Question

Find the value of cosAcos2Acos3A...........cos998Acos999A where A=2π/1999

Answer 1

Hello,

Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.

Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)

As sin x *cos x=sin (2x) /2

P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
         *1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
 as sin(x)=-sin(2pi-x) and 2pi=1999a

sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)

So  sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".

Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)

       

Answer 2

Answer:

The value of given expression is $\frac{1}{2^{999}}$.

Step-by-step explanation:

The given expression is

$\cos A\cos 2A\cos 3A...........\cos 998A\cos 999A$

where, $A=\frac{2\pi}{1999}$

Let as assume,

$P=\cos A\cos 2A\cos 3A...........\cos 998A\cos 999A$

$Q=\sin A\sin 2A\sin 3A...........\sin 998A\sin 999A$

$2^{999}PQ=2^{999}(\cos A\cos 2A.........\cos 999A)$$(\sin A\sin 2A........\sin 999A)$

$2^{999}PQ=(2\cos A\sin A)(2\cos 2A\sin 2A)...........(2\cos 998A\sin 998A)(2\cos 999A\sin 999A)$

Using the formula, $2\sin x\cos x=\sin 2x$, we get

$2^{999}PQ=\sin 2A\sin 4A......\sin 1998A$

$2^{999}PQ=[\sin 2A\sin 4A......\sin 998A][-\sin(2\pi -1000A)][-\sin(2\pi -1002A)]...[-\sin(2\pi -1998A)]$             .... (1)

Now,

$-\sin(2\pi -1000A)=-\sin(2\pi -1000(\frac{2\pi}{1999}))$

$-\sin(2\pi -1000A)=-\sin(\frac{2\pi\cdot 999}{1999})$

$-\sin(2\pi -1000A)=-\sin 999A$

So, equation (1) can be written as

$2^{999}PQ=[\sin 2A\sin 4A......\sin 998A][\sin 999A\sin 997...\sin A]$

$2^{999}PQ=\sin A\sin 2A\sin 3A...........\sin 998A\sin 999A$

$2^{999}PQ=Q$

Divide both sides by Q.

$2^{999}P=1$

Divide both sides by $2^{999}$

$P=\frac{1}{2^{999}}$

Therefore the value of given expression is $\frac{1}{2^{999}}$.