<s>[INST] 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? 
 [/INST] 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 $\theta$.
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