How can you use trigonometric functions to simplify  2 e^( ( 7 pi)/12 i )  into a non-exponential complex number?

Dec 24, 2015

Explanation is given below.

Explanation:

Euler's formula ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$

Our question $2 {e}^{\frac{7 \pi}{12} i}$ can be simplified to using the Euler's formula as

$2 \left(\cos \left(\frac{7 \pi}{12}\right) + i \sin \left(\frac{7 \pi}{12}\right)\right)$

Now we need to evaluate this.

$\cos \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{4} + \frac{\pi}{3}\right)$
$\cos \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{3}\right) - \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{3}\right)$
$\cos \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$
$\cos \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}$
$\cos \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$

$\sin \left(\frac{7 \pi}{12}\right) = \sin \left(\frac{\pi}{4} + \frac{\pi}{3}\right)$
$\sin \left(\frac{7 \pi}{12}\right) = \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{3}\right) + \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{3}\right)$
$\sin \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$
$\sin \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}$
$\sin \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2} + \sqrt{6}}{4}$

The complex number would be
$2 \left(\frac{\sqrt{2} - \sqrt{6}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4}\right)$
$\frac{1}{2} \left(\left(\sqrt{2} - \sqrt{6}\right) + i \left(\sqrt{2} + \sqrt{6}\right)\right)$

That should do for an answer, further simplification is possible depending on how the answer needs to be represented.