# How can you use trigonometric functions to simplify  5 e^( ( 13 pi)/12 i )  into a non-exponential complex number?

Aug 11, 2017

The answer is $= - \frac{5}{4} \left(\sqrt{6} + \sqrt{2}\right) + i \frac{5}{4} \left(\sqrt{2} - \sqrt{6}\right)$

#### Explanation:

We apply Euler's relation

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

$5 {e}^{\frac{13}{12} i \pi} = 5 \left(\cos \left(\frac{13}{12} \pi\right) + i \sin \left(\frac{13}{12} \pi\right)\right)$

$\cos \left(\frac{13}{12} \pi\right) = \cos \left(\frac{1}{3} \pi + \frac{3}{4} \pi\right)$

$= \cos \left(\frac{1}{3} \pi\right) \cos \left(\frac{3}{4} \pi\right) - \sin \left(\frac{1}{3} \pi\right) \sin \left(\frac{3}{4} \pi\right)$

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

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

$\sin \left(\frac{13}{12} \pi\right) = \sin \left(\frac{1}{3} \pi + \frac{3}{4} \pi\right)$

$= \sin \left(\frac{1}{3} \pi\right) \cos \left(\frac{3}{4} \pi\right) + \cos \left(\frac{1}{3} \pi\right) \sin \left(\frac{3}{4} \pi\right)$

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

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

Therefore,

$5 {e}^{\frac{13}{12} i \pi} = - \frac{5}{4} \left(\sqrt{6} + \sqrt{2}\right) + i \frac{5}{4} \left(\sqrt{2} - \sqrt{6}\right)$