# How can you use trigonometric functions to simplify  15 e^( ( 3 pi)/8 i )  into a non-exponential complex number?

Jan 15, 2016

$15 {e}^{\frac{3 \pi}{8} i} = \frac{15}{2} \left(\sqrt{2 - \sqrt{2}} + \sqrt{2 + \sqrt{2}} i\right)$

#### Explanation:

Using the identity ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$ we have:

$15 {e}^{\frac{3 \pi}{8} i} = 15 \left(\cos \left(\frac{3 \pi}{8}\right) + i \sin \left(\frac{3 \pi}{8}\right)\right)$

If we do not want trig functions either, we can simplify further using the half angle formulas.

$\cos \left(\frac{3 \pi}{8}\right) = \cos \left(\frac{1}{2} \left(\frac{3 \pi}{4}\right)\right) = \sqrt{\frac{1 + \cos \left(\frac{3 \pi}{4}\right)}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

$\sin \left(\frac{3 \pi}{8}\right) = \sin \left(\frac{1}{2} \left(\frac{3 \pi}{4}\right)\right) = \sqrt{\frac{1 - \cos \left(\frac{3 \pi}{4}\right)}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

Thus

$15 {e}^{\frac{3 \pi}{8} i} = 15 \left(\frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2} i\right)$

$= \frac{15}{2} \left(\sqrt{2 - \sqrt{2}} + \sqrt{2 + \sqrt{2}} i\right)$