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.