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If $x_n=\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}$, then $\prod_{n=1}^{\infty} x_n$ is equal to
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$-1$
$\begin{aligned} & \text { Given that, }=\frac{9+6 i}{13} \\ & \qquad \begin{aligned} X_n & =\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}=\operatorname{cis} \frac{\pi}{2^n} \\ \therefore \quad \prod_{n=1}^{\infty} X_n & =\prod_{n=1}^{\infty} \operatorname{cis} \frac{\pi}{2^n} \\ & =\operatorname{cis} \frac{\pi}{2} \cdot \operatorname{cis} \frac{\pi}{2^2} \cdot \operatorname{cis} \frac{\pi}{2^3} \ldots . \\ & =\operatorname{cis}\left(\frac{\pi}{2}+\frac{\pi}{2^2}+\frac{\pi}{2^3}+\ldots . .\right) \\ & =\operatorname{cis}\left(\frac{\frac{\pi}{2}}{1-\frac{1}{2}}\right)=\operatorname{cis} \pi \\ & =-1\end{aligned}\end{aligned}$
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