8.392. Find $\sin \alpha$, if $\cos \alpha=\operatorname{tg} \beta, \cos \beta=\operatorname{tg} \gamma, \cos \gamma=\operatorname{tg} \alpha\left(0<\alpha<\frac{\pi}{2}\right.$, $\left.0<\beta<\frac{\pi}{2}, 0<\gamma<\frac{\pi}{2}\right)$.
## Solution.

From the condition we have

$$
\left\{\begin{array} { l } 
{ \operatorname { cos } \alpha = \operatorname { tan } \beta , } \\
{ \operatorname { cos } \beta = \operatorname { tan } \gamma , } \\
{ \operatorname { cos } \gamma = \operatorname { tan } \alpha , }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \operatorname { cos } \alpha = \frac { \operatorname { sin } \beta } { \operatorname { cos } \beta } , } \\
{ \operatorname { cos } \beta = \frac { \operatorname { sin } \gamma } { \operatorname { cos } \gamma } , \Leftrightarrow } \\
{ \operatorname { cos } \gamma = \frac { \operatorname { sin } \alpha } { \operatorname { cos } \alpha } , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\cos ^{2} \alpha=\frac{\sin ^{2} \beta}{\cos ^{2} \beta}, \\
\cos ^{2} \beta=\frac{\sin ^{2} \gamma}{\cos ^{2} \gamma},
\end{array} \Leftrightarrow\right.\right.\right.
$$

$$
\Leftrightarrow\left\{\begin{array}{l}
\cos ^{2} \alpha=\frac{1-\cos ^{2} \beta}{\cos ^{2} \beta} \\
\cos ^{2} \beta=\frac{1-\cos ^{2} \gamma}{\cos ^{2} \gamma} \\
\cos ^{2} \gamma=\frac{1-\cos ^{2} \alpha}{\cos ^{2} \alpha}
\end{array}\right.
$$

Since $\cos ^{2} \gamma=\frac{1-\cos ^{2} \alpha}{\cos ^{2} \alpha}$, we have

$$
\cos ^{2} \beta=\frac{1-\frac{1-\cos ^{2} \alpha}{\cos ^{2} \alpha}}{\frac{1-\cos ^{2} \alpha}{\cos ^{2} \alpha}}=\frac{2 \cos ^{2} \alpha-1}{1-\cos ^{2} \alpha}
$$

Substituting the value $\cos ^{2} \beta=\frac{2 \cos ^{2} \alpha-1}{1-\cos ^{2} \alpha}$ into the first equation of the system, we get

$$
\cos ^{2} \alpha=\frac{1-\frac{2 \cos ^{2} \alpha-1}{1-\cos ^{2} \alpha}}{\frac{2 \cos ^{2} \alpha-1}{1-\cos ^{2} \alpha}}=\frac{2-3 \cos ^{2} \alpha}{2 \cos ^{2} \alpha-1} \Leftrightarrow \cos ^{4} \alpha+\cos ^{2} \alpha-1=0
$$

Solving this equation as a quadratic equation in terms of $\cos ^{2} \alpha$, we have $\cos ^{2} \alpha=\frac{-1-\sqrt{5}}{2}<0, \varnothing$ or $\cos ^{2} \alpha=\frac{-1+\sqrt{5}}{2}$.

## Therefore

$$
\begin{aligned}
& 1-\sin ^{2} \alpha=\frac{-1+\sqrt{5}}{2}, \sin ^{2} \alpha=1-\frac{-1+\sqrt{5}}{2}=\frac{3-\sqrt{5}}{2} \\
& \sin \alpha=\sqrt{\frac{3-\sqrt{5}}{2}}=\sqrt{\frac{6-2 \sqrt{5}}{4}}=\frac{\sqrt{6-2 \sqrt{5}}}{2}=\frac{\sqrt{5-2 \sqrt{5}+1}}{2}= \\
& =\frac{\sqrt{(\sqrt{5})^{2}-2 \sqrt{5}-1}}{2}=\frac{\sqrt{(\sqrt{5}-1)^{2}}}{2}=\frac{\sqrt{5}-1}{2} \\
& \text { Answer: } \sin \alpha=\frac{\sqrt{5}-1}{2}
\end{aligned}
$$