Question

The value of $\cos \frac{\pi }{65}\cos \frac{2\pi }{65}\cos \frac{4\pi }{65}\cos \frac{8\pi }{65}\cos \frac{16\pi }{65}\cos \frac{32\pi }{65}$ is
(a) $\frac{1}{8}$

(b) $\frac{1}{16}$

(c) $\frac{1}{32}$

(d) none of these

Answer 1

(d) none of these

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \cos \frac{\mathrm{\pi }}{65}\cos \frac{2\mathrm{\pi }}{65}\cos \frac{4\mathrm{\pi }}{65}\cos \frac{8\mathrm{\pi }}{65}\cos \frac{16\mathrm{\pi }}{65}\cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}{2\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\cos \frac{\mathrm{\pi }}{65}\cos \frac{2\mathrm{\pi }}{65}\cos \frac{4\mathrm{\pi }}{65}\cos \frac{8\mathrm{\pi }}{65}\cos \frac{16\mathrm{\pi }}{65}\cos \frac{32\mathrm{\pi }}{65} \\ \left(\mathrm{dividing}\mathrm{and}\mathrm{multiplying}\mathrm{by}2\sin \frac{\mathrm{\pi }}{65}\right) \\ =\frac{2\sin {\displaystyle \frac{2\mathrm{\pi }}{65}}}{2\times 2\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\cos \frac{2\mathrm{\pi }}{65}\cos \frac{4\mathrm{\pi }}{65}\cos \frac{8\mathrm{\pi }}{65}\cos \frac{16\mathrm{\pi }}{65}\cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\sin {\displaystyle \frac{4\mathrm{\pi }}{65}}}{2\times 4\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\cos \frac{4\mathrm{\pi }}{65}\cos \frac{8\mathrm{\pi }}{65}\cos \frac{16\mathrm{\pi }}{65}\cos \frac{32\mathrm{\pi }}{65} \end{gathered}$$
$$\begin{gathered} =\frac{2\sin {\displaystyle \frac{8\mathrm{\pi }}{65}}}{2\times 8\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\cos \frac{8\mathrm{\pi }}{65}\cos \frac{16\mathrm{\pi }}{65}\cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\sin {\displaystyle \frac{16\mathrm{\pi }}{65}}}{2\times 16\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\cos \frac{16\mathrm{\pi }}{65}\cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\sin {\displaystyle \frac{32\mathrm{\pi }}{65}}}{2\times 32\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\cos \frac{32\mathrm{\pi }}{65} \\ =\frac{\sin {\displaystyle \frac{64\mathrm{\pi }}{65}}}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}} \\ =\frac{\sin \left(\mathrm{\pi }-{\displaystyle \frac{\mathrm{\pi }}{65}}\right)}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}} \\ =\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}} \\ =\frac{1}{64} \end{gathered}$$