# How can you use trigonometric functions to simplify  8 e^( ( 19 pi)/12 i )  into a non-exponential complex number?

Dec 4, 2017

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

#### Explanation:

Apply Euler's identity

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

$\frac{19}{12} \pi = \frac{5}{4} \pi + \frac{1}{3} \pi$

$\cos \left(\frac{19}{12} \pi\right) = \cos \left(\frac{5}{4} \pi + \frac{1}{3} \pi\right)$

$= \cos \left(\frac{5}{4} \pi\right) \cos \left(\frac{1}{3} \pi\right) - \sin \left(\frac{5}{4} \pi\right) \sin \left(\frac{1}{3} \pi\right)$

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

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

$\sin \left(\frac{19}{12} \pi\right) = \sin \left(\frac{5}{4} \pi + \frac{1}{3} \pi\right)$

$= \sin \left(\frac{5}{4} \pi\right) \cos \left(\frac{1}{3} \pi\right) + \sin \left(\frac{1}{3} \pi\right) \cos \left(\frac{5}{4} \pi\right)$

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

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

Therefore,

$8 {e}^{\frac{19}{12} \pi} = 8 \cos \left(\frac{19}{12} \pi\right) + i 8 \sin \left(\frac{19}{12} \pi\right)$

$= 8 \cdot \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) - 8 i \left(- \frac{\sqrt{2} + \sqrt{6}}{4}\right)$

$= 2 \left(\sqrt{6} - \sqrt{2}\right) - 2 i \left(\sqrt{6} + \sqrt{2}\right)$