# How can you use trigonometric functions to simplify  7 e^( ( 3 pi)/4 i )  into a non-exponential complex number?

Nov 20, 2016

The answer is $= - \frac{7 \sqrt{2}}{2} + \frac{i 7 \sqrt{2}}{2}$

#### Explanation:

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

So, $7 {e}^{3 \pi \frac{i}{4}} = 7 \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

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

$\sin \left(3 \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Therefore,

$7 {e}^{3 \pi \frac{i}{4}} = 7 \left(- \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right)$

$- \frac{7 \sqrt{2}}{2} + \frac{i 7 \sqrt{2}}{2}$