# Thread: A nice identity of Viette and Euler

1. ## A nice identity of Viette and Euler

Show that for $0<\theta<\pi$,

$\theta = \frac{\sin\theta}{\cos\left(\frac{\theta}{2}\right )\cos\left(\frac{\theta}{4}\right)\cos\left(\frac{ \theta}{8}\right)\cos\left(\frac{\theta}{16}\right )...}$

2. Let $$T_n = \cos(\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^n)$$

$= \frac{1}{2\sin(\theta/2^n)}\cos(\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^n) (2\sin(\theta/2^n))$

$= \frac{1}{2\sin(\theta/2^n)}\cos(\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^{n-1}) \sin(\theta/2^{n-1}))$

$= \frac{ T_{n-1} \sin(\theta/2^{n-1})}{2 \sin(\theta/2^n)}$

when n tend to infinity

$T_n = \frac{T_1 \sin(\theta/2)}{2^{n-1} \sin(\theta/2^n)}$

For n is very large $\sin(\theta/2^n) = \theta/2^n$
$T_n = 2 \sin(\theta/2)\cos(\theta/2) /\theta =\sin(\theta)/\theta$

3. Here is my solution : consider the function $f(\theta)=\theta\cos\left(\frac{\theta}{2}\right)\ cos\left(\frac{\theta}{4}\right)\cos\left(\frac{\t heta}{8}\right)\cos\left(\frac{\theta}{16}\right). ..$. Its zeroes are $\theta = 0$ and $\theta=2^n(k\pi+\pi/2)$ for $k\in \mathbb{Z}$, $n\in \mathbb{N^+}$ Thus its zeroes are all $k\pi$, $k \in \mathbb{Z}$, which are the zeroes of $\sin\theta$.

Moreover we know $f(\theta)$ must be an entire function because it is a product of entire functions. By the identity theorem for entire functions, we must have $f(\theta)=m\sin\theta$ for some $m \in \mathbb{C}$. If we consider the coefficient of $\theta$ in the series expansion of $f(\theta)$ about $\theta=0$, we see that it is equal to $\sum_{j=1}^\infty\frac{1}{2^j}=1$, and on the left side as well. Hence we must have $k=1$ and $f(\theta)=\sin\theta$.

4. Originally Posted by Bruno J.
Here is my solution : consider the function $f(\theta)=\theta\cos\left(\frac{\theta}{2}\right)\ cos\left(\frac{\theta}{4}\right)\cos\left(\frac{\t heta}{8}\right)\cos\left(\frac{\theta}{16}\right). ..$. Its zeroes are $\theta = 0$ and $\theta=2^n(k\pi+\pi/2)$ for $k\in \mathbb{Z}$, $n\in \mathbb{N^+}$ Thus its zeroes are all $k\pi$, $k \in \mathbb{Z}$, which are the zeroes of $\sin\theta$.

Moreover we know $f(\theta)$ must be an entire function because it is a product of entire functions. By the identity theorem for entire functions, we must have $f(\theta)=m\sin\theta$ for some $m \in \mathbb{C}$. If we consider the coefficient of $\theta$ in the series expansion of $f(\theta)$ about $\theta=0$, we see that it is equal to $\sum_{j=1}^\infty\frac{1}{2^j}=1$, and on the left side as well. Hence we must have $k=1$ and $f(\theta)=\sin\theta$.

Perfect ! The important thing is the order of its zeros ( $f(\theta) = 0$ ) or the degree of all factors is zero , then $f(\theta) = m \sin(\theta)$