# How do you evaluate  e^( ( 7 pi)/6 i) - e^( ( pi)/3 i) using trigonometric functions?

Dec 18, 2017

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

#### Explanation:

Using Euler's formula (${e}^{x i} = \cos x - i \sin x$) we get:
${e}^{\frac{7 \pi}{6} i} - {e}^{\frac{\pi}{3} i} = \left(\cos \left(\frac{7 \pi}{6}\right) - i \sin \left(\frac{7 \pi}{6}\right)\right) - \left(\cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right)\right)$

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

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

$= - \frac{\sqrt{3}}{2} + \frac{i}{2} - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

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

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