Using $\;\;\sin(\frac{nx}{2^{n-1}})=\sin x\times \sin(x+\pi/n)\times \cdots \times \sin(x+\frac{(n-1)\pi}{n}) $
Simplify the following expression. $$\sum_{k=1}^{n}{\cot(x+\frac{(k-1)\pi}{n})}$$
Choices are { $n\cot(nx)$ }, { $\cot(nx)/n$ }, { $\frac{n\cot(nx)}{2^{n-1}}$ }, and { $\frac{\cot(nx)}{2^{n-1}}$ }.
What I've tried so far are,
Let $a_n=\sin(\frac{nx}{2^{n-1}})$ $n=1,\; \sin(x)=\sin(x) \\ n=2, \;\sin(x)=\sin(x)\cdot \sin(x+1/2\pi) \quad \therefore \sin(x+1/2\pi)=1\\ n=3, \;\sin(3x/4)=\sin(x)\cdot \sin(x+1/2\pi)\cdot \sin(x+2/3\pi) =\sin(x)\cdot \sin(x+2/3\pi) \\ \cdots \\ \;\text{ then,} \quad \sin(x+\frac{(n-1)\pi}{n})=a_n/a_{n-1} $
$\cos(x+\frac{(n-1)\pi}{n})=\sin(x+\pi/2+\frac{(n-1)\pi}{n})$
How should I solve it?