# How do you perform the operation in trigonometric form (12(cos52+isin52))/(3(cos110+isin110))?

Jun 1, 2017

$\frac{12 \left(\cos 52 + i \sin 52\right)}{3 \left(\cos 110 + i \sin 110\right)} = 4 \left(\cos 58 - i \sin 58\right)$

#### Explanation:

$\frac{12 \left(\cos 52 + i \sin 52\right)}{3 \left(\cos 110 + i \sin 110\right)}$

= $\frac{12 \left(\cos 52 + i \sin 52\right) \left(\cos 110 - i \sin 110\right)}{3 \left(\cos 110 + i \sin 110\right) \left(\cos 110 - i \sin 110\right)}$

= $\frac{12 \left(\cos 52 \cos 110 + i \sin 52 \cos 110 - i \cos 52 \sin 110 - {i}^{2} \sin 52 \sin 110\right)}{3 \left({\cos}^{2} 110 - {i}^{2} {\sin}^{2} 110\right)}$

= $\frac{12 \left(\cos 52 \cos 110 + \sin 52 \sin 110 + i \left(\sin 52 \cos 110 - \cos 52 \sin 110\right)\right)}{3 \left({\cos}^{2} 110 + {\sin}^{2} 110\right)}$

= $\frac{12 \left(\cos \left(52 - 110\right) + i \sin \left(52 - 110\right)\right)}{3 \left({\cos}^{2} 110 + {\sin}^{2} 110\right)}$

= $\frac{12 \left(\cos \left(- 58\right) + i \sin \left(- 58\right)\right)}{3}$

= $4 \left(\cos 58 - i \sin 58\right)$

Jun 1, 2017

$\frac{12 \left(\cos 52 + i \sin 52\right)}{3 \left(\cos 110 + i \sin 110\right)} = 4 \left(\cos 58 - i \sin 58\right)$

#### Explanation:

$\frac{12 \left(\cos 52 + i \sin 52\right)}{3 \left(\cos 110 + i \sin 110\right)} = 4 \times \frac{\cos 52 + i \sin 52}{\cos 110 + i \sin 110}$

$\cos 52 + i \sin 52 = {e}^{52 i}$

$\cos 110 + i \sin 110 = {e}^{110 i}$

$\therefore 4 \times \frac{\cos 52 + i \sin 52}{\cos 110 + i \sin 110} = 4 \times {e}^{52 i} / {e}^{110 i} = 4 {e}^{- 58 i}$

$4 {e}^{- 58 i} = 4 \left(\cos - 58 + i \sin - 58\right) = 4 \left(\cos 58 - i \sin 58\right)$