# How can you use trigonometric functions to simplify  6 e^( ( 23 pi)/12 i )  into a non-exponential complex number?

Nov 15, 2016

$6 {e}^{i \frac{23 \pi}{12}} = \frac{3 \left(\sqrt{6} + \sqrt{2}\right)}{2} - i \frac{3 \left(\sqrt{6} - \sqrt{2}\right)}{2}$

#### Explanation:

A complex number $a + i b$ in polar form can be written as $r \cos \theta + i r \sin \theta$

and using series expansion, it can also be written as $r {e}^{i \theta}$

Hence $6 {e}^{i \frac{23 \pi}{12}} = 6 \cos \left(\frac{23 \pi}{12}\right) + i 6 \sin \left(\frac{23 \pi}{12}\right)$

= $6 \cos \left(2 \pi - \frac{\pi}{12}\right) + i 6 \sin \left(2 \pi - \frac{\pi}{12}\right)$

= $6 \cos \left(\frac{\pi}{12}\right) - i 6 \sin \left(\frac{\pi}{12}\right)$

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

= $\frac{3 \left(\sqrt{6} + \sqrt{2}\right)}{2} - i \frac{3 \left(\sqrt{6} - \sqrt{2}\right)}{2}$