How do you multiply e^(( 5 pi )/ 12 i) * e^( 3 pi/2 i )  in trigonometric form?

Jul 8, 2017

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

Explanation:

We apply Euler's formula

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

${e}^{i \frac{5}{12} \pi} = \cos \left(\frac{5}{12} \pi\right) + i \sin \left(\frac{5}{12} \pi\right)$

$\cos \left(\frac{5}{12} \pi\right) = \cos \left(\frac{3}{12} \pi + \frac{2}{12} \pi\right)$

$= \cos \left(\frac{1}{4} \pi + \frac{1}{6} \pi\right)$

$= \cos \left(\frac{1}{4} \pi\right) \cos \left(\frac{1}{6} \pi\right) - \sin \left(\frac{1}{4} \pi\right) \sin \left(\frac{1}{6} \pi\right)$

$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$

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

$\sin \left(\frac{5}{12} \pi\right) = \sin \left(\frac{1}{4} \pi + \frac{1}{6} \pi\right)$

$= \sin \left(\frac{1}{4} \pi\right) \cos \left(\frac{1}{6} \pi\right) + \cos \left(\frac{1}{4} \pi\right) \sin \left(\frac{1}{6} \pi\right)$

$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$

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

Therefore,

${e}^{i \frac{5}{12} \pi} = \frac{\sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{6} + \sqrt{2}}{4}$

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

$= 0 - i$

So,

${e}^{i \frac{5}{12} \pi} . {e}^{i \frac{3}{2} \pi} = \left(\frac{\sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{6} + \sqrt{2}}{4}\right) \cdot - i$

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

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

Verification

${e}^{i \frac{5}{12} \pi} \cdot {e}^{i \frac{3}{2} \pi} = {e}^{i \left(\frac{5}{12} + \frac{3}{2}\right) \pi}$

$= {e}^{i \frac{23}{12} \pi}$

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

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

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

$= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$

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

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

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

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

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

Therefore,

${e}^{i \frac{23}{12} \pi} = \frac{\sqrt{6} + \sqrt{2}}{4} - i \frac{\sqrt{6} - \sqrt{2}}{4}$

The result is the same.