Cebysev (Tschebyscheff-) Approximation I have to find the best Tschebyscheff-Approximation of $x(t) = \sin(t)$ on the Interval $[0,d]$ by a line where $0 < d < 2\pi$. But I have no idea how to perform such an approximation, do you have any resources where it is explained, I can not find anything good?
 A: Posting another answer, because  the other one uses a different approach and has   comments concerning that approach.
Let's begin with a more general situation:   we want the best uniform (Chebyshev) linear fit to a smooth function   $f$ on an interval $[A,B]$. A variational argument (which is standard, and therefore  omitted) shows that if $ax+b$ is the best fit, there  are three points $x_1<x_2<x_3$ at which the difference $f(x)-(ax+b)$  attains values (left to right) $M,-M,M$, with  $|M|=\max_{[A,B]}|f(x)-(ax+b)|$. Note that $M$ can be negative or positive, we don't know. 
Since $x_2$ is an interior extremum of $f(x)-(ax+b)$, we have $a=f'(x_2)$. The other two extrema could be interior or boundary points, we don't know. But we do know that 
$$f(x_1)- f'(x_2) x_1 = f(x_3) - f'(x_2) x_3 \tag1$$
since both sides are equal to $M+b$. 
So there are quite a few cases to check in general. For the concrete situation, we know that $f'(x)=\cos x$ cannot attain  same value three times. Therefore either $x_1$ or $x_3$ is an endpoint, maybe both. More specifically, either $$(x_1,x_2,x_3) = (0, z , 2\pi-z) \tag2 $$
or
$$(x_1,x_2,x_3) = ( z , 2\pi-z, d) \tag3 $$
or 
$$(x_1,x_2,x_3) = (0, z , d) \tag4 $$
In each of these three cases, we have one unknown $z\in(0,2\pi)$ which can be found by solving (1). 
Then compare the maximal deviation you get from $$2M = (f(x_1)-f'(x_2)x_1) -  (f(x_2)-f'(x_2)x_2)$$ and pick the solution with the smallest $|M|$. 
A: Let $at+b$ be the optimal line. Note that the best $b$ is the average of the maximum and minimum of $\sin t-at$. Thus, it remains to find $a\in\mathbb R$ such that the oscillation of $\sin t-at$ on the interval $[0,d]$ is minimized. (Oscillation is the $\max-\min$ difference).  
One can visualize the problem in terms of the graph $G=\{(t,\sin t):t\in [0,d]\}$. The oscillation of $\sin t-at$ is the length of a skew projection of $G$ onto the vertical axis. We want the direction of projection that minimizes this length, "the position of the Sun when $G$ has the smallest shadow."
Unfortunately, the computation is a tedious case-by-case analysis. There are between $2$ and $4$  points $t$ where $\sin t-at$ could attain an extremum:
$$t_1=0, \quad t_2=d, \quad t_3=\arccos a, \quad t_4=2\pi -\arccos a$$
The last two may or may not be in $[0,d]$. So the maximum of $\sin t-at$ is
$$\max(0,\sin d-ad, \sqrt{1-a^2} - at_3, -\sqrt{1-a^4} - at_4)   $$
where some of the last two may have to be excluded. Similarly the minimum. I'd   plot these as functions of $a$ (I assume you have a specific value of $d$) to see what goes on.
