Let p(x) be a linear function interpolating sin(x) at $x=0, x=\frac{\pi}{2}$. Prove that $|p(x)-\sin x| \leq \frac{1}{2}(\frac{\pi}{4})^2$ on $[0, \frac{\pi}{2}]$.
I've already done a bit of work and found that $p(x) = \frac{2}{\pi}x$, therefore $|p(x)-\sin x| = |\frac{2}{\pi}x-\sin x|$.
However, I'm not quite sure how to show that the value $\frac{1}{2}(\frac{\pi}{4})^2$ is the upper bound of this error on our given interval. My thought would be to show that the largest error on this interval is equal to that value, but can't easily see a way to find where that error occurs.
How would I go about proving that the given value is a bound for the error of $|p(x)-\sin x|$?
If you look at Polynomial interpolation at paragraph interpolation error, you’ll find that $$p(x)-\sin x = \frac{\sin^{\prime \prime}\xi}{2!}x(x-\pi/2)$$ with $\xi \in (0,\pi/2)$. You can mimic the proof given in Wikipedia article to your special case.
Based on that you will easily get the requested bound as $\vert \sin^{\prime \prime}\xi \vert \le 1$ and $0 \le \vert x(x-\pi/2) \vert \le \left( \frac{\pi}{4} \right)^2$ for $x \in [0,\pi/2]$.
This is a special case of polynomial interpolation: linear interpolation.