Prove the relation for cos inverse Prove the relation $\cos^{-1}x_0=\dfrac{\sqrt {1-x^2_0}}{x_1\cdot x_2\cdot x_3\cdots \text{ ad inf.}}$ where the successive quantities $x_r$ are connected by the relation $x_{r+1}=\sqrt{\frac{1}{2}(1+x_r)}$
My attempt:
$$x_1=\sqrt{\frac{1}{2}(1+x_0)}$$
$$x_2=\sqrt{\frac{1}{2}(1+x_1)}$$
$$x_3=\sqrt{\frac{1}{2}(1+x_2)}$$
$$x_4=\sqrt{\frac{1}{2}(1+x_3)}$$
Multiplying all these, we get
$$x_1\cdot x_2\cdot x_3\cdots\text{ ad inf.}=\sqrt{\frac{1}{2}(1+x_0)\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf.}}$$
Putting in equation,
$$\cos^{-1}x_0=\frac{\sqrt {1-x^2_0}}{\sqrt{\frac{1}{2}(1+x_0)\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf.}}}$$
$$\cos^{-1}x_0=\frac{\sqrt {1-x_0}}{\sqrt{\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf}}}$$
but i could not solve further. Can someone guide me in this question? 
 A: Let $x_0:=\cos y$. Want to prove that $y = RHS$. Also we know, that in case of convergence, $x_r \to 1$ because $\bar x = \sqrt{\frac12(1+\bar x)} \Rightarrow \bar x^2 - \frac12 \bar x - \frac12 = 0$ and $\bar x \ge 0$. The roots of the quadratic are $1$ and $-\frac12$.  
We now need to show that
$$y = \frac{\sqrt{1-\cos^2 y}}{\prod_{r=1}^\infty x_r}$$
By restricting $y$ to $[0, \pi]$ we can thus write
$$y = \frac{\sin y}{\prod_{r=1}^\infty x_r}$$
Or equivalently
$$\frac{\sin y}y = \prod_{r=1}^\infty x_r$$
This identity might be related to some other product identity for the cardinal sine. Specifically looking at
$$\frac{\sin y}y = \prod_{k=1}^\infty \cos \left(\frac y{2^k}\right)$$
looks promising.
In fact we can prove that
$$x_r = \cos\left(\frac y{2^k}\right)$$
by induction.
Note that
$$\frac12(1+\cos x) = \frac12 (1+\cos(\frac x2 + \frac x2)) = \frac12(1 + \cos^2 \frac x2 - \sin^2 \frac x2) = \frac12 (2\cos^2 \frac x2) = \cos^2 \frac x2$$
Thus
$$x_1 = \sqrt{\frac12(1+\cos y)} = \cos \frac y2$$
Now by induction
$$x_r = \sqrt{\frac12(1+\cos \frac y{2^{r-1}})} = \cos \frac y{2^r}$$
A: Let $\cos(\theta)=x_0$, so that $\sqrt{1-x_0^2}=\sin(\theta)$.
From known trigonometry, we have
$$x_{r+1}=\cos\left(\frac{\arccos(x_r)}2\right),$$
and by recurrence
$$x_r=\cos\left(\frac{\arccos(x_0)}{2^r}\right)=\cos\left(\frac{\theta}{2^r}\right).$$
Hence the denominator is the infinite product
$$\prod_1^\infty\cos\left(\frac\theta{2^r}\right)=\text{sinc}(\theta),$$
and the ratio is $\theta$.
