4. In $\triangle A B C$, if $\sin A=2 \sin C$, and the three sides $a, b, c$ form a geometric sequence, then the value of $\cos A$ is $\qquad$ .
Answer: $-\frac{\sqrt{2}}{4}$.
Solution: According to the Law of Sines, $\frac{a}{c}=\frac{\sin A}{\sin C}=2$, and $b^{2}=a c$, thus $a: b: c=2: \sqrt{2}: 1$. Therefore, by the Law of Cosines, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(\sqrt{2})^{2}+1^{2}-2^{2}}{2 \times \sqrt{2} \times 1}=-\frac{\sqrt{2}}{4}$.