# How do you multiply e^(( 19pi )/ 12 ) * e^( pi/2 i )  in trigonometric form?

Jan 26, 2018

The answer is $= \frac{\sqrt{6} + \sqrt{2}}{4} + i \frac{\sqrt{6} - \sqrt{2}}{4}$

#### Explanation:

Apply Euler's Identity

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

${i}^{2} = - 1$

Therefore,

${e}^{i \frac{\pi}{2}} = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = 0 + i \cdot 1 = i$

${e}^{i \frac{19}{12} \pi} = \cos \left(\frac{19}{12} \pi\right) + i \sin \left(\frac{19}{12} \pi\right)$

$\cos \left(\frac{19}{12} \pi\right) = \cos \left(\frac{5}{4} \pi + \frac{1}{3} \pi\right)$

$= \cos \left(\frac{5}{4} \pi\right) \cos \left(\frac{1}{3} \pi\right) - \sin \left(\frac{5}{4} \pi\right) \sin \left(\frac{1}{3} \pi\right)$

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

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

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

$\sin \left(\frac{19}{12} \pi\right) = \sin \left(\frac{5}{4} \pi + \frac{1}{3} \pi\right)$

$= \sin \left(\frac{5}{4} \pi\right) \cos \left(\frac{1}{3} \pi\right) + \cos \left(\frac{5}{4} \pi\right) \sin \left(\frac{1}{3} \pi\right)$

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

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

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

Finally,

${e}^{i \frac{19}{12} \pi} \cdot {e}^{i \frac{\pi}{2}} = \left(\frac{\sqrt{6} - \sqrt{2}}{4} - i \frac{\sqrt{6} + \sqrt{2}}{4}\right) \cdot \left(i\right)$

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