Question

If $X_r=\cos \frac{\pi }{2^r}+i\sin \frac{\pi }{2^r}$, prove that
$X_1{X}_2{X}_3....\infty$ = -1

Answer 1

$x_1.x_2.x_3\cdots$

$=\left[\cos \right(\pi /2)+i\sin (\pi /2\left)\right].\left[\cos \right(\pi /2^2)+i\sin (\pi /2^2\left)\right]$

$\left[\cos \right(\pi /2^3)+i\sin (\pi /2^3\left)\right]\cdots$

$=\cos (\pi /2+\pi /2^2+\pi /2^3+...)+i\sin (\pi /2+\pi /2^2+\pi /2^3+...)$

$=\cos \left(\frac{\pi /2}{1-1/2}\right)+i\sin \left(\frac{\pi /2}{1-1/2}\right)$

(Summing the infinite G.P)

$=\cos \pi +i\sin \pi =-1$

$[\because cos\pi =-1andsin\pi =0]$