# How do you evaluate e^( ( 13 pi)/8 i) - e^( ( 5 pi)/4 i) using trigonometric functions?

Dec 18, 2017

Euler's theorem: ${e}^{i \theta} = \cos \theta - i \sin \theta$

#### Explanation:

${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{5 \pi}{4} i}$
$= \left(\cos \left(\frac{13 \pi}{8}\right) + i \sin \left(\frac{13 \pi}{8}\right)\right) - \left(\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right)$
$= \sin \left(\frac{\pi}{8}\right) - i \cos \left(\frac{\pi}{8}\right) - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i$
$\cong - 0.32442 - 0.21677 i$

Dec 18, 2017

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

#### Explanation:

Reminder :

Euler's relation

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

Here, we have

$z = {e}^{\frac{13}{8} \pi} - {e}^{\frac{5}{4} \pi} = \cos \left(\frac{13}{8} \pi\right) + i \sin \left(\frac{13}{8} \pi\right) - \cos \left(\frac{5}{4} \pi\right) - i \sin \left(\frac{5}{4} \pi\right)$

Therefore,

$\frac{13}{8} \pi = \frac{5}{8} \pi + \pi = \frac{1}{8} \pi + \frac{3}{2} \pi$

$\frac{5}{4} \pi = \frac{1}{4} \pi + \pi$

$\cos \left(\frac{1}{4} \pi\right) = 1 - 2 {\sin}^{2} \left(\frac{1}{8} \pi\right) = 2 {\cos}^{2} \left(\frac{1}{8} \pi\right) - 1$

$\sin \left(\frac{1}{8} \pi\right) = \sqrt{\frac{1 - \cos \left(\frac{1}{4} \pi\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

$\cos \left(\frac{1}{8} \pi\right) = \sqrt{\frac{1 + \cos \left(\frac{1}{4} \pi\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$#

So,
$z = \cos \left(\frac{1}{8} \pi + \frac{3}{2} \pi\right) + i \sin \left(\frac{1}{8} \pi + \frac{3}{2} \pi\right) - \cos \left(\frac{1}{4} \pi + \pi\right) - i \sin \left(\frac{1}{4} \pi + \pi\right)$

$\cos \left(\frac{1}{8} \pi + \frac{3}{2} \pi\right) = \cos \left(\frac{1}{8} \pi\right) \cos \left(\frac{3}{2} \pi\right) - \sin \left(\frac{1}{8} \pi\right) \sin \left(\frac{3}{2} \pi\right)$

$= \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot 0 - \frac{\sqrt{2 - \sqrt{2}}}{2} \cdot \left(- 1\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$

$\sin \left(\frac{1}{8} \pi + \frac{3}{2} \pi\right) = \sin \left(\frac{1}{8} \pi\right) \cos \left(\frac{3}{2} \pi\right) + \cos \left(\frac{1}{8} \pi\right) \sin \left(\frac{3}{2} \pi\right)$

$= \frac{\sqrt{2 - \sqrt{2}}}{2} \cdot 0 + \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot \left(- 1\right) = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

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

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

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

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

Finally,

$z = \frac{\sqrt{2 - \sqrt{2}}}{2} - i \frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$

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