805. In the theory of concave mirrors, the formula is encountered:

$$
x=r \frac{\sin ^{2} \frac{\alpha}{2}}{\cos x}
$$

For small $\alpha$, this formula is approximated with high accuracy by the formula $x=\frac{r \alpha^{2}}{4}$. Justify this. In what units of measure is the angle $\alpha$ expressed?
805. For $0<\alpha<\frac{\pi}{2}$, there exist inequalities

$$
\begin{aligned}
& 0<\alpha-\sin \alpha<\frac{\alpha^{3}}{2} \\
& 0<1-\cos \alpha<\frac{\alpha^{2}}{2}
\end{aligned}
$$

From the first inequality, it follows that for small $\alpha$, $\sin \alpha$ is almost equal to $\alpha$ and, even more so, $\sin \frac{\alpha}{2}$ is approximately equal to $\frac{\alpha}{2}$. $\cos \alpha$ is approximately equal to 1, so we have with a high degree of approximation

$$
\frac{r \sin ^{2} \frac{\alpha}{2}}{\cos \alpha}=\frac{r\left(\frac{\alpha}{2}\right)^{2}}{1}=\frac{r \alpha^{2}}{4}
$$