Find 2 datapoints too interpolate sin(x) 
I have the function $f:\left [ 0,\pi \right ]\rightarrow \mathbb{R}, x \mapsto \sin(x)$.
  How can I choose two points $x_0, x_1 \in \left [ 0,\pi \right ]$ for my polynomial interpolation such that I get a polynomial $p$ of degree $1$ that minimizes the pointwise-error, i.e. $\min\left \| f-p \right \|_{\infty}$.

My thoughts: 
Just by taking a closer look at the plot of $\sin(x)$, I think it becomes quite obvious that $x_0$ and $x_1$ have to be chosen such that $p(x) = \frac{1}{2}$.
However, I have no idea how to proof this.
I'd really appreciate if someone could show me how to proof this or give me some advice/hints.
Thanks in advance!
 A: Generically speaking the minimax polynomial cannot be explicitly computed, because it typically requires the solution of a transcendental equation. For pedagogical purposes, let's see what that would look like here. The objective function is
$$\| ax+b-\sin(x) \|_\infty = \max \left \{ |b|,|a \pi + b|,|a \arccos(a) + b - \sin(\arccos(a))| \right \} \\= \max \left \{ |b|,|a\pi+b|,\left |a\arccos(a) + b - \sqrt{1-a^2} \right | \right \}$$
by routine calculus. It is not difficult to see by comparing to the case $a=0,b=1/2$ and using some triangle inequality manipulations that the optimum must have $a \in [-1/\pi,1/\pi]$ and $b \in [-1/2,1/2]$. While this gives you a compact domain to search for an optimum in (which is plenty for purposes of a numerical implementation), proceeding any further in this approach by hand is already difficult because of the piecewise character of the objective function.
As you noted originally, we can change our basis so that we are looking at $a(x-\pi/2)+b$. Since $\sin(x)$ is symmetric with respect to the line $x=\pi/2$, for a fixed $b$, the optimal choice of $a$ is necessarily $a=0$. Any nonzero value of $a$ improves the error on one side of this axis at the expense of the other. Similarly, $b=1/2$ is ideal once you have chosen $a=0$; making it any bigger improves the error at $\pi/2$ at the expense of the error at $0$. So the optimal solution of degree at most $1$ is indeed $p(x)=1/2$. 
There is no optimal solution which is actually degree $1$. This is no real surprise, because the polynomials of degree exactly $1$ do not form a closed set. Accordingly, when we talk about optimal polynomials of a specified degree we usually (explicitly or implicitly) mean to refer to optimal polynomials of a specified maximum degree, allowing for it to possibly be smaller.
A: Sry that my answere took so long, but now I got the same task.
First you draw the graph of sin(x) and see that 1/2 has the smallest failure.
My idea:
You want to minimize:$$\left\|sin(x)-p_1(x)\right\|_\infty=\left\|\frac{f''(\xi)}{2!}(x-x_0)(x-x_1)\right\|\leq \left\|\frac{1}{2}(x-x_0)(x-x_1)\right\|\leq \left\|(x-\frac{\pi}{6})(x-\frac{5\pi}{6})\right\|\leq \frac{\pi^2}{18},\forall x\in [0,\pi]. $$
And then you do the interpolation, by Newton for example:
$$ f[x_0]=\frac{1}{2}=c_0$$
$$ f[x_0x_1]=\frac{\frac{1}{2}-\frac{1}{2}}{\frac{2\pi}{3}}=0=c_1 $$
Now it follows:
$$ p_1(x)=\frac{1}{2}+0*(x-\frac{pi}{6})=\frac{1}{2}\in \mathbb{R_{[n\leq 0]}} $$
But $p_1(x)$ is not polynomial of 1st degree. I think:
$$ p^{`}_1(x)=\frac{1}{2}+\epsilon(x-\frac{\pi}{6}),  \epsilon \rightarrow0 $$
is the desired polynom.
