# How do you multiply e^(( 7 pi )/ 12 ) * e^( pi/2 i )  in trigonometric form?

##### 1 Answer
Nov 22, 2017

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

#### Explanation:

Apply Euler's relation

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

Therefore,

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

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

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

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

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

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

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

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

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

Therefore,

${e}^{i \frac{7}{12} \pi} . {e}^{i \frac{\pi}{2}} = \left(\frac{\left(\sqrt{2} - \sqrt{6}\right) + \left(\sqrt{2} + \sqrt{6}\right) i}{4}\right) \left(i\right)$

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