Using the real factorizations…

Using the real factorizations $\ x^{2n}-1=(x-1)(x+1)\displaystyle\prod_{k=1}^{n-1}\left(x^2-2x\cos\frac{k\pi}n+1\right)$ and $\ x^{2n+1}-1=(x-1)\displaystyle\prod_{k=1}^{n}\left(x^2-2x\cos\frac{2k\pi}{2n+1}+1\right)$, prove: $\quad1^{\circ}\,\, \displaystyle\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{2n}\right)=\frac{\sqrt n}{2^{n-1}};\qquad 2^{\circ}\,\, \displaystyle\prod_{k=1}^{n}\cos\left(\frac{k\pi}{2n+1}\right)=\frac1{2^n}.$

Part $1^{\circ}$ · $\displaystyle\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}=\frac{\sqrt n}{2^{n-1}}$

Step 1 · The classical lemma

For any integer $N\ge2$, the roots of $x^N=1$ other than $x=1$ are $e^{2\pi ik/N}$, $k=1,\ldots,N-1$, so $\dfrac{x^N-1}{x-1}=\displaystyle\prod_{k=1}^{N-1}\left(x-e^{2\pi ik/N}\right)$. Setting $x=1$ (as a limit) gives $N=\displaystyle\prod_{k=1}^{N-1}\left(1-e^{2\pi ik/N}\right)$. Writing $1-e^{i\varphi}=-2i\sin\frac\varphi2\,e^{i\varphi/2}$ with $\varphi=\frac{2\pi k}N$ and taking absolute values:

$$ N=\prod_{k=1}^{N-1}2\sin\frac{k\pi}N \qquad\Longrightarrow\qquad \prod_{k=1}^{N-1}\sin\frac{k\pi}N=\frac N{2^{N-1}} $$

Step 2 · Apply the lemma with $N=2n$

$$ \prod_{k=1}^{2n-1}\sin\frac{k\pi}{2n}=\frac{2n}{2^{2n-1}}=\frac n{4^{n-1}} $$

Step 3 · Pair the factors by symmetry

Since $\sin\!\left(\frac{(2n-k)\pi}{2n}\right)=\sin\!\left(\pi-\frac{k\pi}{2n}\right)=\sin\frac{k\pi}{2n}$, the product over $k=1,\ldots,2n-1$ splits into the middle term $k=n$ (giving $\sin\frac\pi2=1$) and $n-1$ repeated pairs:

$$ \prod_{k=1}^{2n-1}\sin\frac{k\pi}{2n}=\left(\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}\right)^2\cdot1 $$

Step 4 · Solve for the half-product

$$ \left(\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}\right)^2=\frac n{4^{n-1}} $$

and since every factor $\sin\frac{k\pi}{2n}$ with $1\le k\le n-1$ satisfies $0<\frac{k\pi}{2n}<\frac\pi2$, all factors are positive, so we may take the positive square root:

$$ \boxed{\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}=\frac{\sqrt n}{2^{n-1}}} $$

Part $2^{\circ}$ · $\displaystyle\prod_{k=1}^{n}\cos\frac{k\pi}{2n+1}=\frac1{2^n}$

Step 1 · Substitute $x=-1$ into the odd factorization

$$ x^{2n+1}-1=(x-1)\prod_{k=1}^{n}\left(x^2-2x\cos\frac{2k\pi}{2n+1}+1\right) $$

At $x=-1$: the left side is $(-1)^{2n+1}-1=-2$, and $x-1=-2$, so:

$$ -2=-2\prod_{k=1}^{n}\left(1+2\cos\frac{2k\pi}{2n+1}+1\right)=-2\prod_{k=1}^{n}2\left(1+\cos\frac{2k\pi}{2n+1}\right) $$

Step 2 · Solve for the product of $(1+\cos)$ terms

$$ \prod_{k=1}^{n}\left(1+\cos\frac{2k\pi}{2n+1}\right)=\frac1{2^n} $$

Step 3 · Convert to a cosine using the half-angle identity

Using $1+\cos2\theta=2\cos^2\theta$ with $\theta=\dfrac{k\pi}{2n+1}$:

$$ \prod_{k=1}^{n}2\cos^2\left(\frac{k\pi}{2n+1}\right)=\frac1{2^n} \qquad\Longrightarrow\qquad 2^n\prod_{k=1}^n\cos^2\left(\frac{k\pi}{2n+1}\right)=\frac1{2^n} $$

$$ \prod_{k=1}^n\cos^2\left(\frac{k\pi}{2n+1}\right)=\frac1{4^n} $$

Step 4 · Take the square root

For $1\le k\le n$, the angle $\dfrac{k\pi}{2n+1}<\dfrac{n\pi}{2n+1}<\dfrac\pi2$, so every factor $\cos\frac{k\pi}{2n+1}$ is positive, and the square root is taken without a sign ambiguity:

$$ \boxed{\prod_{k=1}^n\cos\left(\frac{k\pi}{2n+1}\right)=\frac1{2^n}} $$

This holds for every positive integer $n$ — there is no alternating sign or parity restriction on $n$.

Verification

Both formulas were checked numerically for $n=1,\ldots,7$, to full floating-point precision.