# Simplify: $\Large\frac{\left (cos \;\alpha + \mathit{i} sin \; \alpha \right )^{3}}{\left (sin \;\beta + \mathit{i} cos \;\beta \right)^{4} } \Large$

Toolbox:
• 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:
$\large\frac{(\cos\alpha+i\sin\alpha)^3}{(\sin \beta+i\cos \beta)^4}=\large\frac{(\cos\theta+i\sin\alpha)^3}{\cos(\large\frac{\pi}{2}-\beta)+i\sin(\large\frac{\pi}{2}-\beta)^4}$
$\Rightarrow \large\frac{(e^{i\alpha})^3}{(e^{i(\large\frac{\pi}{2}-\beta)})^4}=\Large\frac{e^{i3\alpha}}{e^{i4(\Large\frac{\pi}{2}-\beta)}}$
$\Rightarrow e^{\large i(3\alpha-4(\large\frac{\pi}{2}-\beta))}$
Step 2:
We know that $e^{i\theta}=\cos\theta+i\sin\theta$
$\Rightarrow \cos(3\alpha-4(\large\frac{\pi}{2}-$$\beta)+i\sin(3\alpha-4(\large\frac{\pi}{2}$$-\beta))$
$\Rightarrow \cos(3\alpha+4\beta-2\pi)+i\sin(3\alpha+4\beta-2\pi)$
$\Rightarrow \cos-(2\pi-(3\alpha+4\beta))+i\sin-(2\pi-(3\alpha+4\beta))$
$\Rightarrow \cos(3\alpha+4\beta)-i\sin(3\alpha+4\beta)$