Question

Prove that:
$\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{16\pi }{15}=\frac{1}{16}$

Answer 1

$\mathrm{LHS}=\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\cos \frac{16\mathrm{\pi }}{15}$

On dividing and multiplying by $2\sin \frac{2\mathrm{\pi }}{15}$, we get

$$\begin{gathered} =\frac{1}{2\sin \frac{2\mathrm{\pi }}{15}}\times \left(2\sin \frac{2\mathrm{\pi }}{15}\times \cos \frac{2\mathrm{\pi }}{15}\right)\times \cos \frac{4\mathrm{\pi }}{15}\times \cos \frac{8\mathrm{\pi }}{15}\times \cos \frac{16\mathrm{\pi }}{15} \\ =\frac{1}{2\times 2\sin \frac{2\mathrm{\pi }}{15}}\times \left(2\sin \frac{4\mathrm{\pi }}{15}\times \cos \frac{4\mathrm{\pi }}{15}\right)\times \cos \frac{8\mathrm{\pi }}{15}\times \cos \frac{16\mathrm{\pi }}{15} \\ =\frac{1}{2\times 4\sin \frac{2\mathrm{\pi }}{15}}\left(2\sin \frac{8\mathrm{\pi }}{15}\times \cos \frac{8\mathrm{\pi }}{15}\right)\times \cos \frac{16\mathrm{\pi }}{15} \\ =\frac{1}{2\times 8\sin \frac{2\mathrm{\pi }}{15}}\left(2\sin \frac{16\mathrm{\pi }}{15}\times \cos \frac{16\mathrm{\pi }}{15}\right) \\ =\frac{1}{16\sin \frac{2\mathrm{\pi }}{15}}\left(\sin \frac{32\mathrm{\pi }}{15}\right) \end{gathered}$$

$$\begin{gathered} =-\frac{1}{16\sin \frac{2\mathrm{\pi }}{15}}\left(\sin 2\mathrm{\pi }-\frac{32\mathrm{\pi }}{15}\right)\left[\because \sin \left(2\mathrm{\pi }-\mathrm{\theta }\right)=-\mathrm{sin\theta }\right] \\ =-\frac{1}{16\sin \frac{2\mathrm{\pi }}{15}}\sin \left(-\frac{2\mathrm{\pi }}{15}\right) \\ =\frac{1}{16}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$