# How do you evaluate  e^( ( 7 pi)/6 i) - e^( ( 7 pi)/8 i) using trigonometric functions?

$\frac{\left(\sqrt{2 + \sqrt{2}} - \sqrt{3}\right) - i \left(\sqrt{2 - \sqrt{2}} + 1\right)}{2.}$
We have, ${e}^{i \theta} = \cos \theta + i \sin \theta .$
$\therefore {e}^{\left(7 \frac{\pi}{6}\right) i} - {e}^{\left(7 \frac{\pi}{8}\right) i} = \cos 7 \frac{\pi}{6} + i \sin 7 \frac{\pi}{6} - \cos 7 \frac{\pi}{8} - i \sin 7 \frac{\pi}{8} = \cos \left(\pi + \frac{\pi}{6}\right) + i \sin \left(\pi + \frac{\pi}{6}\right) - \cos \left(\pi - \frac{\pi}{8}\right) - i \sin \left(\pi - \frac{\pi}{8}\right) = - \cos \left(\frac{\pi}{6}\right) - i \sin \left(\frac{\pi}{6}\right) + \cos \left(\frac{\pi}{8}\right) - i \sin \left(\frac{\pi}{8}\right) = - \frac{\sqrt{3}}{2} - \frac{i}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2} - i \frac{\sqrt{2 - \sqrt{2}}}{2} = \frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{3}}{2} - i \frac{\sqrt{2 - \sqrt{2}}}{2} - \frac{i}{2} = \frac{\left(\sqrt{2 + \sqrt{2}} - \sqrt{3}\right) - i \left(\sqrt{2 - \sqrt{2}} + 1\right)}{2.}$