Check the identity:
$$\prod\limits_{0\le k\le n-1}\sin\left(\frac{k\pi}{n}+x\right)=\frac{\sin(nx)}{2^{n-1}}$$
$$t^n-1=\prod\limits_{0\le k\le n-1}\big(t-e^{i\frac{2k\pi}{n}}\big)$$
Let $t=e^{i\theta}$. In the following I have used that $|1-e^{i\theta}|=2|\sin\frac{\theta}{2}|$
$\text{LHS}=e^{in\theta}-1\Rightarrow|\text{LHS}|=2|\sin\frac{n\theta}{2}|$
Now we will work on the expression
$A=e^{i\theta}-e^{i\theta}e^{i(\frac{2k\pi}{n}-\theta)}=e^{i\theta}-e^{i\theta}e^{i(\frac{2k\pi}{n}-\theta)}=e^{i\theta}\big(1-e^{i(\frac{2k\pi}{n}-\theta)}\big)$
$\Rightarrow |A|=2|\sin\big(\frac{k\pi}{n}-\frac{\theta}{2}\big)|$
$\Rightarrow|\text{RHS}|=2^n\big|\prod\limits_{0\le k\le n-1}\sin\big(\frac{k\pi}{n}-\frac{\theta}{2}\big)\big|$
$|\text{LHS}|=|\text{RHS}|\Rightarrow\big|\prod\limits_{0\le k\le n-1}\sin\big(\frac{k\pi}{n}-\frac{\theta}{2}\big)\big|=\frac{|\sin\frac{n\theta}{2}|}{2^{n-1}}$
Taking $\theta=-2x\Rightarrow\big|\prod\limits_{0\le k\le n-1}\sin\big(\frac{\pi}{n}+x\big)\big|=\frac{|\sin(nx)|}{2^{n-1}}$
How do I get rid of the absolute value?
Thanks for the help.
P.S. I suggest viewing with ×2 magnifier because the exponentials get tiny.