# How do you simplify 4(cos(pi/3)+isin(pi/3))*7(cos((2pi)/3)+isin((2pi)/3)) and express the result in rectangular form?

##### 1 Answer
Dec 31, 2016

$4 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right) \cdot 7 \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right) = - 28$

#### Explanation:

Note that:

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

So we have:

$\left(\cos \alpha + i \sin \alpha\right) \left(\cos \beta + i \sin \beta\right) = {e}^{i \alpha} \cdot {e}^{i \beta}$

$\textcolor{w h i t e}{\left(\cos \alpha + i \sin \alpha\right) \left(\cos \beta + i \sin \beta\right)} = {e}^{i \left(\alpha + \beta\right)}$

$\textcolor{w h i t e}{\left(\cos \alpha + i \sin \alpha\right) \left(\cos \beta + i \sin \beta\right)} = \cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)$

So in our example:

$4 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right) \cdot 7 \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)$

$= 28 \left(\cos \left(\frac{\pi}{3} + \frac{2 \pi}{3}\right) + i \sin \left(\frac{\pi}{3} + \frac{2 \pi}{3}\right)\right)$

$= 28 \left(\cos \left(\pi\right) + i \sin \left(\pi\right)\right)$

$= 28 \left(- 1 + i \cdot 0\right)$

$= - 28$