# How do you evaluate  e^( (3 pi)/2 i) - e^( ( 5 pi)/6 i) using trigonometric functions?

Mar 4, 2018

${e}^{\frac{3 \pi}{2} i} - {e}^{\frac{5 \pi}{6} i} = \frac{\sqrt{3} - 3 i}{2}$

#### Explanation:

Recall the identity :

${e}^{\textcolor{red}{i} \delta} = \cos \delta + \textcolor{red}{i} \sin \delta$

Given the values of $\delta$, we have :

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

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

These are fairly small values of $\delta$, so you can figure them out using the unit circle and properties of the trigonometric functions.

$\cos \left(\frac{3 \pi}{2}\right) = 0$
$\sin \left(\frac{3 \pi}{2}\right) = - 1$

For $\delta = \frac{5 \pi}{6}$, we have to use the equalities $\sin \left(\textcolor{red}{\pi} - \delta\right) = \sin \delta$ and $\cos \left(\textcolor{red}{\pi} - \delta\right) = \textcolor{red}{-} \cos \delta$:

$\sin \left(\frac{5 \pi}{6}\right) = \sin \left(\textcolor{red}{\pi} - \frac{5 \pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

$\cos \left(\frac{5 \pi}{6}\right) = \textcolor{red}{-} \cos \left(\frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$

Plugging in these values for ${e}^{\frac{3 \pi}{2} i} - {e}^{\frac{5 \pi}{6} i}$, we get :

${e}^{\frac{3 \pi}{2} i} - {e}^{\frac{5 \pi}{6} i} = 0 + i \left(- 1\right) - \left(- \frac{\sqrt{3}}{2}\right) - i \left(\frac{1}{2}\right)$

$\textcolor{red}{{e}^{\frac{3 \pi}{2} i} - {e}^{\frac{5 \pi}{6} i} = \frac{\sqrt{3} - 3 i}{2}}$.