Q49 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

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
    $\sin (6 a+8 b)$
  • Correct answer
    D
    $\cos (6 a+8 b)$

Solution

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 $=\cos (6 a+8 b)$.

Hence, the correct answer is option: 'D'.

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