Respuesta :
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)
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:
The value of given expression is [tex]\frac{1}{2^{999}}[/tex].
Step-by-step explanation:
The given expression is
[tex]\cos A\cos 2A\cos 3A...........\cos 998A\cos 999A[/tex]
where, [tex]A=\frac{2\pi}{1999}[/tex]
Let as assume,
[tex]P=\cos A\cos 2A\cos 3A...........\cos 998A\cos 999A[/tex]
[tex]Q=\sin A\sin 2A\sin 3A...........\sin 998A\sin 999A[/tex]
[tex]2^{999}PQ=2^{999}(\cos A\cos 2A.........\cos 999A)[/tex][tex](\sin A\sin 2A........\sin 999A)[/tex]
[tex]2^{999}PQ=(2\cos A\sin A)(2\cos 2A\sin 2A)...........(2\cos 998A\sin 998A)(2\cos 999A\sin 999A)[/tex]
Using the formula, [tex]2\sin x\cos x=\sin 2x[/tex], we get
[tex]2^{999}PQ=\sin 2A\sin 4A......\sin 1998A[/tex]
[tex]2^{999}PQ=[\sin 2A\sin 4A......\sin 998A][-\sin(2\pi -1000A)][-\sin(2\pi -1002A)]...[-\sin(2\pi -1998A)][/tex] .... (1)
Now,
[tex]-\sin(2\pi -1000A)=-\sin(2\pi -1000(\frac{2\pi}{1999}))[/tex]
[tex]-\sin(2\pi -1000A)=-\sin(\frac{2\pi\cdot 999}{1999})[/tex]
[tex]-\sin(2\pi -1000A)=-\sin 999A[/tex]
So, equation (1) can be written as
[tex]2^{999}PQ=[\sin 2A\sin 4A......\sin 998A][\sin 999A\sin 997...\sin A][/tex]
[tex]2^{999}PQ=\sin A\sin 2A\sin 3A...........\sin 998A\sin 999A[/tex]
[tex]2^{999}PQ=Q[/tex]
Divide both sides by Q.
[tex]2^{999}P=1[/tex]
Divide both sides by [tex]2^{999}[/tex]
[tex]P=\frac{1}{2^{999}}[/tex]
Therefore the value of given expression is [tex]\frac{1}{2^{999}}[/tex].