Question

The value of $\cos \frac{\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}$ is ______________.

Answer 1

$$\begin{gathered} \cos \frac{\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5} \\ \mathrm{multiply}\mathrm{and}\mathrm{divide}\mathrm{above}\mathrm{equation}\mathrm{by}2\sin \frac{\mathrm{\pi }}{5} \\ =\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(2\sin \frac{\mathrm{\pi }}{5}\cos \frac{\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right) \\ =\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \frac{2\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right)\left(\mathrm{using}\mathrm{identity}2\sin \theta \cos \theta =\sin 2\theta \right) \\ =\frac{1}{2^2\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(2\sin \frac{2\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right) \\ =\frac{1}{2^2\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \frac{4\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right) \\ =\frac{1}{2^3\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(2\sin \frac{4\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right) \\ =\frac{1}{2^3\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \frac{8\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right) \\ =\frac{1}{2^4\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(2\sin \frac{8\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right) \\ =\frac{1}{2^4\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \frac{16\mathrm{\pi }}{5}\right) \\ =\frac{1}{16\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \left(3\mathrm{\pi }+\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)\right) \\ =\frac{1}{16\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \frac{\mathrm{\pi }}{5}\right)\left(\because \sin \left(3\mathrm{\pi }+\mathrm{\theta }\right)=-\sin \theta \right) \\ \therefore \cos \frac{\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}=-\frac{1}{16} \end{gathered}$$