determine the error bound for the interpolation error at $x = \frac{\pi}{4}.$

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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

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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 $${\rm err}=\frac{1}{3!}1\left(\frac\pi2\right)^2$$