By-products of PRODUCT // Ürün yan ürünleri

               Here we will demonstrate the formulas

$$\sin \frac{\pi}{2n+1}\cdot\sin \frac{2\pi}{2n+1}\cdot_{\dots} \cdot\sin \frac{n\pi}{2n+1}=\frac{\sqrt{2n+1}}{2^n} \tag{1}$$

$$\sin\frac{\pi}{2n}\cdot\sin \frac{2\pi}{2n}\cdot_{\dots}\cdot\sin \frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}} \tag{2}$$

              We will start from a previously demonstrated formula, written for $m\in \mathbb{N} \setminus \{0,\;1\}$ instead of $n$:

$$\sin \frac{\pi}{m}\cdot\sin \frac{2\pi}{m}\cdot_{\dots} \cdot \sin \frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}} \tag{3}$$

          Let's put in (3): $m$-odd, i.e. $m=2n+1,\;n\in \mathbb{N}\setminus\{0\}$. We will now write down a few more of the $m-1=2n$ factors of (3):

$\sin \frac{\pi}{2n+1} \cdot \sin \frac{2\pi}{2n+1}\cdot_{\dots} \cdot \sin \frac{n\pi}{2n+1}  \cdot$

$\cdot \sin \frac{(n+1)\pi}{2n+1} \cdot_{\dots}\cdot \sin \frac{(2n-1)\pi}{2n+1} \cdot \sin \frac{2n\pi}{2n+1}=\frac{2n+1}{2^{2n}} \tag{4}$

Considering the formula $\sin (\pi-\alpha)=\sin \alpha$ and noting that

$$\pi-\frac{2n\pi}{2n+1}=\frac{\pi}{2n+1} \quad \pi-\frac{(2n-1)\pi}{2n+1}=\frac{2\pi}{2n+1} \quad \dots \quad \pi-\frac{(n+1)\pi}{2n+1}=\frac{n\pi}{2n+1}$$

we see that in (4) the factors are two by two equal, one from each row. Then

$$\left (  \sin \frac{\pi}{2n+1} \cdot \sin \frac{2\pi}{2n+1} \cdot_{\dots} \cdot \sin \frac{n\pi}{2n+1}\right )^2=\frac{2n+1}{2^{2n}}$$

and taking the square root of both members results (1).

$\blacksquare$

          Let's put in (3) $m$-even, i.e. $m=2n,\;n\in \mathbb{N}\setminus \{0\}$. We now have

$\sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n}\cdot_{\dots} \cdot \sin \frac{(n-1)\pi}{2n} \cdot \overset{=1}{\overbrace{\sin \frac{n\pi}{2n}}} \cdot$

$\cdot \sin \frac{(n+1)\pi}{2n} \cdot_{\dots} \cdot \sin \frac{(2n-2)\pi}{2n} \cdot \sin \frac{(2n-1)\pi}{2n}=\frac{2n}{2^{2n-1}} \tag{5}$

Noting that 

$$\pi-\frac{(2n-1)\pi}{2n}=\frac{\pi}{2n}\quad \pi-\frac{(2n-2)\pi}{2n}=\frac{2\pi}{2n}\quad \dots \quad \pi-\frac{(n+1)\pi}{2n}=\frac{(n-1)\pi}{2n}$$

and considering again the formula  $\sin (\pi-\alpha)=\sin \alpha$ we have

$$\left ( \sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n}\cdot_{\dots} \cdot \sin \frac{(n-1)\pi}{2n}\right )^2=\frac{n}{2^{2n-2}}.$$

     Taking the square root of both members results (2).

$\blacksquare\;\blacksquare$