**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:

$\large\frac{(\cos 2\theta-i\sin 2\theta)^7(\cos 3\theta+i\sin 3\theta)^{-5}}{(\cos 4\theta+i\sin 4\theta)^{12}(\cos 5\theta-i\sin 5\theta)^{-6}}=\large\frac{(e^{-i2\theta})^7(e^{i3\theta})^{-5}}{(e^{i4\theta})^{12}(e^{-i5\theta})^{-6}}$

$\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;= \large\frac{e^{-i4\theta}e^{-i15\theta}}{e^{i48\theta}e^{i30\theta}}$

$\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;=e^{\large-i(14+15+48+30)}$

Step 2:

$\Rightarrow e^{\large-i107\theta}$

We know that $e^{i\theta}=\cos\theta+i\sin\theta$

$\Rightarrow \cos 107\theta-i\sin 107\theta$