12.098. A cone is inscribed in a sphere. The area of the axial section of the cone is $S$, and the angle between the height and the slant height is $\alpha$. Find the volume of the sphere.
Solution.

The volume of the sphere is: \( V = \frac{4}{3} \pi R^{3} \). The area of the axial section of the cone is: \( S = \frac{1}{2} l^{2} \sin 2 \alpha \). Hence, \( l = \sqrt{\frac{2 S}{\sin 2 \alpha}} \). The following relationship holds between the elements of the sphere and the inscribed cone: \( l = 2 R \sin \left(90^{\circ} - \alpha \right) = 2 R \cos \alpha \). Therefore, \( R = \frac{l}{2 \cos \alpha} \). Substituting, we get: \( R = \sqrt{\frac{2 S}{4 \sin 2 \alpha \cos^{2} \alpha}} \). The volume of the sphere \( V = \frac{4}{3} \pi \frac{2 S}{4 \sin 2 \alpha \cos^{2} \alpha} \sqrt{\frac{2 S}{4 \sin 2 \alpha \cos^{2} \alpha}} = \frac{2}{3} \pi \frac{S \sqrt{2 S \sin 2 \alpha}}{2 \sin^{2} 2 \alpha \cos^{3} \alpha} = \frac{1}{3} \pi S \frac{\sqrt{2 S \sin 2 \alpha}}{\sin^{2} 2 \alpha \cos^{3} \alpha} \).

Answer: \( \frac{1}{3} \pi S \frac{\sqrt{2 S \sin 2 \alpha}}{\sin^{2} 2 \alpha \cos^{3} \alpha} \).