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Complex Numbers

Section: Mathematics

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Question : 47 of 50

Marks: +1, -0

Real part of

\frac{(\cos a + i \sin a)^6}{(\sin b + i \cos b)^8}

[AP EAPCET 19 May 2024 Shift 2]

$\sin(6a-8b)$
$\cos(6a-8b)$
$\sin(6a+8b)$
$\cos(6a+8b)$

Solution:

\text{We have,}\; \frac{(\cos a + i \sin a)^6}{(\sin b + i \cos b)^8}

= \frac{(\cos a + i \sin a)^6}{(-i^2 \sin b + i \cos b)^8} \;\; [\because 1=-i^2]

= \frac{(\cos a + i \sin a)^6}{i^8 (\cos b - i \sin b)^8}

= \frac{(\cos a + i \sin a)^6}{(\cos b - i \sin b)^8}

= \frac{\cos 6a + i \sin 6a}{\cos 8b - i \sin 8b}

= \frac{(\cos 6a + i \sin 6a)(\cos 8b + i \sin 8b)}{(\cos 8b - i \sin 8b)(\cos 8b + i \sin 8b)}

\therefore \text{Required real part}

= \frac{\cos 6a \cos^{8b- \sin^8 6a \sin 8b}}{\cos^2 8b + \sin^2 8b}

= \cos(6a+8b)

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