Question

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

• A. $\frac{1}{{16}^{16}}$
• B. $\frac{1}{8^8}$
• C. $\frac{1}{{32}^{32}}$
• D. $\frac{1}{{64}^{64}}$

Answer 1

The correct option is A $\frac{1}{{16}^{16}}$

Let $P=\cos \left(\frac{2\pi }{2^{64}-1}\right)\cos \left(\frac{2^2\pi }{2^{64}-1}\right)\cdots \cos \left(\frac{2^{64}\pi }{2^{64}-1}\right)$
$\Rightarrow P=\frac{1}{2\sin \left(\frac{2\pi }{2^{64}-1}\right)}2\sin \left(\frac{2\pi }{2^{64}-1}\right)\cos \left(\frac{2\pi }{2^{64}-1}\right)\cos \left(\frac{2^2\pi }{2^{64}-1}\right)\cdots \cos \left(\frac{2^{64}\pi }{2^{64}-1}\right)$
$\Rightarrow P=\frac{1}{4\sin \left(\frac{2\pi }{2^{64}-1}\right)}2\sin \left(\frac{2^2\pi }{2^{64}-1}\right)\cos \left(\frac{2^2\pi }{2^{64}-1}\right)\cdots \cos \left(\frac{2^{64}\pi }{2^{64}-1}\right)$
$\Rightarrow P=\frac{1}{2^2\sin \left(\frac{2\pi }{2^{64}-1}\right)}\sin \left(\frac{2^3\pi }{2^{64}-1}\right)\cdots \cos \left(\frac{2^{64}\pi }{2^{64}-1}\right)$
$\Rightarrow P=\frac{1}{2^{64}\sin \left(\frac{2\pi }{2^{64}-1}\right)}\sin \left(\frac{2^{65}\pi }{2^{64}-1}\right)$
$\Rightarrow P=\frac{1}{2^{64}\sin \left(\frac{2\pi }{2^{64}-1}\right)}\sin (2\pi +\frac{2\pi }{2^{64}-1})$
$\Rightarrow P=\frac{1}{2^{64}\sin \left(\frac{2\pi }{2^{64}-1}\right)}\sin \left(\frac{2\pi }{2^{64}-1}\right)$
$\Rightarrow P=\frac{1}{2^{64}}=\frac{1}{{16}^{16}}$