Question

State whether the given statement is true or false

If $x_{n+1}=\sqrt{\frac{1+x_n}{2}}$, then $\cos \left(\frac{\sqrt{1-{x_0}^2}}{x_1{x}_2{x}_3...}\right)(-1<x_0<1)$ is equal to $x_0$

• A. True
• B. False

Answer 1

The correct option is A True

Let $x_0=\cos \theta$, then
$x_1=\sqrt{\frac{1}{2}(1+\cos \theta )}=\cos \frac{\theta }{2},x_2=\cos \left(\frac{\theta }{2^2}\right),x_3=\cos \left(\frac{\theta }{2^3}\right)$ ...and so on
$\therefore \left[\frac{\sqrt{1-{x_0}^2}}{x_1{x}_2{x}_3...}\right]=\frac{\sin \theta }{\cos \frac{\theta }{2}\cos \left(\frac{\theta }{2^2}\right)...\cos \left(\frac{\theta }{2^n}\right)...}$
$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{\cos \frac{\theta }{2}\cos \left(\frac{\theta }{2^2}\right)...\cos \left(\frac{\theta }{2^n}\right)...}=\frac{2^2\sin \frac{\theta }{2^2}\cos \frac{\theta }{2^2}}{\cos \left(\frac{\theta }{2^2}\right)...\cos \left(\frac{\theta }{2^n}\right)...}$
$=\underset{n\to \infty }{lim}\frac{2^n\sin \frac{\theta }{2^n}}{\cos \frac{\theta }{2^{n+1}}}=\underset{n\to \infty }{lim}\theta \left(\frac{\sin \frac{\theta }{2^n}}{\frac{\theta }{2^n}}\right)\frac{1}{\cos \frac{\theta }{2^{n+1}}}=\theta$
$\therefore \cos \left(\frac{\sqrt{1-{x_0}^2}}{x_1{x}_2{x}_3...}\right)=\cos \theta =x_0$