Prove that $$\prod_{k=0}^{n-1} \cos \left( x+\frac{k}{n} \right)\pi= \frac{1}{2^{n-1}} \sin n\left( x+\frac{1}{2} \right)\pi $$
given that $$\prod_{k=0}^{n-1} \sin \left( x+\frac{k}{n} \right)\pi= \frac{\sin n\pi x}{2^{n-1}}$$ and $$\sin \left( \phi+\frac{\pi}{2} \right)= \cos \phi$$
From my understanding,
$$\frac{\sin n\left( x+\dfrac{1}{2} \right)\pi}{2^{n-1}}= \frac{\sin n\pi x}{2^{n-1}}$$
Because $\sin \pi x$ has period $\dfrac{1}{2}$, I can say
$$\frac{\sin n\left( x+ \dfrac{1}{2} \right)\pi}{2^{n-1}}= \frac{\sin n\pi x}{2^{n-1}}= \prod_{k=0}^{n-1} \sin \left( x+\frac{k}{n} \right)\pi$$
So I want to use the given $\sin \left( \phi+\dfrac{\pi}{2} \right)= \cos \phi$, but I can't prove that
$$\sin \left( x+\frac{k}{n} \right)\pi = \sin \left( \phi +\frac{\pi}{2} \right),$$
specifically
$$\frac{k}{n}=\frac{\pi}{2}$$
Perhaps I'm missing some tools for me to answer this problem. How would you solve this problem? Is my logic flawed?