# How can you use trigonometric functions to simplify  16 e^( ( pi)/6 i )  into a non-exponential complex number?

Dec 31, 2015

Euler formula ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$ using this we can represent the given complex number as .

$16 {e}^{\frac{\pi}{6} i} = 8 \sqrt{3} + 8 i$

#### Explanation:

$16 {e}^{\frac{\pi}{6} i}$
Comparing to the Euler's formula ${e}^{i \theta}$
We can see $\theta = \frac{\pi}{6}$

Therefore,

$16 {e}^{\frac{\pi}{6} i} = 16 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)$

$16 {e}^{\frac{\pi}{6} i} = 16 \left(\sqrt{\frac{3}{2}} + i \left(\frac{1}{2}\right)\right)$

$16 {e}^{\frac{\pi}{6} i} = \frac{16}{2} \left(\sqrt{3} + i\right)$

$16 {e}^{\frac{\pi}{6} i} = 8 \left(\sqrt{3} + i\right)$

$16 {e}^{\frac{\pi}{6} i} = 8 \sqrt{3} + 8 i$ Answer