AP EAMCET · Maths · Complex Number
Real part of \(\frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}\) is
- A \(\sin (6 a-8 b)\)
- B \(\cos (6 a-8 b)\)
- C \(\sin (6 a+8 b)\)
- D \(\cos (6 a+8 b)\)
Answer & Solution
Correct Answer
(D) \(\cos (6 a+8 b)\)
Step-by-step Solution
Detailed explanation
Since, \(\frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}=\frac{\left(e^{i a}\right)^6}{(\cos b-i \sin b)^8}\) \(=\frac{e^{i 6 a}}{\left(e^{-i b}\right)^8}=e^{i(6 a+8 b)}=\cos (6 a+8 b)+i \sin (6 a+8 b)\) So, real part \(=\cos (6 a+8 b)\).
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