Q94 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics
$\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right) \times\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$ $\times\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right) \times \ldots \ldots \ldots \infty=$
- A1
- B0
- Correct answerC–1
- D2
Solution
$$
\begin{aligned}
& \text { }\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \\
& \left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right) \ldots \infty \\
& e^{i \frac{\pi}{2}} \cdot e^{i \frac{\pi}{4}} \cdot e^{i \frac{\pi}{8}} \ldots \infty \\
& e^{i(\pi / 2+\pi / 4+\pi / 8+\ldots \ldots \infty)} \\
& =e^{i\left(\frac{\frac{\pi}{2}}{1-\frac{1}{2}}\right)}\left[\begin{array}{l}
\because \frac{\pi}{2}+\frac{\pi}{4}+\frac{\pi}{8}+\ldots \infty \text { in GP } \\
\therefore a=\frac{\pi}{4}, r=\frac{1}{2} \\
S_{\infty}=\frac{a}{1-r}
\end{array}\right] \\
& =e^{i(\pi)}=\cos \pi+i \sin \pi=-1+i .0=-1 \\
&
\end{aligned}
$$
Hence, option (3) is correct.
Hence, the correct answer is option: 'C'.