SOLUTION: Prove that for any n >= 2 , given {{{(sin(pi/n))*(sin(2pi/n))*"..."*(sin((n-1)pi/n)) = n/(2^(n-1) )}}}.

Prove: for 

.
 
Euler's equation: 

From which we can get: 


 
So each of the factors in the original expression is:
 

 
where k goes from 1 to n-1
 
Factor out the first term:
 

 
The original expression becomes a product of these three things:
 
#1.  
 
#2. 
 
#3. 
 
where k goes from 1 to n-1
 
Adding the exponents in #1.
 
#1.  
 
The sum of the first n-1 integers is , so the
exponent of e becomes
 

 
and #1 is now 
 
#1. 
 
Since #3  
it gives the product:
 


Putting #1 and #3 together:

#1*#3  

Now we look at #2 again.

#2. 

Those factors are all (1 - an nth root of 1 other than 1 itself. To show that

1 = cis(0 + 2k*pi) = e^(2k*pi) 

and by DeMoivre's theorem the n nth roots of unity are 

cis(2k*pi/n) and

that's what the second terms in those parenthetical expressions are.

and we can use any n-1 consecutive even integers for k, so here we are using
the n-1 negative consecutive even integers -2, -4, -6,  -2(n-1). And they
won't include 1 itself because we are not including integers 0 or 2n.

The nth roots of 1 can also be gotten by solving the equation





The first parentheses give us the root of one which is 1 itself.
The n-1 degree polynomial in the second parentheses has solutions
which are all the n-1 nth roots of 1 other than 1.

Therefore the polynomial

#4. x^(n-1)+x^(n-2)+x^(n-3)+"..."+1 

is equivalent to a polynomial which is like #2 with
x's placed where the 1's are.

#5. 

because they have the same solution and both have leading
coefficient 1.

So when we substitute 1 for x in #5, we get #2 and when
we substitute 1 for x in #4 we get n because there are n terms.
Therefore #2 simplifies to just n
 
Therefore #1*#2*#3 = 

Edwin