Prove that: \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64} \]

Using the identity \[ 2\sin\theta\cos\theta=\sin2\theta \]

Start with \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} \]

Multiply and divide by \[ \sin\frac{\pi}{65} \]

\[ = \frac{ \sin\frac{\pi}{65} \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} }{ \sin\frac{\pi}{65} } \]

Now, \[ 2\sin\frac{\pi}{65}\cos\frac{\pi}{65} = \sin\frac{2\pi}{65} \]

\[ = \frac{ \frac{1}{2} \sin\frac{2\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} }{ \sin\frac{\pi}{65} } \]

Again, \[ 2\sin\frac{2\pi}{65}\cos\frac{2\pi}{65} = \sin\frac{4\pi}{65} \]

\[ = \frac{ \frac{1}{4} \sin\frac{4\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} }{ \sin\frac{\pi}{65} } \]

Again, \[ 2\sin\frac{4\pi}{65}\cos\frac{4\pi}{65} = \sin\frac{8\pi}{65} \]

\[ = \frac{ \frac{1}{8} \sin\frac{8\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} }{ \sin\frac{\pi}{65} } \]

Again, \[ 2\sin\frac{8\pi}{65}\cos\frac{8\pi}{65} = \sin\frac{16\pi}{65} \]

\[ = \frac{ \frac{1}{16} \sin\frac{16\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} }{ \sin\frac{\pi}{65} } \]

Again, \[ 2\sin\frac{16\pi}{65}\cos\frac{16\pi}{65} = \sin\frac{32\pi}{65} \]

\[ = \frac{ \frac{1}{32} \sin\frac{32\pi}{65} \cos\frac{32\pi}{65} }{ \sin\frac{\pi}{65} } \]

Again, \[ 2\sin\frac{32\pi}{65}\cos\frac{32\pi}{65} = \sin\frac{64\pi}{65} \]

\[ = \frac{ \frac{1}{64} \sin\frac{64\pi}{65} }{ \sin\frac{\pi}{65} } \]

Now, \[ \sin\frac{64\pi}{65} = \sin\left(\pi-\frac{\pi}{65}\right) \]

\[ = \sin\frac{\pi}{65} \]

Therefore,

\[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64} \]