# Find the value of $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$ [duplicate]

Find the value of $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$

$$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$ $$=\sin\frac{2\pi}{14}\sin\frac{4\pi}{14}\sin\frac{6\pi}{14}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$ $$=\sin\frac{\pi}{14}\sin\frac{2\pi}{14}\sin\frac{3\pi}{14}\sin\frac{4\pi}{14}\sin\frac{5\pi}{14}\sin\frac{6\pi}{14}$$

How can I compute $$\prod\limits_{r=1}^{n}\sin\frac{r\pi}{c}$$

## marked as duplicate by Jyrki LahtonenMar 3 '16 at 5:47

• If $c=2N$ and $n=N-1$ then from the Chebyshev polynomials you get the formula $$\prod_{r=1}^{N-1}\sin\frac{r\pi}{2N}=\frac{\sqrt{N}}{2^{N-1}}.$$ – Jyrki Lahtonen Mar 2 '16 at 17:36

It can be seen as an instance of the well-known identity: $$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{2n}{2^n} \tag{1}$$ that for $n=14$ gives: $$\left[\prod_{k=1}^{6}\sin\left(\frac{k\pi}{14}\right)\right]^2 = \frac{14}{2^{13}} \tag{2}$$ from which: $$\prod_{k=1}^{6}\sin\left(\frac{k\pi}{14}\right) = \color{red}{\frac{\sqrt{7}}{64}}.\tag{3}$$

By the basic properties of Chebyshev polynomials the numbers $x_k=\sin(k\pi/14)$, $k=-6,-5,\ldots,5,6, k\neq0$ are the twelve zeros of the polynomial $$\frac{U_{13}(x)}x=8192 x^{12}-24576 x^{10}+28160 x^8-15360 x^6+4032 x^4-448 x^2+14.$$

By the usual relations (aka Vieta relations) of the zeros of a polynomial and its coefficients it follows that $$A=\prod_{k=-6..6, k\neq0}x_k=\frac{14}{8192}=\frac{7}{2^{12}}.$$

Because sine is known to be an odd function, the products $N=\prod_{k=-6}^{-1}x_k$ and $P=\prod_{k=1}^6x_k$ are equal. You are asking about the product $P$. It is clearly a positive number, so from the equation $$A=NP=P^2=\frac{7}{2^{12}}$$ we can solve $$P=\frac{\sqrt7}{2^6}.$$

The method generalizes for other products of sines as long as you take the product of ALL the numbers $\sin (k\pi/(2N))$, $k=1,2,\ldots,N-1$.

Use $\sin\left(\dfrac\pi2-A\right)=\cos A$ and $\sin2B=2\sin B\cos B$

to find the product $$=\dfrac{\sin2t\sin4t\sin6t}8$$ where $7t=\pi$

Now use Method to find $\sin (2\pi/7)$, to find

$$\sin2t\sin4t\sin6t=\sqrt{\dfrac7{64}}$$
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