AP EAMCET · Maths · Complex Number
\(\frac{\left(\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}\right)^8}{\left(\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}\right)^8}\) is equal to
- A \(i\)
- B \(-i\)
- C 1
- D 2
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
\text { } \begin{aligned} & \frac{\left(\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}\right)^8}{\left(\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}\right)^8} \\ & =\frac{i^8\left(\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}\right)^8}{(-i)^8\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right)^8}…
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