# How do you evaluate  e^( ( 7 pi)/6 i) - e^( ( 19 pi)/8 i) using trigonometric functions?

Nov 4, 2016

$2 \sin \left(\frac{29 \pi}{24}\right) {e}^{\left(\frac{25 \pi}{24}\right) i}$

#### Explanation:

we know euler's theorem states that ${e}^{i \theta} = \cos \theta + i \sin \theta$
so,${e}^{7 \frac{\pi}{6} i} - {e}^{19 \frac{\pi}{8} i}$
=$\cos \left(7 \frac{\pi}{6}\right) + i \sin \left(7 \frac{\pi}{6}\right) - \left(\cos \left(19 \frac{\pi}{8}\right) + i \sin \left(19 \frac{\pi}{8}\right)\right)$
=$\cos \left(7 \frac{\pi}{6}\right) - \cos \left(19 \frac{\pi}{8}\right) + i \left(\sin \left(7 \frac{\pi}{6}\right) - \sin \left(19 \frac{\pi}{8}\right)\right)$
now $\cos c - \cos d = 2 \sin \left(\frac{c + d}{2}\right) \sin \left(\frac{d - c}{2}\right)$
$\sin c - \sin d = 2 \cos \left(\frac{c + d}{2}\right) \sin \left(\frac{c - d}{2}\right)$
so we get $\cos \left(\frac{7 \pi}{6}\right) - \cos \left(\frac{19 \pi}{8}\right)$
=$2 \sin \left(\frac{85 \pi}{24}\right) \sin \left(\frac{29 \pi}{24}\right)$
again similarly we get $i \left(\sin \left(\frac{7 \pi}{6}\right) - \sin \left(\frac{19 \pi}{8}\right)\right)$
=$2 i \cos \left(\frac{85 \pi}{24}\right) \sin \left(\frac{29 \pi}{24}\right)$
so,${e}^{\frac{7 \pi}{6}} - {e}^{\frac{19 \pi}{8}}$=$2 \sin \left(\frac{29 \pi}{24}\right) \left(\sin \left(\frac{85 \pi}{24}\right) - i \cos \left(\frac{85 \pi}{24}\right)\right)$
=$2 \sin \left(\frac{29 \pi}{24}\right) \left[\sin \left(2 \pi + \left(\frac{37 \pi}{24}\right)\right) - i \cos \left(2 \pi + \left(\frac{37 \pi}{24}\right)\right)\right]$
=$2 \sin \left(\frac{29 \pi}{24}\right) \left[\sin \left(\frac{37 \pi}{24}\right) - i \cos \left(\frac{37 \pi}{24}\right)\right]$
=$2 \sin \left(\frac{29 \pi}{24}\right) \left[\cos \left(\frac{25 \pi}{24}\right) + i \sin \left(\frac{25 \pi}{24}\right)\right]$
=$2 \sin \left(\frac{29 \pi}{24}\right) {e}^{\left(\frac{25 \pi}{24}\right) i}$.