# How do you evaluate  e^( ( 11 pi)/6 i) - e^( ( 15 pi)/8 i) using trigonometric functions?

Mar 30, 2017

The general equation is:

${r}_{1} {e}^{{\theta}_{1} i} - {r}_{2} {e}^{{\theta}_{1} i} = {r}_{1} \cos \left({\theta}_{1}\right) - {r}_{2} \cos \left({\theta}_{2}\right) + i \left({r}_{1} \sin \left({\theta}_{1}\right) - {r}_{2} \sin \left({\theta}_{2}\right)\right)$

#### Explanation:

You can derive the general equation, using Euler's Formula

${e}^{\theta i} = \cos \left(\theta\right) + i \sin \left(\theta\right)$

$r {e}^{\theta i} = r \cos \left(\theta\right) + i \left(r \sin \left(\theta\right)\right)$

For the given expression:

${e}^{\frac{11 \pi}{6} i} - {e}^{\frac{15 \pi}{8} i} = \cos \left(\frac{11 \pi}{6}\right) - \cos \left(\frac{15 \pi}{8}\right) + i \left(\sin \left(\frac{11 \pi}{6}\right) - \sin \left(\frac{15 \pi}{8}\right)\right)$