interpolation polynom $L$ is given for $x_1 \le x \le x_n$ by
$$R(x) = {f^{n}(\xi) \over n!} \prod_{i=1}^n (x - x_i)$$ where $x_1 < \xi < x_n $ if $x_i$ is sorted in a ascending order. Find (two) sampling points $x_i$ such that sin$( x)$ is on $[0, \pi]$ interpolated with a first-degree polynomial as precise as possible (w.r.p. to the maximum-norm).
My approach was basically to minimize $$R(x) = {-sin(\xi) \over 2} (x-x_1)(x-x_2)$$ (ignoring the first term at first). I received that $f(x) = (x-x_1)(x-x_2)$ has a local minimum at $x_0 = {x_1 + x_2 \over 2}$. Nevertheless $x_1=0$ and $x_2 = \pi$ on the boundary should also be taken into consideration.
I don't see right know how I can continue from that.
Your idea in general fails because $\xi$ is a function of $x$.
The equioscillation theorem tells you that the minimizer is uniquely characterized by the requirement that there be three points a,b,c where the error is the same value but alternating in sign. By the symmetry this can be achieved by simply taking p to be 1/2, so that the requirement is satisfied by $0,\pi/2,\pi$.