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

Jan 31, 2016

${e}^{i \frac{7 \pi}{4}} - {e}^{i \frac{\pi}{6}} = \frac{\sqrt{2} - \sqrt{3}}{2} - i \left(\frac{1 + \sqrt{2}}{2}\right)$

#### Explanation:

Use the Euler's Formula, which states that ${e}^{i \theta} \equiv \cos \left(\theta\right) + i \sin \left(\theta\right)$. (Proof omitted)

Therefore,

${e}^{i \frac{7 \pi}{4}} - {e}^{i \frac{\pi}{6}} = \left(\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)\right) - \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)$

$= \left(\frac{\sqrt{2}}{2} + i \left(- \frac{\sqrt{2}}{2}\right)\right) - \left(\frac{\sqrt{3}}{2} + i \left(\frac{1}{2}\right)\right)$

$= \frac{\sqrt{2} - \sqrt{3}}{2} - i \left(\frac{1 + \sqrt{2}}{2}\right)$