Value of cos(π/65) cos(2π/65) cos(4π/65) cos(8π/65) cos(16π/65) cos(32π/65)
Question
Find the value of
\[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} \]
(a) \(\frac{1}{8}\)
(b) \(\frac{1}{16}\)
(c) \(\frac{1}{32}\)
(d) none of these
Solution
Use the standard identity:
\[ \sin(2^n x) = 2^n \sin x \cos x \cos 2x \cos 4x \cdots \cos(2^{\,n-1}x) \]
Taking
\[ x=\frac{\pi}{65}, \qquad n=6 \]
\[ \sin\frac{64\pi}{65} = 64\sin\frac{\pi}{65} \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} \]
Since
\[ \sin\frac{64\pi}{65} = \sin\left(\pi-\frac{\pi}{65}\right) = \sin\frac{\pi}{65} \]
Substituting,
\[ \sin\frac{\pi}{65} = 64\sin\frac{\pi}{65} \left( \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} \right) \]
Cancelling \(\sin\frac{\pi}{65}\),
\[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64} \]
Final Answer
\[ \boxed{\frac{1}{64}} \]
Since \(\frac{1}{64}\) is not among the given options, the correct answer is:
(d) none of these