# How do you evaluate  e^( ( 13 pi)/8 i) - e^( (5 pi)/6 i) using trigonometric functions?

Apr 26, 2018

${e}^{i \left(\frac{13 \pi}{8}\right)} - {e}^{i \left(\frac{5 \pi}{6}\right)}$

$= \cos \left(\frac{3 \pi}{8}\right) - i \sin \left(\frac{3 \pi}{8}\right) + \cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)$

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

#### Explanation:

Pet peeve: every trig problem is 30,60,90 or 45.45.90. This one's both.

${e}^{i \left(\frac{13 \pi}{8}\right)} - {e}^{i \left(\frac{5 \pi}{6}\right)}$

$= {e}^{i \left(\frac{13 \pi}{8}\right)} - {e}^{i \left(\frac{5 \pi}{6}\right)}$

$= \cos \left(\frac{13 \setminus \pi}{8}\right) + i \setminus \sin \left(\frac{13 \setminus \pi}{8}\right) - \left(\cos \left(\frac{5 \setminus \pi}{6}\right) + i \setminus \sin \left(\frac{5 \pi}{6}\right)\right)$

$= \cos \left(2 \pi - \frac{13 \pi}{8}\right) - i \sin \left(2 \pi - \frac{13 \pi}{8}\right) - \left(- \cos \left(\pi - \frac{5 \pi}{6}\right) + i \sin \left(\pi - \frac{5 \pi}{6}\right)\right)$

$= \cos \left(\frac{3 \pi}{8}\right) - i \sin \left(\frac{3 \pi}{8}\right) + \cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)$

We've evaluated the expression using trigonometric functions. We can go further and evaluate those trigonometric functions.

Let's start by noting $\frac{3 \pi}{4}$ is ${135}^{\circ}$ so $\frac{3 \pi}{8}$ is ${135}^{\circ} / 2 = {67.5}^{\circ}$

That's in the first quadrant, so we can use the half angle formula with a + sign:

$\cos {67.5}^{\circ} = + \setminus \sqrt{\frac{1}{2} \left(1 + \cos {135}^{\circ}\right)} = \sqrt{\frac{1}{2} \left(1 - \frac{\sqrt{2}}{2}\right)}$

$\sin {67.5}^{\circ} = + \setminus \sqrt{\frac{1}{2} \left(1 - \cos {135}^{\circ}\right)} = \sqrt{\frac{1}{2} \left(1 + \frac{\sqrt{2}}{2}\right)}$

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

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

That may be right. I'll leave it here.