The value of cos

$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}$

$=\cos 2\left(\frac{\pi}{15}\right) \cos 2^{2}\left(\frac{\pi}{15}\right) \cos 2^{3}\left(\frac{\pi}{15}\right) \cos \left(\pi+\frac{\pi}{15}\right)$

$[$ since, $\cos (\pi+\theta)=-\cos \theta]$

$=\cos 2\left(\frac{\pi}{15}\right) \cos 2^{2}\left(\frac{\pi}{15}\right) \cos 2^{3}\left(\frac{\pi}{15}\right)\left(-\cos \frac{\pi}{15}\right)$

$=-\left[\cos \frac{\pi}{15} \cos 2\left(\frac{\pi}{15}\right) \cos 2^{2}\left(\frac{\pi}{15}\right) \cos 2^{3}\left(\frac{\pi}{15}\right)\right]$

multiply and divide by $2 \sin \pi / 15$

$=\frac{-\left[2 \sin \pi / 15 \cos \pi / 15 \cos 2(\pi / 15) \cos 2^{2}(\pi / 15) \cos 2^{3}(\pi / 15)\right]}{2 \sin \pi / 15}$

$=\frac{-1}{2 \sin \pi / 15}\left[\sin 2 \frac{\pi}{15} \cos 2(\pi / 15) \cos 2^{2}(\pi / 15) \cos 2^{3}(\pi / 15)\right] \quad($ using identity, $2 \sin \theta \cos \theta=\sin 2 \theta)$

$=\frac{-1}{2^{2} \sin \pi / 15}\left[\sin 4(\pi / 15) \cos 4(\pi / 15) \cos 2^{3}(\pi / 15)\right]$

$=\frac{-1}{2^{3} \sin \pi / 15}[\sin 8(\pi / 15) \cos 8(\pi / 15)]$

$=\frac{-1}{2^{4} \sin (\pi / 15)}[\sin 16(\pi / 15)]$

$=\frac{-1}{16} \frac{\sin (\pi+\pi / 15)}{\sin \pi / 15}$

$=\frac{-1}{16} \frac{(-\sin \pi / 15)}{\sin \pi / 15} \quad($ since, $\sin (\pi+0)=-\sin \theta)$

$=\frac{-1}{16} \quad(-1)$

$=\frac{1}{16}$

Hence, value of $\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}$ is $\frac{1}{16}$