# How can you use trigonometric functions to simplify  15 e^( ( 17 pi)/12 i )  into a non-exponential complex number?

Aug 4, 2018

#### Answer:

The answer is $= \frac{15}{4} \left(\left(\sqrt{2} - \sqrt{6}\right) - i \left(\sqrt{2} + \sqrt{6}\right)\right)$

#### Explanation:

Apply the identity

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

And

$\frac{17}{12} \pi = \frac{3}{4} \pi + \frac{2}{3} \pi$

Therefore,

$15 {e}^{\frac{17}{12} i \pi} = 15 \left(\cos \left(\frac{3}{4} \pi + \frac{2}{3} \pi\right) + i \sin \left(\frac{3}{4} \pi + \frac{2}{3} \pi\right)\right)$

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

$= \left(- \frac{\sqrt{2}}{2} \cdot - \frac{1}{2}\right) - \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right)$

$= \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}$

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

$= \left(\frac{\sqrt{2}}{2} \cdot - \frac{1}{2}\right) + \left(- \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right)$

$= - \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}$

Finally,

$15 {e}^{\frac{17}{12} i \pi} = \frac{15}{4} \left(\left(\sqrt{2} - \sqrt{6}\right) + i \left(- \sqrt{2} - \sqrt{6}\right)\right)$