# How do you perform the operation in trigonometric form (0.45(cos(310)+isin(310)))(0.6(cos(200)+isin(200)))?

##### 1 Answer
Jan 5, 2017

$\left(0.45 \left(\cos {310}^{\circ} + i \sin {310}^{\circ}\right)\right) \left(0.6 \left(\cos {200}^{\circ} + i \sin {200}^{\circ}\right)\right) = 0.27 \left(- \frac{\sqrt{3}}{2} + \frac{1}{2} i\right)$

#### Explanation:

Two complex numbers in polar form ${z}_{1} = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and ${z}_{2} = {r}_{2} \left(\cos \beta + i \sin \beta\right)$ can be multiplied as under,

${z}_{1} \cdot {z}_{2} = {r}_{1} \cdot {r}_{2} \left[\cos \alpha \cos \beta + i \cos \alpha \sin \beta + i \sin \alpha \cos \beta + {i}^{2} \sin \alpha \sin \beta\right]$

= ${r}_{1} \cdot {r}_{2} \left[\left(\cos \alpha \cos \beta - \sin \alpha \sin \beta\right) + i \left(\cos \alpha \sin \beta + \sin \alpha \cos \beta\right)\right]$

= ${r}_{1} \cdot {r}_{2} \left[\cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)\right]$

Hence $\left(0.45 \left(\cos {310}^{\circ} + i \sin {310}^{\circ}\right)\right) \left(0.6 \left(\cos {200}^{\circ} + i \sin {200}^{\circ}\right)\right)$

= $0.45 \times 0.6 \left(\cos \left({310}^{\circ} + {200}^{\circ}\right) + i \sin \left({310}^{\circ} + {200}^{\circ}\right)\right)$

= $0.27 \left(\cos {510}^{\circ} + i \sin {510}^{\circ}\right)$

= $0.27 \left(\cos \left({360}^{\circ} + {150}^{\circ}\right) + i \sin \left({360}^{\circ} + {150}^{\circ}\right)\right)$

= $0.27 \left(\cos {150}^{\circ} + i \sin {150}^{\circ}\right)$

= $0.27 \left(- \frac{\sqrt{3}}{2} + \frac{1}{2} i\right)$