The real part of frac{(cos a+i sin a)^6}{(sin | vedclass

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Question

(D) Given expression: $Z = \frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$ Using De Moivre's theorem,the numerator is $(\cos a+i \sin a)^6 = \cos(6a)

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$\cos (6 a+8 b)$

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(D) Given expression: $Z = \frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$ Using De Moivre's theorem,the numerator is $(\cos a+i \sin a)^6 = \cos(6a) + i \sin(6a) = e^{i6a}$. For the denominator,note that $\sin b + i \cos b = i(\cos b - i \sin b) = i e^{-ib}$. Thus,$(\sin b + i \cos b)^8 = i^8 (e^{-ib})^8 = 1 \cdot e^{-i8b} = e^{-i8b}$. Substituting these back: $Z = \frac{e^{i6a}}{e^{-i8b}} = e^{i(6a+8b)}$. Expanding using Euler's formula: $Z = \cos(6a+8b) + i \sin(6a+8b)$. The real part is $\cos(6a+8b)$.

The real part of $\frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$ is

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