Question

The product $\cos \left(\frac{2\pi }{2^{64}-1}\right)\cdot \cos \left(\frac{2^2\pi }{2^{64}-1}\right)\cdots ,\cos \left(\frac{2^{64}\pi }{2^{64}-1}\right)$ is

Answer 1

We have,

$\cos \left(\frac{2\pi }{2^{64}-1}\right)\cos \left(\frac{2^2\pi }{2^{64}-1}\right)....,\cos \left(\frac{2^{64}\pi }{2^{64}-1}\right)$

$=\frac{\sin \left(\frac{2\pi }{2^{64}-1}\right)\cos \left(\frac{2\pi }{2^{64}-1}\right)\cos \left(\frac{2^2\pi }{2^{64}-1}\right)....\cos \left(\frac{2^{64}\pi }{2^{64}-1}\right)}{\sin \left(\frac{2\pi }{2^{64}-1}\right)}$

$=\frac{\sin \left(\frac{2^2\pi }{2^{64}-1}\right)\cos \left(\frac{2^2\pi }{2^{64}-1}\right)....\cos \left(\frac{2^{64}\pi }{2^{64}-1}\right)}{\sin \left(\frac{2\pi }{2^{64}-1}\right)}$

$=\frac{\sin \left(\frac{2^{65}\pi }{2^{64}-1}\right)}{\sin \left(\frac{2\pi }{2^{64}-1}\right)}=\frac{\sin \left(\frac{2^{64}\pi }{2^{64}-1}\right)}{\sin \left(\frac{2\pi }{2^{64}-1}\right)}$

$=\frac{\sin \left[\frac{(2.2^{64}-2)\pi +2\pi }{(2^{64}-1)}\right]}{\sin \left(\frac{2\pi }{2^{64}-1}\right)}$

$=\frac{\sin [\frac{2(2^{64}-1)\pi }{(2^{64}-1)}+\frac{2\pi }{2^{64}-1}]}{\sin \left(\frac{2\pi }{2^{64}-1}\right)}=\frac{\sin (2\pi +\frac{2\pi }{2^{64}-1})}{\sin (2\pi +\frac{2\pi }{2^{64}-1})}=1$

Hence, this is the answer.