# How do you calculate (3(cos14^circ+isin14^circ))/(2(cos121^circ+isin121^circ)?

Mar 22, 2017

$\frac{3 \left(\cos {14}^{\circ} + i \sin {14}^{\circ}\right)}{2 \left(\cos {121}^{\circ} + i \sin {121}^{\circ}\right)} = \frac{3}{2} \left(\cos {107}^{\circ} - i \sin {107}^{\circ}\right)$

#### Explanation:

(3(cos14^@+isin14^@))/(2(cos121^@+isin121^@)

= 3/2xx((cos14^@+isin14^@))/((cos121^@+isin121^@))xx((cos121^@-isin121^@))/((cos121^@-isin121^@)

= $\frac{3}{2} \frac{\cos {14}^{\circ} \cos {121}^{\circ} - {i}^{2} \sin {14}^{\circ} \sin {121}^{\circ} - i \cos {14}^{\circ} \sin {121}^{\circ} + i \sin {14}^{\circ} \cos {121}^{\circ}}{{\cos}^{2} {121}^{\circ} - {i}^{2} \sin {121}^{\circ}}$

= $\frac{3}{2} \frac{\cos {14}^{\circ} \cos {121}^{\circ} + \sin {14}^{\circ} \sin {121}^{\circ} + i \left(\sin {14}^{\circ} \cos {121}^{\circ} - \cos {14}^{\circ} \sin {121}^{\circ}\right)}{{\cos}^{2} {121}^{\circ} + \sin {121}^{\circ}}$

= $\frac{3}{2} \frac{\cos \left({14}^{\circ} - {121}^{\circ}\right) + i \sin \left({14}^{\circ} - {121}^{\circ}\right)}{1}$

= $\frac{3}{2} \left(\cos \left(- {107}^{\circ}\right) + i \sin \left(- {107}^{\circ}\right)\right)$

= $\frac{3}{2} \left(\cos {107}^{\circ} - i \sin {107}^{\circ}\right)$