# How can you use trigonometric functions to simplify  4 e^( ( 11 pi)/6 i )  into a non-exponential complex number?

Jul 2, 2016

We get $- 2 \setminus \sqrt{3} + 2 i$.

#### Explanation:

Euler's formula ${e}^{i x} = \cos x + i \sin x$ is the key. Here $x = \frac{11 \setminus \pi}{6}$, then:

${e}^{i \frac{11 \setminus \pi}{6}} = \cos \left(\frac{11 \setminus \pi}{6}\right) + i \sin \left(\frac{11 \setminus \pi}{6}\right)$
$= - \cos \left(\setminus \frac{\pi}{6}\right) + i \sin \left(\setminus \frac{\pi}{6}\right)$
$\left(\frac{- \setminus \sqrt{3}}{2}\right) + i \left(\frac{1}{2}\right)$

Multiplying by 4 then gives the answer.