# How do you simplify 6[cos(-pi/3)+isin(-pi/3)]*3(cos((5pi)/6)+isin((5pi)/6)) and express the result in rectangular form?

Jan 1, 2017

18i

#### Explanation:

Using Euler's formula, it would be

$\cos \left(- \frac{\pi}{3}\right) + i \sin \left(- \frac{\pi}{3}\right) = \cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right) = {e}^{- i \frac{\pi}{3}}$

$\cos \left(5 \frac{\pi}{6}\right) + i \sin \left(5 \frac{\pi}{6}\right) = {e}^{i} \frac{5 \pi}{6}$

The given expression is thus equivalent to 18 e^(-ipi/3)*e^(i (5pi)/6

= 18 e^(i(-pi/3 +5pi/6) = $18 {e}^{i \frac{\pi}{2}}$ = $18 \left[\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right]$= 18i.