Proving the formula for $\prod_{k=1}^n \cos(kx)$

I'm trying to find a closed-form for the following product:

$$ \prod_{k=1}^n \cos(kx)$$

Here is my attempt at it:

$$ \prod_{k=1}^n \cos(2kx) = \prod_{k=1}^n \frac{ e^{i2kx} + e^{-i2kx} }{2} = \frac{(e^{2ix} - e^{-2ix}) \prod_{k=1}^n e^{i2kx} + e^{-i2kx} }{2^n (e^{i2x}- e^{-i2x})}= \frac{1}{2^n (e^{i2x}- e^{-i2x})} ( e^{4inx} - e^{-4inx} )= \frac{1}{2^n} \frac{\sin(4nx)}{\sin(2x)} $$

The above was simplified by using the algebraic identity $(a+b)(a-b)=a^2 -b^2$ repeatedly in the product. Now sub:$ x \to \frac{x}{2}$

$$ \prod_{k=1}^n \cos( kx)= \frac{1}{2^n} \frac{\sin(2nx)}{\sin x}$$

I'm not sure if I'm doing something wrong, but when I plug $ x= \frac{\pi}{n}$, I'm not getting the result as shown in this MSE post

Edit: Whoops! The algebraic manipulation doesn't hold after k=3 as correctly pointe od out by @leon. However, still am looking for a closed form.

Solution 1:

Your result seems to be wrong for $n>1$, because the limit as $x\to 0$ of the left hand side is $1$, whereas the limit as $x\to 0$ of the r.h.s is $\frac{2n}{2^n}$