A Product of Many Sinuses // Birçok Sinüsün Ürünü

          I will look for some proofs of the trigonometric identity

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

          Solution 1[based on Marcel ȚENA's book "Rădăcinile Unității" (The Roots 

                         of Unity) from the solution on pages 112-113 of Exercise #6 (page 16)]

           We consider the polynomial $X^n-1$ whose roots are $\zeta=\cos \frac{2\pi}{n}+\imath \sin \frac{2\pi}{n}$

$\zeta ^2,\;\zeta^3,\;\dots \zeta^{n-1},\;\zeta^n=1.$ We have equality

$$X^n-1=\prod \limits_{k=0}^{k=n-1}(X-\zeta^k) \tag{1.1}$$

from which, simplifying by the factor $X-1$, it results

$$X^{n-1}+X^{n-2}+\dots+X+1=\prod \limits_{k=1}^{k=n-1}(X-\zeta^k).$$

     For $X=1$ we obtain

$$n=\prod \limits_{k=1}^{k=n-1}(1-\zeta^k). \tag{1.2}$$

But, using the trigonometric formulas

$$1-\cos 2\alpha=2\sin^2 \alpha,\;\sin 2\alpha =2\sin \alpha \cdot \cos \alpha$$

 we have $1-\zeta^k=1-\cos \frac{2k\pi}{n}-\imath\sin \frac{2k\pi}{n}=-2\imath\sin \frac{k\pi}{n}\left(\cos\frac{k\pi}{n}+\imath \sin \frac{k\pi}{n}\right )$

and then (1.2) is written 

$n=2^{n-1} \cdot \imath^{n-1} \cdot \prod \limits_{k=1}^{k=n-1}\sin \frac{k\pi}{n} \cdot \prod \limits_{k=1}^{k=n-1}\left ( \cos \frac{k\pi}{n}+\imath \sin \frac{k\pi}{n}\right ).$

But then the moduli of the two members are also equal, so

$$n=2^{n-1} \cdot 1^{n-1} \cdot \prod \limits_{k=1}^{k=n-1} \sin \frac{k\pi}{n} \cdot \prod \limits_{k=1}^{k=n-1}1$$

from which the relation (1) immediately follows.

$\blacksquare$

          Solution 2 

           From the identity$^{\color{Red}{citation\; needed}}$  $\prod\limits_{k=0}^{k=n-1} \sin \left ( x+\frac{k\pi}{n}\right )=\frac{\sin nx}{2^{n-1}}$  we obtain, passing the first factor to the right-hand side

$\sin \left (x+\frac{\pi}{n} \right ) \cdot \sin \left ( x+\frac{2\pi}{n} \right ) \cdot _{\dots} \cdot \sin \left ( x+\frac {(n-1)\pi}{n} \right )=\frac{1}{2^{n-1}} \cdot \frac{\sin nx}{\sin x} \tag{2.1}$

Going in (2.1) to the limit when $x \rightarrow 0$, in the left-hand side we obtain the product that appears on the left at (1), and in the right-hand side we consider that 

$$\lim \limits_{x \to 0} \frac{\sin nx}{\sin x}=\lim \limits_{x \to 0} \frac{\sin nx}{nx} \cdot \lim \limits_{x \to 0} \frac{x}{\sin x} \cdot n=1 \cdot 1 \cdot n$$

and we obtain the value on the right of (1)

$\blacksquare$