Question

$x_n=\cos \frac{\pi }{2^n}+i\sin \frac{\pi }{2^n}$ then $x_1{x}_2{x}_3{x}_4...........x_{\infty }=$

• A. $1$
• B. $-1$
• C. $0$
• D. $Noneofthese$

Answer 1

The correct option is B $-1$

A] $x_n=cos\left(\frac{x}{2^n}\right)+isin\left(\frac{\pi }{2^n}\right)=e^{ix/2n}$

$\therefore x_1=e^{i\pi /2}{x}_2=e^{i\pi 1}$ etc ...

Let $y=e^{i\pi /2}{e}^{i\pi /4}{e}^{i\pi /8}$ ...

Tacking ln ,

$lny=ln{e}^{i\pi /2}+ln{e}^{ix/2^2}$ + . . . .

$lny=i\frac{\pi }{2}+i\frac{\pi }{2^2}$ + . . . .

$lny=i[\frac{\pi }{2}+\frac{\pi }{2^2}+....]$

$lny=i\left[\frac{\pi }{1-\pi /2}\right]$

$y=e^{i\pi }=cos\pi +isin\pi$

$\therefore y=-1$