Alternate proof for Viète's infinite product of nested radicals I am looking for alternate proof for Viete's infinite product of nested radicals. (Reference - Wikipedia)
Basically we need to find $\lim_{n\to \infty}\prod_{k=1}^{n} T_k$ where $$T_{k+1} = \sqrt{\left(\frac{T_k + 1}{2}\right)}$$ and $T_1 = \sqrt{\frac{1}{2}}$. Series looks like 
$$\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}...$$
Miss gave a solution treating $\cos(\theta) = \frac{1}{\sqrt2}$ that is $\theta = 45^\circ$. The series result is given easily using the identity $\cos(\theta) + 1 = 2 \cos^2(\theta/2)$ and using $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. The final result is $\frac{\sin(2\theta)}{2\theta} = \frac{2}{\pi}$.
I look for alternate ways to get to this! I am open to calculus methods.
 A: 
Answer: we have $$\lim_{n\rightarrow\infty}\prod_{k=1}^{n} T_k =\lim_{n\rightarrow\infty}\prod_{k=1}^{n} \cos\left(\frac{\pi}{2^{k+1}}\right) =\lim_{n\rightarrow\infty}\frac{\sin (\pi/2)}{2^{n}\sin\left(\frac{\sqrt{2}}{2^{n}}\right)} = \color{blue}{\frac{\sin (\pi/2)}{\pi/2}}$$
  



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*First check that for all $k$ we have $0\le T_k\le 1$ this is obvious by induction since $$0\le T_1= \frac{\sqrt{2}}{2}\le 1$$

*Hence there exists $a_k \in [0,\frac{\pi}{2} ]$ such that $$T_k = \cos a_k, $$

*Easily, $T_1 =\frac{\sqrt{2}}{2}\implies a_1 =\frac{\pi}{4}$ and $$\cos (a_{k+1})= T_{k+1} =  \sqrt{\left(\frac{T_k + 1}{2}\right)}= \sqrt{\left(\frac{\cos a_k + 1}{2}\right)} = \cos\left(\frac{ a_k }{2}\right)$$

*Since $x\mapsto \cos x$ realize a bijection in $[0,\frac{\pi}{2} ],$ It turn out that $a_k$ is geometric sequence with ratio $1/2$
that is we have $$\color{red}{a_{k+1} = \frac{ a_k }{2}\implies a_k = \frac{a_1}{2^{k-1}}=\frac{\pi}{2^{k+1}}}$$

*Therefore, we have 
$$\color{red}{T_{k} = \cos\left(\frac{\pi}{2^{k+1}}\right)}$$
6.By double angle formula we have $$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)=4\sin\left(\frac{x}{4}\right)\cos\left(\frac{x}{4}\right)\cos\left(\frac{x}{2}\right)\\=\dots=2^{n}\sin\left(\frac{x}{2^{n}}\right)\prod_{k\leq n}\cos\left(\frac{x}{2^{k}}\right)$$ now remains to note that $$\lim_{n\rightarrow\infty}2^{n}\sin\left(\frac{x}{2^{n}}\right)=  \lim_{n\rightarrow\infty} x\frac{\sin\left(\frac{x}{2^{n}}\right)}{\frac{x}{2^{n}}}=\lim_{h\rightarrow 0} x\frac{\sin h}{h}= x .$$

*Thus $$\lim_{n\rightarrow\infty}\prod_{k=1}^{n} T_k =\lim_{n\rightarrow\infty}\prod_{k=1}^{n} \cos\left(\frac{\pi}{2^{k+1}}\right) =\lim_{n\rightarrow\infty}\frac{\sin (\pi/2)}{2^{n}\sin\left(\frac{\pi}{2^{n}}\right)} = \color{blue}{\frac{\sin (\pi/2)}{\pi/2}}$$
