Question

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

(ii) $\cos \frac{\pi }{65}\cos \frac{2\pi }{65}\cos \frac{4\pi }{65}\cos \frac{8\pi }{65}\cos \frac{16\pi }{65}\cos \frac{32\pi }{65}=\frac{1}{64}$

Answer 1

(i)

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

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

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

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

(ii)

$\mathrm{LHS}=\cos \frac{\mathrm{\pi }}{65}\cos \frac{2\mathrm{\pi }}{65}\cos \frac{4\mathrm{\pi }}{65}\cos \frac{8\mathrm{\pi }}{65}\cos \frac{16\mathrm{\pi }}{65}\cos \frac{32\mathrm{\pi }}{65}$

On dividing and multiplying by $2\sin \frac{\mathrm{\pi }}{65}$, we get

$$\begin{gathered} =\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times 2\sin \frac{\mathrm{\pi }}{65}\times \cos \frac{\mathrm{\pi }}{65}\times \cos \frac{2\mathrm{\pi }}{65}\times \cos \frac{4\mathrm{\pi }}{65}\times \cos \frac{8\mathrm{\pi }}{65}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\times \sin 2{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 2\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{2\mathrm{\pi }}{65}\times \cos \frac{4\mathrm{\pi }}{65}\times \cos \frac{8\mathrm{\pi }}{65}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\times \sin 4{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 4\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{4\mathrm{\pi }}{65}\times \cos \frac{8\mathrm{\pi }}{65}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\times \sin 8{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 8\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{8\mathrm{\pi }}{65}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \end{gathered}$$

$$\begin{gathered} =\frac{2\times \sin 16{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 16\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\times \sin 32{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 32\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{\sin 64{\displaystyle \frac{\mathrm{\pi }}{65}}}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}=\frac{\sin \left(\mathrm{\pi }-{\displaystyle \frac{\mathrm{\pi }}{65}}\right)}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}} \\ =\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\left[\because \sin \left(\mathrm{\pi }-\mathrm{\theta }\right)=\mathrm{sin\theta }\right] \\ =\frac{1}{64}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$