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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?

Ng Chung Tak
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A.Fidel
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  • Set $x=y+\dfrac12$ in $$ \frac{\sin\pi n(x+1/2)}{2^{n-1}}=\frac{\sin\pi n(x)}{2^{n-1}} $$ – lab bhattacharjee Oct 09 '14 at 06:37
  • Can you give me more of an explanation? Here's what I'm doing. $$ Let :x=y+1/2:then\ \frac{sin\pi n(x+1/2)}{2^{n-1}}=\frac{sin\pi n(y+1/2+1/2)}{2^{n-1}}=\frac{sin\pi n(y)}{2^{n-1}}:then\\prod_{k=0}^{n-1}sin\pi (y+k/n)=\prod_{k=0}^{n-1}sin\pi (y):because:periodity.:Then\let :y=(\phi+\pi/2).:Then\prod_{k=0}^{n-1}sin\pi (y)=\prod_{k=0}^{n-1}sin\pi(\phi+\pi/2)=\prod_{k=0}^{n-1}cos(\phi)=\prod_{k=0}^{n-1}cos(\phi+k/n) $$ Obviously, this is wrong. My final notation ends using phi as the independent variable instead of x. I feel like I'm doing something obviously wrong, and simple too. – A.Fidel Oct 09 '14 at 22:22

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