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Q.
For $n \in N$, If $A _{n}=\cos \left(\frac{\pi}{2^{n}}\right)+i \sin \left(\frac{\pi}{2^{n}}\right)$, then $\left( A _{1} A _{2} A _{3} A _{4}\right)^{4}=$
TS EAMCET 2020
Solution:
It is given that,
$A_{n}=\cos \left(\frac{\pi}{2^{n}}\right)+i \sin \left(\frac{\pi}{2^{n}}\right), n \in N=e^{i\left(\frac{\pi}{2^{n}}\right)}$
$\therefore \left(A_{1} A_{2} A_{3} A_{4}\right)^{4}=e^{i \pi\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)^{4}}$
$=e^{i \frac{15 \pi}{4}}=e^{i\left(4 \pi-\frac{\pi}{4}\right)}$
$=\cos \left(4 \pi-\frac{\pi}{4}\right)+i \sin \left(4 \pi-\frac{\pi}{4}\right)$
$=\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}=\frac{1-i}{\sqrt{2}}$