# How do you perform the operation in trigonometric form (5/3(cos(140)+isin(140)))(2/3(cos(60)+isin(60)))?

Dec 20, 2016

$\left(\frac{5}{3} \left(\cos {140}^{\circ} + i \sin {140}^{\circ}\right)\right) \left(\frac{2}{3} \left(\cos {60}^{\circ} + i \sin {60}^{\circ}\right)\right) = \frac{10}{9} \left(\cos {200}^{\circ} + i \sin {200}^{\circ}\right)$

#### Explanation:

$\left(\frac{5}{3} \left(\cos {140}^{\circ} + i \sin {140}^{\circ}\right)\right) \left(\frac{2}{3} \left(\cos {60}^{\circ} + i \sin {60}^{\circ}\right)\right)$

= $\frac{10}{9} \left[\cos {140}^{\circ} \cos {60}^{\circ} + i \cos {140}^{\circ} \sin {60}^{\circ} + i \sin {140}^{\circ} \cos {60}^{\circ} + {i}^{2} \sin {140}^{\circ} \sin {60}^{\circ}\right]$

= $\frac{10}{9} \left(\cos {140}^{\circ} \cos {60}^{\circ} - \sin {140}^{\circ} \sin {60}^{\circ} + i \cos {140}^{\circ} \sin {60}^{\circ} + i \sin {140}^{\circ} \cos {60}^{\circ}\right]$

= $\frac{10}{9} \left(\cos \left({140}^{\circ} + {60}^{\circ}\right) + i \sin \left({140}^{\circ} + {60}^{\circ}\right)\right)$

= $\frac{10}{9} \left(\cos {200}^{\circ} + i \sin {200}^{\circ}\right)$