# How do you evaluate e^( ( pi)/12 i) - e^( ( 9 pi)/8 i) using trigonometric functions?

Jul 16, 2017

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

#### Explanation:

We apply Euler's relation

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

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

${e}^{\frac{9}{8} \pi i} = \cos \left(\frac{9}{8} \pi\right) + i \sin \left(\frac{9}{8} \pi\right)$

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

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

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

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

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

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

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

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

$\cos \left(\frac{9}{8} \pi\right) = \cos \left(\pi + \frac{1}{8} \pi\right)$

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

$= - 1 \cdot \cos \left(\frac{1}{8} \pi\right) - 0 \cdot \cos \left(\frac{1}{8} \pi\right)$

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

$C o s \left(2 x\right) = 2 {\cos}^{2} x - 1$

${\cos}^{2} x = \frac{1 + \cos \left(2 x\right)}{2}$

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

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

$\cos \left(\frac{9}{8} \pi\right) = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\sin \left(\frac{9}{8} \pi\right) = \sin \left(\pi + \frac{1}{8} \pi\right) = \sin \left(\pi\right) \cos \left(\frac{1}{8} \pi\right) + \sin \left(\frac{1}{8}\right) \pi \cos \left(\pi\right)$

$= 0 - \sin \left(\frac{1}{8} \pi\right)$

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

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

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

$\sin \left(\frac{9}{8} \pi\right) = - \frac{\sqrt{2 - \sqrt{2}}}{2}$

Therefore,

${e}^{\frac{1}{12} \pi i} - {e}^{\frac{9}{8} \pi i} = \cos \left(\frac{1}{12} \pi\right) + i \sin \left(\frac{1}{12} \pi\right) - \cos \left(\frac{9}{8} \pi\right) + i \sin \left(\frac{9}{8} \pi\right)$

$= \frac{\sqrt{2} + \sqrt{6}}{4} + \frac{\sqrt{2 + \sqrt{2}}}{2} + i \left(\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{2 - \sqrt{2}}}{2}\right)$