Using roots of unity to prove that $\cos\frac{\pi}{2n}\cos\frac{2\pi}{2n}\cdots\cos\frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$

Question

I have proven this using trigonometry: $$\cos\frac{\pi}{2n}\cos\frac{2\pi}{2n}\cdots\cos\frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$$

Can someone please help me to prove the result using the concept of "roots of unity"?

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My trigonometric proof:

Answer 1

$x_k,\bar x_k$ are $(2n-2)$ roots of $x^{2n}=1,$ such that $x_k=e^{ik\pi/(2n)}, k=1,2,3,...n $, $x=\pm 1$ are other roots then $$\frac{x^{2n}-1}{x^2-1}=\prod_{k=1}^{n}(x-\bar x_k) (x-x_k)=(x^2-2x\cos(\pi/n)+1)~$$ $$Ta (x^2-2x\cos(2\pi/n)+1)~(x^2-2x\cos(3\pi/n)+1)......(x^2-2x\cos((n-1)\pi/n)+1)~~~(1) $$ Take limit of (1) as $x\to -1$, we get $$\implies n= 4^{n-1} \cos^2(\pi/(2n))~\cos^2(2\pi/(2n))~\cos^2(3\pi/(2n))...\cos^2((n-1)\pi/(2n))$$ $$\implies \cos(\pi/(2n))~\cos(2\pi/(2n))~\cos(3\pi/(2n))...\cos((n-1)\pi/(2n))=\frac{\sqrt{n}}{2^{n-1}}$$

Answer 2

$$\Pi_{k=1}^{[n/2]} \sin(\frac{k \pi}{n}) =\frac{\sqrt{n}}{2^{(n-1)/2}}$$

(for all n)

$$\Pi_{k=1}^{[n/2]} \cos(\frac{k \pi}{n}) = \frac{1}{2^{(n-1)/2}}$$ (for odd n)

$$\Pi_{k=1}^{[n/2]} \cos(\frac{k \pi}{n}) = \frac{\sqrt{n}}{2^{(n-1)/2}}$$(for even n)

[.] Represents the greatest integer function

Proof: $z^n-1=(z-1)(z-z_1)(z-z_2)\ldots(z-z_{n-1})$.(1) Shifting (z-1) to other side and taking Lim z tending to 1 we get $n=(1-z_1)(1-z_2)\ldots(1-z_{n-1})$.(2) Multiplying this equation with its conjugate gives $n^2=(2-2\cos(2\pi/n))...(2-2cos(2(n-1)\pi/n))=2^{2(n-1)} \sin^2(\pi/n)\ldots \sin^2((n-1)\pi/n)$. Taking square root both sides and using $\sin(x)=\sin(π-x)$ proves the first result. This is true for the even case(OP's question) cos product as well (Replacing x with π/2-x where n=even (this result is valid for all n, but we perform this only for n=even,as for n=odd,we would get another sequence (not mentioned above) of cos on doing so)$\cos(x)=\sin(π/2-x)$, in the case where n)

Similarly, for the cosine product when n is odd, multiplying conjugate of equation (1) (when z=-1) with itself gives:

$(1+(-1)^{n-1})^2=4$(as n is odd)$=4*2^{2(n-1)} \cos(\pi/n)\cos(2\pi/n)\ldots \cos((n-1)\pi/n) $, using $\cos(x)=-\cos(π-x))$ proves the last result as well.