Question

The value of $\cos A\cos 2A\cos 3A...\cos 999A$$=\frac{1}{2^a}$, where $A=\frac{2\pi }{1999}$. Find the value of $a.$

Answer 1

Let, $P=\cos a\cos 2a\cos 3a...\cos 999a$
$Q=\sin a\sin 2a\sin 3a...\sin 999a$.

Then,

$2^{999}PQ=\left(2\sin a\cos a\right)\left(2\sin 2a\cos 2a\right)...\left(2\sin 999a\cos 999a\right)$

$=\sin 2a\sin 4a...\sin 1998a$

$=(\sin 2a\sin 4a...\sin 998a)[-\sin (2\pi -1000a)][-\sin (2\pi -1002a)]...[-\sin (2\pi -1998a)]$

$=\sin 2a\sin 4a...\sin 998a\sin 999a\sin 997a...\sin a=Q$.

It is easy to see that $Q\ne 0$. Hence, the desired product is $P=\frac{1}{2^{999}}$
Ans: $a=999$