Question

Prove that: $\cos \frac{\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}=\frac{-1}{16}$

Answer 1

$$\begin{gathered} \cos \frac{\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5} \\ =\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(2\sin \frac{\mathrm{\pi }}{5}\cos \frac{\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right)\left(\mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{by}\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\right) \\ =\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \frac{2\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right)\left(\sin 2A=2\sin A\cos A\right) \\ =\frac{1}{4\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(2\sin \frac{2\mathrm{\pi }}{5}\cos \frac{2\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right)\left(\mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{by}2\right) \end{gathered}$$
$$\begin{gathered} =\frac{1}{4\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \frac{4\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right) \\ =\frac{1}{8\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(2\sin \frac{4\mathrm{\pi }}{5}\cos \frac{4\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right)\left(\mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{by}2\right) \\ =\frac{1}{8\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(\sin \frac{8\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right) \\ =\frac{1}{16\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left(2\sin \frac{8\mathrm{\pi }}{5}\cos \frac{8\mathrm{\pi }}{5}\right)\left(\mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{by}2\right) \end{gathered}$$
$$\begin{gathered} =\frac{\sin {\displaystyle \frac{16\mathrm{\pi }}{5}}}{16\sin {\displaystyle \frac{\mathrm{\pi }}{5}}} \\ =\frac{\sin \left(3\mathrm{\pi }+{\displaystyle \frac{\mathrm{\pi }}{5}}\right)}{16\sin {\displaystyle \frac{\mathrm{\pi }}{5}}} \\ =\frac{-\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}{16\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\left[\sin \left(3\mathrm{\pi }+\theta \right)=-\sin \theta \right] \\ =\frac{-1}{16} \end{gathered}$$