# How can you use trigonometric functions to simplify  9 e^( ( 17 pi)/12 i )  into a non-exponential complex number?

Jul 26, 2018

The answer is $= \frac{9}{4} \left(\left(- \sqrt{6} + \sqrt{2}\right) - i \left(\sqrt{2} + \sqrt{6}\right)\right)$

#### Explanation:

The Euler's Identity is

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

Therefore,

$z = 9 {e}^{\frac{17}{12} \pi i}$

$= 9 \cos \left(\frac{17}{12} \pi\right) + i \sin \left(\frac{17}{12} \pi\right)$

$\cos \left(\frac{17}{12} \pi\right) = \cos \left(\frac{7}{6} \pi + \frac{1}{4} \pi\right)$

$= \cos \left(\frac{7}{6} \pi\right) \cos \left(\frac{1}{4} \pi\right) - \sin \left(\frac{7}{6} \pi\right) \sin \left(\frac{1}{4} \pi\right)$

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

$= \frac{1}{4} \left(- \sqrt{6} + \sqrt{2}\right)$

$\sin \left(\frac{17}{12} \pi\right) = \sin \left(\frac{7}{6} \pi + \frac{1}{4} \pi\right)$

$= \sin \left(\frac{7}{6} \pi\right) \cos \left(\frac{1}{4} \pi\right) + \cos \left(\frac{7}{6} \pi\right) \sin \left(\frac{1}{4} \pi\right)$

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

$= - \frac{1}{4} \left(\sqrt{2} + \sqrt{6}\right)$

Finally,

$z = \frac{9}{4} \left(\left(- \sqrt{6} + \sqrt{2}\right) - i \left(\sqrt{2} + \sqrt{6}\right)\right)$