determine the error bound for the interpolation error at $x = \frac{\pi}{4}.$ I have been able to derive the interpolation polynomial $P_2(x)$ of degree two which interpolates $f(x) = \sin x$, given the points $(0,0), \left(\frac{\pi}{2}, 1\right), (\pi, 0).$
Solution:
$$P_2(x) = \frac{4}{\pi ^2}x(\pi - x)$$
Here is the question below I am having trouble with 

Calculate $P_2\left(\frac{\pi}{4}\right)$, an approximation for $f\left( \frac{\pi}{4}\right) = \sin \left( \frac{\pi}{4} \right)$ and determine the error bound for the interpolation error at $x = \frac{\pi}{4}.$

Calculating  $P_2\left(\frac{\pi}{4}\right)$
$$P_2\left(\frac{\pi}{4}\right) = \frac{4}{\pi ^2} \times \frac{\pi}{4}\left( \pi - \frac{\pi}{4} \right)=  \frac{3}{4}$$
If we plug $x = \frac{\pi}{4}$ into $f(x) = \sin x$ we get $f(\pi /4) = \sin (\pi /4) \approx 0.7071 $
Here is my attempt below at finding the error bound
Writing the error as $err(x) = \sin x - P_2(x) = \sin x - \frac{4}{\pi ^2}x(\pi - x)$ 
Differentiating once gives,
$$\cos (x) - \frac{4(\pi - 2x)}{\pi ^2}$$
Differentiating twice gives,
$$\frac{8}{\pi ^2}-\sin(x)$$
Finally differentiating a third time gives, 
$$- \cos x $$
But how do I use this to get my error bound

For a polynomial interpolation of order $n$, the maximum error is given by
$${\rm err}=\frac{1}{(n+1)!}\max |f^{(n+1)}(x)|\max|\prod_{p=0}^n(x-x_p)|$$
Here $x_p$ are the roots of your polynomial.
Let's suppose that you are interested in finding the maximum only in the interval from $0$ to $\pi$.
The maximum of the derivative in your case is $1$, the maximum for the product $x(\pi-x)$ occurs at $\pi/2$ so $${\rm err}=\frac{1}{3!}1\left(\frac\pi2\right)^2$$
