Prove that: cos(2π/15) cos(4π/15) cos(8π/15) cos(16π/15) = 1/16

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

Start with \[ \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} \]

Multiply and divide by \[ \sin\frac{2\pi}{15} \]

\[ = \frac{ \sin\frac{2\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} } { \sin\frac{2\pi}{15} } \]

Now, \[ 2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15} = \sin\frac{4\pi}{15} \]

\[ = \frac{ \frac{1}{2} \sin\frac{4\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} } { \sin\frac{2\pi}{15} } \]

Again, \[ 2\sin\frac{4\pi}{15}\cos\frac{4\pi}{15} = \sin\frac{8\pi}{15} \]

\[ = \frac{ \frac{1}{4} \sin\frac{8\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} } { \sin\frac{2\pi}{15} } \]

Again, \[ 2\sin\frac{8\pi}{15}\cos\frac{8\pi}{15} = \sin\frac{16\pi}{15} \]

\[ = \frac{ \frac{1}{8} \sin\frac{16\pi}{15} \cos\frac{16\pi}{15} } { \sin\frac{2\pi}{15} } \]

Again, \[ 2\sin\frac{16\pi}{15}\cos\frac{16\pi}{15} = \sin\frac{32\pi}{15} \]

\[ = \frac{ \frac{1}{16} \sin\frac{32\pi}{15} } { \sin\frac{2\pi}{15} } \]

Now, \[ \sin\frac{32\pi}{15} = \sin\left(2\pi+\frac{2\pi}{15}\right) \]

\[ = \sin\frac{2\pi}{15} \]

Therefore,

\[ \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} = \frac{1}{16} \]