# How do you perform the operation in trigonometric form (6(cos40+isin40))/(7(cos100+isin100))?

Sep 17, 2016

$\frac{3}{7} \left(1 - i \sqrt{3}\right)$.

#### Explanation:

Let ${z}_{1} = 6 \left(\cos 40 + i \sin 40\right) , \mathmr{and} , {z}_{2} = 7 \left(\cos 100 + i \sin 100\right)$

Knowing that, $\cos \theta + i \sin \theta = {e}^{i \theta}$, we have,

${z}_{1} = {r}_{1} {e}^{i {\theta}_{1}} , \mathmr{and} , {z}_{2} = {r}_{2} {e}^{i {\theta}_{2}}$, then,

${r}_{1} = 6 , {\theta}_{1} = 40 , {r}_{2} = 7 , {\theta}_{2} = 100$.

$\therefore \text{ The Expression=} {z}_{1} / {z}_{2} = \frac{{r}_{1} {e}^{i {\theta}_{1}}}{{r}_{2} {e}^{i {\theta}_{2}}}$

$= \left({r}_{1} / {r}_{2}\right) \cdot {e}^{i {\theta}_{1} - i {\theta}_{2}}$

$= \left({r}_{1} / {r}_{2}\right) \cdot {e}^{i \left({\theta}_{1} - {\theta}_{2}\right)}$

$= \left(\frac{6}{7}\right) \cdot {e}^{i \left(40 - 100\right)}$

$= \left(\frac{6}{7}\right) \cdot {e}^{i \left(- 60\right)}$

$= \left(\frac{6}{7}\right) \left(\cos \left(- 60\right) + i \sin \left(- 60\right)\right)$

$= \left(\frac{6}{7}\right) \left(\cos 60 - i \sin 60\right)$

$= \left(\frac{6}{7}\right) \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right)$

$= \frac{3}{7} \left(1 - i \sqrt{3}\right)$.

Enjoy maths.!