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

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)