# How can you use trigonometric functions to simplify  6 e^( ( pi)/8 i )  into a non-exponential complex number?

Mar 25, 2016

$6 {e}^{i \frac{\pi}{8}} = 6 \left(\cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)\right)$

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

#### Explanation:

Euler showed that for a complex number $z = r {e}^{i \theta}$,

$r {e}^{i \theta} = r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$

In this case

• $r = 6$
• $\theta = \frac{\pi}{8}$

Therefore,

$6 {e}^{i \frac{\pi}{8}} = 6 \left(\cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)\right)$

You can use half angle formulas to find $\sin \left(\frac{\pi}{8}\right)$ and $\cos \left(\frac{\pi}{8}\right)$.

$6 \left(\cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)\right) = 6 \left(\frac{\sqrt{2 + \sqrt{2}}}{2} + i \left(\frac{\sqrt{2 - \sqrt{2}}}{2}\right)\right)$

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