Estimating error linear interpolation

Question

Let $p(x)$ be the linear interpolation polynomial of $\sin(x)$ at the points $x_0$ and $x_0 + h$. We know that $$ \sin(x) - p(x) = \frac{-\sin(\xi)}{2}(x - x_0)(x - x_0 - h)$$ for some $\xi \in (x_0, x_0 + h)$. Because of $\sin(\xi) \approx \sin(x) \approx p(x)$, we can approximate the error made in the above estimation by $$\frac{-p(x)}{2}(x - x_0)(x - x_0 - h) $$ Now, let our new approximation of $\sin(x)$ be given by $$p(x)(1 - \frac{1}{2} (x - x_0)(x - x_0 - h))$$ I have already shown that $\forall x \in [x_0, x_0 + h], \exists \eta \in [x_0, x_0 + h]$ such that $p(x) = \sin(\eta)$ and I am supposed to use this to prove that the absolute error given by the approximation $$p(x)(1 - \frac{1}{2} (x - x_0)(x - x_0 - h))$$ is $\leq \displaystyle\frac{h^3}{8}$. So far I've got $$|\sin(x) - p(x)| \leq \max_{x \in [x_0, x_0 + h]} |p(x)| |(x - x_0)(x - x_0 - h)|$$ which gives $$|\sin(x) - p(x)| \leq \frac{h^2}{8} max_{x \in [x_0, x_0 + h]} |p(x)|$$ and now, if we use the fact that $\forall x \in [x_0, x_0 + h], \exists \eta \in [x_0, x_0 + h]$ such that $p(x) = \sin(\eta)$ we get that $\exists \eta_1 \in [x_0, x_0 + h]$ such that $|\sin(\eta_1)| = |p(x)|$ which implies $$|\sin(x) - p(x)| \leq \frac{h^2}{8}$$ which is not the answer I am looking for. What am I missing here?