# Yagna's Mathematics Blog

## A New Trigonometry Angle Identity

18 November, 2015

Proposition : $$(2n+1)\sin(\theta) \prod_{k=1}^n \left(1 -\frac{\sin^2(\theta)}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right) = \sin((2n+1)\theta)$$

Lemma 1: $$\prod_{r=1}^{n-1}\sin\left(\frac{r\pi}{n}\right)=\frac{n}{2^{n-1}}$$

Since, $$\sin(x) = \frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\displaystyle\Rightarrow\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \left(\frac{1}{2i}\right)^{n-1}\prod_{k=1}^{n-1} \left(e^{\frac{k\pi i}{n}} - e^{\frac{-k\pi i}{n}}\right)\displaystyle= \left(\frac{1}{2i}\right)^{n-1} \ \left(\prod_{k=1}^{n-1} e^{\frac{k\pi i}{n}} \right) \prod_{k=1}^{n-1} \left(1-e^{\frac{-2k\pi i}{n}} \right)$$ $$\displaystyle= \left(\frac{1}{2i}\right)^{n-1} \times i^{n-1} \prod_{k=1}^{n-1} \left(1-e^{\frac{-2k\pi i}{n}} \right)\displaystyle= \frac{1}{2^{n-1}} \prod_{k=1}^{n-1} \left(1-e^{\frac{-2k\pi i}{n}} \right)$$ Now, consider the following factorization, $$x^{n-1}+x^{n-2}+ \ldots + x + 1 = \prod_{k=1}^{n-1}\left(x-e^{\frac{-2\pi k}{n}}\right)$$ Substituting $x=1$ in this above identity: $$\displaystyle \prod_{k=1}^{n-1} \left(1-e^{\frac{-2k\pi i}{n}} \right)=n \displaystyle \Rightarrow \prod_{r=1}^{n-1}\sin\left(\frac{r\pi}{n}\right)=\frac{n}{2^{n-1}}$$ Now, consider $$\text{S}=\sum_{r=1}^{n}\log \left(\Gamma{\left(\frac{r}{n+1}\right)}\right)\displaystyle = \sum_{r=1}^{n}\log \left(\Gamma{\left(\frac{n+1-r}{n+1}\right)}\right) \displaystyle = \sum_{r=1}^{n}\log \left(\Gamma{\left(1-\frac{r}{n+1}\right)}\right)$$ Using Euler's Reflection Formula, $$\displaystyle \text{S} = \sum_{r=1}^{n}\log \left(\frac{\pi}{\Gamma{\left(\frac{r}{n+1}\right)}\sin \left(\frac{r\pi}{n+1}\right)}\right)\displaystyle =\log ({\pi}^n) - \sum_{r=1}^{n}\log\left(\Gamma{\left(\frac{r}{n+1}\right)}\right) - \sum_{r=1}^{n} \log \left(\sin \left(\frac{r\pi}{n+1}\right)\right)$$ $$\displaystyle \Rightarrow 2\text{S} = \log({\pi}^n) - \sum_{r=1}^{n} \log \left(\sin \left(\frac{r\pi}{n+1}\right)\right)$$ Using the Lemma, we have, $$2\text{S}=\log({\pi}^n) - \log\left(\frac{n+1}{2^n}\right)\displaystyle \Rightarrow \text{S} = \log \left(\sqrt{\frac{(2\pi)^n}{n+1}}\right)\displaystyle \Rightarrow \sum_{r=1}^{n}\log \left(\Gamma{\left(\frac{r}{n+1}\right)}\right) = \log \left(\sqrt{\frac{(2\pi)^n}{n+1}}\right)$$ $$\displaystyle \Rightarrow \log\left(\prod_{r=1}^{n}\Gamma{\left(\frac{r}{n+1}\right)}\right) = \log \left(\sqrt{\frac{(2\pi)^n}{n+1}}\right){\Rightarrow \displaystyle \prod_{r=1}^{n}\Gamma{\left(\frac{r}{n+1}\right)}=\sqrt{\frac{(2\pi)^n}{n+1}}}$$

Lemma 2: $$\prod_{k=1}^{n} \sin^2 \left(\frac{k\pi}{2n+1}\right) = \frac{2n+1}{2^{2n}}$$

$$\displaystyle \prod_{k=1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right) = \prod_{k=1}^{n} \sin \left(\frac{k\pi}{2n+1}\right) \cdot \prod_{k=n+1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right) \displaystyle = \prod_{k=1}^{n} \sin \left(\frac{k\pi}{2n+1}\right) \cdot \prod_{k=1}^{n} \sin \left(\frac{(n+k)\pi}{2n+1}\right)$$ $$\displaystyle = \prod_{k=1}^{n} \sin \left(\frac{k\pi}{2n+1}\right) \cdot \prod_{k=1}^{n} \sin \left(\frac{k\pi}{2n+1}\right) \displaystyle = \prod_{k=1}^{n} \sin^2 \left(\frac{k\pi}{2n+1}\right)$$ Using Lemma 1, we can conclude the following: $$\displaystyle \prod_{k=1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right) = \frac{2n+1}{2^{2n}} \displaystyle \Rightarrow \prod_{k=1}^{n} \sin^2 \left(\frac{k\pi}{2n+1}\right) = \frac{2n+1}{2^{2n}}$$

Now, let $$\displaystyle P = (2n+1)\sin(\theta) \prod_{k=1}^n \left(1 -\frac{\sin^2(\theta)}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right) \displaystyle = (2n+1)\sin(\theta) \left(\frac{ \displaystyle \prod_{k=1}^n \left( \sin^2\left(\frac{k\pi}{2n+1}\right) - \sin^2(\theta) \right)}{ \displaystyle \prod_{k=1}^{n} \sin^2\left(\frac{k\pi}{2n+1}\right) }\right)$$ $$\displaystyle = 2^{2n} \sin(\theta) \prod_{k=1}^n \left( \cos^2(\theta) - \cos^2\left(\frac{k\pi}{2n+1}\right) \right)$$ $$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) + \cos \left(\frac{k\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right)$$ $$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) - \cos \left(\frac{(2n+1 - k)\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) \$$ $$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) - \cos \left(\frac{(n + k)\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) \$$ $$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=n+1}^{2n} \left( \cos (\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{ k \pi}{2n+1}\right) \right) \right)$$ $$\displaystyle = 2^{2n} \sin(\theta) \prod_{k=1}^{2n} \left( \cos (\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right)$$ Also, $U_{n} (x) = 2^{n} \prod_{k=1}^{n} \left(x - \cos \left(\frac{k\pi}{n+1}\right) \right)$ where $\displaystyle U_{n} (x)$ denotes the Chebyshev Polynomial of the Second kind. So from this we have the following: $$P = 2^{2n} \sin(\theta) \cdot 2^{-2n} \cdot U_{2n} (\cos \theta) \displaystyle = \sin ((2n+1) \theta)$$

Why is this identity important? We can essentially use this identity to prove Euler's infinite product for $\frac{\sin x}{x}$. In the above proposition, all we do is set $(2n+1)\theta = x$ s.t. $x$ is a constant. Doing so, we will get: $$\sin x = (2n+1)\sin \left( \frac{x}{2n+1} \right) \prod_{k=1}^n \left(1 -\frac{\sin^2 \left( \frac{x}{2n+1} \right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right)$$ Taking $\lim_{n \to \infty}$ and noting that $x$ is a constant, we get, $$\displaystyle \frac{\sin{x}}{x} = \prod_{k=1}^{\infty} \left( 1 - \frac{x^2}{(k\pi)^{2}}\right)$$

tags: math.AG, math.GM, Geometry, Trigonometry-Identities