# How do you evaluate  e^( ( 7 pi)/6 i) - e^( ( 14 pi)/3 i) using trigonometric functions?

Mar 30, 2016

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

#### Explanation:

${e}^{\frac{7 \pi i}{6}} = \cos \left(\frac{7 \pi}{6}\right) + i \sin \left(\frac{7 \pi}{6}\right)$

= $- \cos \frac{\pi}{6} - i \sin \frac{\pi}{6}$ [(7pi)/6 = pi +pi/6, In the third quadrant both sin and cos would b negative]

Like wise ${e}^{\frac{14 \pi i}{3}} = \cos \left(\frac{14 \pi}{3}\right) + i \sin \left(\frac{14 \pi}{3}\right)$

=$\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$ $\left[\frac{14 \pi}{3} = 4 \pi + \frac{2 \pi}{3} = \frac{2 \pi}{3}\right]$

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

Now combining both expressions, it would be $- \frac{\sqrt{3}}{2} - \frac{i}{2} + \frac{1}{2} - i \frac{\sqrt{3}}{2}$