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Simplify: $\Large\frac{\left ( cos2 \theta -\mathit{i}sin2 \theta \right )^{7} \left ( cos 3 \theta +\mathit{i} sin3 \theta \right )^{-5}}{\left (cos 4 \theta + \mathit{i} sin4 \theta \right )^{12} \left ( cos 5 \theta -\mathit{i}sin5 \theta \right )^{-6}}\Large$

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  • From De moivre's theorem we have
  • (i) $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q$
  • (ii) $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
  • (iii) $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
  • (iv) $(\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta)$
  • $e^{i\theta}=\cos\theta+i\sin\theta$
  • $e^{-i\theta}=\cos\theta-i\sin\theta$,also written as $\cos\theta$ and $\cos(-\theta)$
Step 1:
$\large\frac{(\cos 2\theta-i\sin 2\theta)^7(\cos 3\theta+i\sin 3\theta)^{-5}}{(\cos 4\theta+i\sin 4\theta)^{12}(\cos 5\theta-i\sin 5\theta)^{-6}}=\large\frac{(e^{-i2\theta})^7(e^{i3\theta})^{-5}}{(e^{i4\theta})^{12}(e^{-i5\theta})^{-6}}$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;= \large\frac{e^{-i4\theta}e^{-i15\theta}}{e^{i48\theta}e^{i30\theta}}$
Step 2:
$\Rightarrow e^{\large-i107\theta}$
We know that $e^{i\theta}=\cos\theta+i\sin\theta$
$\Rightarrow \cos 107\theta-i\sin 107\theta$


answered Jun 11, 2013 by sreemathi.v
edited Jul 19, 2013 by balaji.thirumalai

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