Q51 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

If $x_n=\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}$, then $\prod_{n=1}^{\infty} x_n=$

  • A
    $0$
  • B
    $1$
  • Correct answer
    C
    $-1$
  • D
    i

Solution

$x_n=\left(\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}\right)$ $\begin{aligned} & =(\cos \pi+i \sin \pi)^{\frac{1}{2} n} \\ & \prod_{n=1}^{\infty} x_n=(\cos \pi+i \sin \pi)^{\frac{1}{2}} \cdot(\cos \pi+i \sin \pi)^{\frac{1}{2^2}} \ldots \infty \\ & =(\cos \pi+i \sin \pi)^{\frac{\frac{1}{2}}{1-\frac{1}{2}}} \\ & =(\cos \pi+i \sin \pi)^1=-1\end{aligned}$

Hence, the correct answer is option: 'C'.

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