$\cos {24}^{\circ }\cos {48}^{\circ }\cos {96}^{\circ }\cos {168}^{\circ }=$

$$\begin{gathered} \text{Let }\mathrm{A}={24}^{\circ }=\frac{2\mathrm{\pi }}{15}\text{ then }\cos \mathrm{acos}2\mathrm{Acos}{2}^2\mathrm{Acos}{2}^3\mathrm{A}\cdots \cos 2^{\mathrm{n}-1}\mathrm{A}=\frac{\sin 2^{\mathrm{n}}\mathrm{A}}{2^{\mathrm{n}}\sin \mathrm{A}}\text{. } \\ \Rightarrow \cos {24}^{\circ }\cdot \cos {48}^{\circ }\cdot \cos {96}^{\circ }\cdot \cos {168}^{\circ }=\cos {24}^{\circ }\cos {48}^{\circ }\cos {96}^{\circ }\cos {192}^{\circ }=\frac{\sin 32\mathrm{\pi }/15}{2^4\sin 2\mathrm{\pi }/15}=\frac{1}{{16}^{.}} \end{gathered}$$