Question

If $P_{n+1}=\sqrt{\frac{1}{2}(1+p_n)}$ then cos $\left(\frac{\sqrt{1-P_0^2}}{p_1{p}_2{p}_3.......to\infty }\right)=p_0$

Answer 1

Putting n = 0,1,2,3,.... in the given relation, we get
$P_1=\sqrt{\frac{1}{2}(1+p_0)}=\sqrt{\frac{1}{2}(1+cos\theta )}$
= cos $\frac{\theta }{2}$ where $p_0$ = cos $\theta$
$P_2=\sqrt{\frac{1}{2}(1+p_1)}=\sqrt{\frac{1}{2}(1+cos\frac{\theta }{2})}$
$\therefore P_3=cos\frac{\theta }{2^3},......,P_n=cos\frac{\theta }{2^n}$
$\therefore P_1{P}_2{P}_3......\infty ={lim}_{n\to \infty }{P}_1{P}_2....P_n$
$={lim}_{n\to \infty }cos\frac{\theta }{2^2}.cos\frac{\theta }{2^2}...cos\frac{\theta }{2^n}$
Choose $\frac{\theta }{2^n}=A,\therefore 2^{n}A=\theta$
as $n\to \infty ,A\to 0$
$\therefore {lim}_{a\to 0}$ cos A cos 2A Cos $2^2$ A...n terms
${lim}_{a\to 0}\frac{sin{2}^n\left(A\right)}{2^{n}sinA}={lim}_{a\to 0}\frac{sin\theta }{2^{n}A}=\frac{sin\theta }{\theta }$
Also $\sqrt{1-P_2^0}=\sqrt{1-co{s}^2\theta }=sin\theta$
$\therefore cos\left(\frac{\sqrt{1-P_0^2}}{P_1{P}_2{P}_3...\infty }\right)=cos\left[\frac{sin\theta }{\left(sin\theta \right)/\theta }\right]=cos\theta =P_0$