Question

The value of $\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\cos \frac{16\mathrm{\pi }}{15}$ is ___________.

Answer 1

$$\begin{gathered} \cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\cos \frac{16\mathrm{\pi }}{15} \\ =\cos 2\left(\frac{\mathrm{\pi }}{15}\right)\cos 2^2\left(\frac{\mathrm{\pi }}{15}\right)\cos 2^3\left(\frac{\mathrm{\pi }}{15}\right)\cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{15}\right) \\ \left[\mathrm{since},\cos (\mathrm{\pi }+\mathrm{\theta })=-\cos \theta \right] \\ =\cos 2\left(\frac{\mathrm{\pi }}{15}\right)\cos 2^2\left(\frac{\mathrm{\pi }}{15}\right)\cos 2^3\left(\frac{\mathrm{\pi }}{15}\right)\left(-\cos \frac{\mathrm{\pi }}{15}\right) \\ =-\left[\cos \frac{\mathrm{\pi }}{15}\cos 2\left(\frac{\mathrm{\pi }}{15}\right)\cos 2^2\left(\frac{\mathrm{\pi }}{15}\right)\cos 2^3\left(\frac{\mathrm{\pi }}{15}\right)\right] \\ \mathrm{multiply}\mathrm{and}\mathrm{divide}\mathrm{by}2\sin \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right. \\ =\frac{-\left[2\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15\cos 2\left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}\right)\cos 2^2\left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}\right)\cos 2^3\left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}\right)$}\right.}\right]}{2\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}} \end{gathered}$$
$$\begin{gathered} =\frac{-1}{2\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}}\left[\sin 2\frac{\mathrm{\pi }}{15}\cos 2\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\cos 2^2\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\cos 2^3\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\right]\left(\mathrm{using}\mathrm{identity}\mathit{,}2\sin \theta \cos \theta =\sin 2\theta \right) \\ =\frac{-1}{2^2\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}}\left[\sin 4\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\cos 4\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\cos 2^3\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\right] \\ =\frac{-1}{2^3\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}}\left[\sin 8\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\cos 8\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\right] \\ =\frac{-1}{2^4\sin \left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}\right)}\left[\sin 16\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.\right)\right] \\ =\frac{-1}{16}\frac{\sin \left(\mathrm{\pi }+{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}\right)}{\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}} \\ =\frac{-1}{16}\frac{\left(-\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}\right)}{\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}}\left(\mathrm{since},\sin \left(\mathrm{\pi }+0\right)=-\mathrm{sin\theta }\right) \\ =\frac{-1}{16}\left(-1\right) \\ =\frac{1}{16} \\ \mathrm{Hence},\mathrm{value}\mathrm{of}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\cos \frac{16\mathrm{\pi }}{15}\mathrm{is}\frac{1}{16} \end{gathered}$$