Show that for $\displaystyle 0<\theta<\pi$,
$\displaystyle \theta = \frac{\sin\theta}{\cos\left(\frac{\theta}{2}\right )\cos\left(\frac{\theta}{4}\right)\cos\left(\frac{ \theta}{8}\right)\cos\left(\frac{\theta}{16}\right )...}$
Show that for $\displaystyle 0<\theta<\pi$,
$\displaystyle \theta = \frac{\sin\theta}{\cos\left(\frac{\theta}{2}\right )\cos\left(\frac{\theta}{4}\right)\cos\left(\frac{ \theta}{8}\right)\cos\left(\frac{\theta}{16}\right )...}$
Let [tex] T_n = \cos(\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^n)[/maTH]
$\displaystyle = \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)) $
$\displaystyle = \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})) $
$\displaystyle = \frac{ T_{n-1} \sin(\theta/2^{n-1})}{2 \sin(\theta/2^n)}$
when n tend to infinity
$\displaystyle T_n = \frac{T_1 \sin(\theta/2)}{2^{n-1} \sin(\theta/2^n)}$
For n is very large $\displaystyle \sin(\theta/2^n) = \theta/2^n$
$\displaystyle T_n = 2 \sin(\theta/2)\cos(\theta/2) /\theta =\sin(\theta)/\theta$
Here is my solution : consider the function $\displaystyle f(\theta)=\theta\cos\left(\frac{\theta}{2}\right)\ cos\left(\frac{\theta}{4}\right)\cos\left(\frac{\t heta}{8}\right)\cos\left(\frac{\theta}{16}\right). ..$. Its zeroes are $\displaystyle \theta = 0 $ and $\displaystyle \theta=2^n(k\pi+\pi/2)$ for $\displaystyle k\in \mathbb{Z}$, $\displaystyle n\in \mathbb{N^+}$ Thus its zeroes are all $\displaystyle k\pi$, $\displaystyle k \in \mathbb{Z}$, which are the zeroes of $\displaystyle \sin\theta$.
Moreover we know $\displaystyle f(\theta)$ must be an entire function because it is a product of entire functions. By the identity theorem for entire functions, we must have $\displaystyle f(\theta)=m\sin\theta$ for some $\displaystyle m \in \mathbb{C}$. If we consider the coefficient of $\displaystyle \theta$ in the series expansion of $\displaystyle f(\theta)$ about $\displaystyle \theta=0$, we see that it is equal to $\displaystyle \sum_{j=1}^\infty\frac{1}{2^j}=1$, and on the left side as well. Hence we must have $\displaystyle k=1$ and $\displaystyle f(\theta)=\sin\theta$.