Prove that: cos(π/5) cos(2π/5) cos(4π/5) cos(8π/5) = -1/16
Question
Prove that \[ \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} = -\frac{1}{16} \]
Solution
Using the identity \[ 2\sin\theta\cos\theta=\sin2\theta \]
Start with \[ \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} \]
Multiply and divide by \[ \sin\frac{\pi}{5} \]
\[ = \frac{ \sin\frac{\pi}{5} \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} } { \sin\frac{\pi}{5} } \]
Now, \[ 2\sin\frac{\pi}{5}\cos\frac{\pi}{5} = \sin\frac{2\pi}{5} \]
\[ = \frac{ \frac{1}{2} \sin\frac{2\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} } { \sin\frac{\pi}{5} } \]
Again, \[ 2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5} = \sin\frac{4\pi}{5} \]
\[ = \frac{ \frac{1}{4} \sin\frac{4\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} } { \sin\frac{\pi}{5} } \]
Again, \[ 2\sin\frac{4\pi}{5}\cos\frac{4\pi}{5} = \sin\frac{8\pi}{5} \]
\[ = \frac{ \frac{1}{8} \sin\frac{8\pi}{5} \cos\frac{8\pi}{5} } { \sin\frac{\pi}{5} } \]
Again, \[ 2\sin\frac{8\pi}{5}\cos\frac{8\pi}{5} = \sin\frac{16\pi}{5} \]
\[ = \frac{ \frac{1}{16} \sin\frac{16\pi}{5} } { \sin\frac{\pi}{5} } \]
Now, \[ \sin\frac{16\pi}{5} = \sin\left(2\pi+\frac{6\pi}{5}\right) \]
\[ = \sin\frac{6\pi}{5} \]
\[ = -\sin\frac{\pi}{5} \]
Therefore,
\[ \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} = \frac{1}{16} \times \frac{-\sin\frac{\pi}{5}} {\sin\frac{\pi}{5}} \]
\[ = -\frac{1}{16} \]
Final Answer
\[ \boxed{ \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} = -\frac{1}{16} } \]