# How do you multiply e^(( 5 pi )/ 8 i) * e^( 3 pi/2 i )  in trigonometric form?

Mar 13, 2018

The answer is $= \frac{1}{2} \left(\left(\sqrt{2 + \sqrt{2}}\right) - i \left(\sqrt{2 - \sqrt{2}}\right)\right)$

#### Explanation:

Apply Euler's Identity

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

${e}^{\frac{5}{8} \pi i} = \cos \left(\frac{5}{8} \pi\right) + i \sin \left(\frac{5}{8} \pi\right)$

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

$\cos 2 \theta = 2 {\cos}^{2} \theta - 1 = 1 - 2 {\sin}^{2} \theta$

$\cos \theta = \sqrt{\frac{1 + \cos 2 \theta}{2}}$

$\sin \theta = \sqrt{\frac{1 - \sin 2 \theta}{2}}$

$\cos \left(\frac{5}{8} \pi\right) = \sqrt{\left(1 + \cos \frac{\frac{10}{8} \pi}{2}\right)}$

$\cos \left(\frac{10}{8} \pi\right) = \cos \left(\frac{5}{4} \pi\right) = \cos \left(\pi + \frac{1}{4} \pi\right)$

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

$= - 1 \cdot \frac{\sqrt{2}}{2} - 0$

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

$\cos \left(\frac{5}{8} \pi\right) = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

sin(5/8pi)=sqrt((1-sin(10/8pi)/2)

$\sin \left(\frac{10}{8} \pi\right) = \sin \left(\frac{5}{4} \pi\right) = 2 \cdot - \frac{\sqrt{2}}{2} = - \sqrt{2}$

$\sin \left(\frac{5}{8} \pi\right) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \frac{1}{2} \sqrt{2 + \sqrt{2}}$

${e}^{\frac{3}{2} \pi i} = \cos \left(\frac{3}{2} \pi\right) + i \sin \left(\frac{3}{2} \pi\right) = 0 - i$

And finally,

${e}^{\frac{5}{8} \pi i} \cdot {e}^{\frac{3}{2} \pi i} = \left(\frac{\sqrt{2 - \sqrt{2}}}{2} + i \frac{1}{2} \sqrt{2 + \sqrt{2}}\right) \cdot \left(- i\right)$

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