I am trying to understand how to show that the best approximation of $\sin(x)$ over $[0,\pi]$ and polynomial of degree $n=0$ (so a constant $c$) by using the maximum norm, i.e $\min_{c} \lVert \sin(x) - c\rVert_\infty$ is with $c=1/2$.
Visually it is clear that the maximal distance to both extreme values of $\sin(x)$ is at $1/2$ but I have no idea how to deduct it from the maximum morm.
Any hints are appreciated.
Yes, it is $\dfrac12$, since $\|\sin-c\|_\infty=c$ if $c\geqslant\dfrac12$ and $\|\sin-c\|_\infty=1-c$ if $c\leqslant\dfrac12$. All you need to know to prove this is that $\max_{x\in[0,\pi]}\sin(x)=1$ and that $\min_{x\in[0,\pi]}\sin(x)=0$.
Therefore, the minimum of $\|\sin-c\|_\infty$ is reached when $c=\dfrac12$.