3.48. The height of the cone is $H$, the angle between the slant height and the height is $\alpha$. A smaller cone is inscribed in this cone such that the vertex of the smaller cone coincides with the center of the base of the larger cone, and the corresponding slant heights of both cones are perpendicular to each other. Find the volume of the inscribed cone.
3.48. According to the condition, $S O=H, O C \perp A S$, $O D \perp B S, \angle O S B=\alpha$ (Fig. 3.50); it is required to find the volume of the cone $O D C$. Obviously, $\angle O D O_{1}=\angle O S B$ as acute angles with mutually perpendicular sides. From $\triangle S O D$ we find $O D=H \sin \alpha$; further, from $\triangle O O_{1} D$ we get

$$
\begin{gathered}
O O_{1}=O D \sin \alpha=H \sin ^{2} \alpha \\
O_{1} D=O D \cos \alpha=H \sin \alpha \cos \alpha
\end{gathered}
$$

Finally, we have

$$
\begin{gathered}
V=\frac{1}{3} \pi O_{1} D^{2} \cdot O O_{1}=\frac{1}{3} \pi H^{2} \sin ^{2} \alpha \cos ^{2} \alpha \cdot H \sin ^{2} \alpha= \\
=\frac{1}{3} \pi H^{3} \sin ^{4} \alpha \cos ^{2} \alpha
\end{gathered}
$$