Linear interpolation with two points Question: $p(x)$ is the linear function that interpolates $\sin(x)$ at $0$ and $\frac{\pi}{2}$. And I need to show that $\ |p(x) - \sin(x)|\le\ \frac{1}{2}(\frac{\pi}{4})^2$ 
My attempt:
$\ |f(x)-p(x)| \le (K_n/(n+1)!) * |W(x)|$ 
where $\ K_2 = -\sin(x)$ and  $\max|W(x)| = ||W(x)||\inf$ where x is in $\ [0, \pi/2]$
I ended up with $\ \frac {1}{2}(x^2 - (\frac{ \pi}{2})x)$ which I cannot find a connection with the given relationship in the problem. Am I missing something?
Sorry in advanced for the poor syntax.  
 A: $p(x)=ax+b$ such that $p(0)=0$ and $p(\frac{\pi}2)=1$
That is $p(x) = \frac2{\pi} x$
The error of linear interpolation is given by
$$|E(x)|=|p(x)-\sin x|=\max_{x\in[0,\pi/2]}\left|\frac{ (x-0)(x-\pi/2)}2 \left\{\frac{d^2}{dx^2}\sin\right\} (c)\right|
=\max_{x\in[0,\pi/2]}\left|\frac{ (x-0)(x-\pi/2)}2 \sin(c)\right|\le
\max_{x\in[0,\pi/2]}\frac{ (x-0)(x-\pi/2)}2 = \frac{1}{2} \left(\frac{\pi }{4}\right)^2
$$
To show the last equality note that 
$\frac{d}{dx}x(x-\frac\pi2)=(x-\frac\pi2)+x=0
$ iff $x=\frac\pi4$, this is of course a maximum (why)? Finally evaluate the polynomial at this point, to get the required result.
A: Hint:
As $p(x)$ is linear through $(0,0)$ and $(\frac\pi2,1)$, we have $p(x)=\frac{2x}\pi$.
Then consider $|p(x)-\sin(x)|^2=(\frac{2x}\pi-\sin(x))^2$.
We find the maximum of this function by derivation
$$2(\frac2\pi-\cos(x))(\frac{2x}\pi-\sin(x)).$$
The first factor cancels for $\cos(x)=\frac2\pi$. It corresponds to a single extremum of the second factor. The latter is known to cancel for $x=0$ and $x=\frac\pi2$ and has no other root.
Hence the maximum occurs at $x=\arccos(\frac2\pi)$, with value (of the square)
$$\left(\frac2\pi\arccos(\frac2\pi)-\sqrt{1-\frac4{\pi^2}}\right)^2.$$
A: The maximum of the difference $\sin(x)-p(x)$ is reached when $\cos(x)=\frac2\pi,$ so the maximum difference is actually
$$\sqrt{1-\frac4{\pi^2}}-\frac2{\pi}\arccos\frac2{\pi}.$$
