# How do you evaluate  e^( ( 7 pi)/6 i) - e^( ( 17 pi)/8 i) using trigonometric functions?

Jul 22, 2017

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

#### Explanation:

We need

$\cos 2 \theta = 2 {\cos}^{2} \theta - 1$, $\implies$, $\cos \theta = \sqrt{\frac{1 + \cos 2 \theta}{2}}$

$\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$, $\implies$, $\sin \theta = \sqrt{\frac{1 - \cos 2 \theta}{2}}$

We apply Euler's relation

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

${e}^{\frac{7}{6} \pi i} = \cos \left(\frac{7}{6} \pi\right) + i \sin \left(\frac{7}{6} \pi\right)$

${e}^{\frac{17}{8} \pi i} = \cos \left(\frac{17}{8} \pi\right) + i \sin \left(\frac{17}{8} \pi\right)$

So,

$\cos \left(\frac{7}{6} \pi\right) = \cos \left(\pi + \frac{1}{6} \pi\right) = - \cos \left(\frac{1}{6} \pi\right) = - \frac{\sqrt{3}}{2}$

$\sin \left(\frac{7}{6} \pi\right) = \sin \left(\pi + \frac{1}{6} \pi\right) = - \sin \left(\frac{1}{6} \pi\right) = - \frac{1}{2}$

$\cos \left(\frac{17}{8} \pi\right) = \cos \left(2 \pi + \frac{1}{8} \pi\right) = \cos \left(\frac{1}{8} \pi\right) = \sqrt{\frac{1 + \cos \left(\frac{\pi}{4}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\sin \left(\frac{17}{8} \pi\right) = \sin \left(2 \pi + \frac{1}{8} \pi\right) = \sin \left(\frac{1}{8} \pi\right) = \sqrt{\frac{1 - \cos \left(\frac{\pi}{4}\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

Therefore,

${e}^{\frac{7}{6} \pi i} - {e}^{\frac{17}{8} \pi i} = - \frac{\sqrt{3}}{2} - \frac{1}{2} i - \left(\frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{2} i\right)$

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