Express frac{(cos 2theta - isin 2theta)^4 (co | vedclass

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Question

(A) Using De Moivre's theorem,we know that $(\cos 	heta + i\sin 	heta)^n = \cos n	heta + i\sin n	heta$. Let $z = \cos 	heta + i\sin 	heta$. Then

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$\cos 49	heta - i\sin 49	heta$

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(A) Using De Moivre's theorem,we know that $(\cos 	heta + i\sin 	heta)^n = \cos n	heta + i\sin n	heta$. Let $z = \cos 	heta + i\sin 	heta$. Then $\cos n	heta - i\sin n	heta = z^{-n}$ and $\cos n	heta + i\sin n	heta = z^n$. The expression becomes: $\frac{(z^{-2})^4 (z^4)^{-5}}{(z^3)^{-2} (z^{-3})^{-9}}$ $= \frac{z^{-8} \cdot z^{-20}}{z^{-6} \cdot z^{27}}$ $= \frac{z^{-28}}{z^{21}}$ $= z^{-28-21} = z^{-49}$ $= \cos(-49	heta) + i\sin(-49	heta)$ $= \cos 49	heta - i\sin 49	heta$.

Express $\frac{(\cos 2\theta - i\sin 2\theta)^4 (\cos 4\theta + i\sin 4\theta)^{-5}}{(\cos 3\theta + i\sin 3\theta)^{-2} (\cos 3\theta - i\sin 3\theta)^{-9}}$ in the form $x + iy$.

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