# How do you evaluate  e^( ( 7 pi)/6 i) - e^( ( 4 pi)/3 i) using trigonometric functions?

Feb 13, 2016

$2 \left[\sin \left(\frac{5 \pi}{4}\right) \sin \left(\frac{\pi}{12}\right) + i \cos \left(\frac{5 \pi}{4}\right) \sin \left(\frac{\pi}{12}\right)\right]$

#### Explanation:

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

${e}^{\frac{4 \pi i}{3}} = \cos \left(\frac{4 \pi}{3}\right) - i \sin \left(\frac{4 \pi}{3}\right)$

The required simplification would be

$\cos \left(\frac{7 \pi}{6}\right) - \cos \left(\frac{4 \pi}{3}\right) + i \left[\sin \left(\frac{4 \pi}{3}\right) - \sin \left(\frac{7 \pi}{6}\right)\right]$

= $2 \sin \left(\frac{\frac{7 \pi}{6} + \frac{4 \pi}{3}}{2}\right) \sin \left(\frac{\frac{4 \pi}{3} - \frac{7 \pi}{6}}{2}\right) + i \left[2 \cos \left(\frac{\frac{4 \pi}{3} + \frac{7 \pi}{6}}{2}\right) \sin \left(\frac{\frac{4 \pi}{3} - \frac{7 \pi}{6}}{2}\right)\right]$

= $2 \left[\sin \left(\frac{5 \pi}{4}\right) \sin \left(\frac{\pi}{12}\right) + i \cos \left(\frac{5 \pi}{4}\right) \sin \left(\frac{\pi}{12}\right)\right]$