Let $O$ be the center of the sphere of radius $R$ circumscribed around a regular triangular pyramid $ABCD$ with vertex $D$. The point $O$ lies on the line $DM$, where $M$ is the center of the base $ABC$, and $K$ is the midpoint of $BC$. By the problem's condition, $OA=R$, $\angle BDC=\phi$. Denote $AB=BC=AC=a$. Then $BM=\frac{a \sqrt{3}}{3}$. From the right triangles $DBK$ and $BMD$, we find that

$$\begin{aligned} & BD=\frac{BK}{\sin \angle BDK}=\frac{\frac{a}{2}}{\sin \frac{\Psi}{2}}=\frac{a}{2 \sin \frac{\Psi}{2}} \\ & DM=\sqrt{BD^{2}-BM^{2}}=\sqrt{\frac{a^{2}}{4 \sin ^{2} \frac{\Psi}{2}}-\frac{a^{2}}{3}}=\frac{a \sqrt{3-4 \sin ^{2} \frac{\varphi}{2}}}{3 \sqrt{3} \sin ^{\frac{4}{2}}}. \end{aligned}$$

Consider the section of the pyramid and the sphere by a plane passing through the points $A, D$, and $M$. We obtain a circle of radius $R$ with center $O$ on the line $MD$. Extend the segment $DM$ beyond point $M$ to intersect the circle at point $P$. Since $AM$ is the altitude of the right triangle $DAP$ drawn from the right angle vertex, we have

$$AM^2 = DM \cdot MP = DM \cdot (2R - DM) = 2R \cdot DM - DM^2$$

or

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$$\begin{aligned} & \Leftrightarrow \frac{\underline{a}}{3} + \frac{\underline{a}\left(3-4 \sin ^{2} \frac{\varphi}{2}\right)}{12 \sin ^{2} \frac{\varphi}{2}} = \frac{R \sqrt{3-4 \sin ^{2} \frac{\varphi}{2}}}{\sqrt{3} \sin _{\frac{\varphi}{2}}^{2}}, \end{aligned}$$

from which we find that

$$a = \frac{4 R \sin \frac{\varphi}{2} \sqrt{3-4 \sin ^{2} \varphi}}{\sqrt{3}}.$$

Then

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Therefore,

$$\begin{gathered} V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DM = \frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot DM = \frac{\sqrt{3}}{12} \cdot a^2 \cdot DM = \\ = \frac{\sqrt{3}}{12} \cdot \frac{16 \lambda^{2} \sin ^{2} \frac{\varphi\left(3-4 \sin ^{2} \frac{\varphi}{2}\right)}{3}}{3} \cdot \frac{2 R\left(3-4 \sin ^{2} \frac{\varphi}{2}\right)}{3} = \frac{\frac{b}{27}}{27} R^3 \sin^2 \frac{\varphi}{2}\left(3-4 \sin^2 \frac{\varphi}{2}\right)^2 \end{gathered}$$

## Answer

$$\frac{\frac{8}{27}}{27} 3 \sin^2 \frac{\varphi}{2}\left(3-4 \sin^2 \frac{\varphi}{2}\right)^2 = \frac{8}{27} R^3 \sin^2 \frac{\varphi}{2}(1+2 \cos \phi)^2.$$