# How can you use trigonometric functions to simplify  7 e^( ( 13 pi)/12 i )  into a non-exponential complex number?

Jul 25, 2017

The answer is $= - \frac{7}{4} \left(\left(\sqrt{6} + \sqrt{2}\right)\right) + i \frac{7}{4} \left(\left(\sqrt{2} - \sqrt{6}\right)\right)$

#### Explanation:

We apply Euler's relation

${e}^{i \theta} = \cos \theta + i \sin \theta$

Therefore,

$7 {e}^{i \frac{13}{12} \pi} = 7 \left(\cos \left(\frac{13}{12} \pi\right) + i \sin \left(\frac{13}{12} \pi\right)\right)$

$\cos \left(\frac{13}{12} \pi\right) = \cos \left(\pi + \frac{1}{12} \pi\right) = - \cos \left(\frac{1}{12} \pi\right)$

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

$= - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$

$= - \frac{\sqrt{6} + \sqrt{2}}{4}$

$\sin \left(\frac{13}{12} \pi\right) = \sin \left(\pi + \frac{1}{12} \pi\right) = - \sin \left(\frac{1}{12} \pi\right)$

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

$= - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2}$

$= \frac{\sqrt{2} - \sqrt{6}}{4}$

Therefore,

$7 {e}^{i \frac{13}{12} \pi} = - 7 \cdot \frac{\left(\sqrt{6} + \sqrt{2}\right)}{4} + i \cdot 7 \frac{\left(\sqrt{2} - \sqrt{6}\right)}{4}$