# How do you evaluate  e^( ( 5 pi)/4 i) - e^( ( 17 pi)/12 i) using trigonometric functions?

Jul 6, 2017

The answer is $= \frac{\sqrt{6} - 3 \sqrt{2}}{4} - i \frac{\sqrt{6} - \sqrt{2}}{4}$

#### Explanation:

We need

$\cos \left(a + b\right) = \cos a \cos b - \sin a \sin b$

$\sin \left(a + b\right) = \sin a \cos b + \sin b \cos a$

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

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

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

We apply Euler's Formula

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

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

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

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

${e}^{\frac{17}{12} \pi i} = \cos \left(\frac{17}{12} \pi\right) + i \sin \left(\frac{17}{12} \pi\right)$

$= \cos \left(\frac{15}{12} \pi + \frac{2}{12} \pi\right) + i \sin \left(\frac{15}{12} \pi + \frac{2}{12} \pi\right)$

$= \cos \left(\frac{5}{4} \pi\right) \cos \left(\frac{1}{6} \pi\right) - \sin \left(\frac{5}{4} \pi\right) \sin \left(\frac{1}{6} \pi\right) + i \left(\sin \left(\frac{5}{4} \pi\right) \cos \left(\frac{1}{6} \pi\right) + \cos \left(\frac{5}{4} \pi\right) \sin \left(\frac{1}{6} \pi\right)\right)$

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

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

Therefore,

${e}^{\frac{5}{4} \pi i} - {e}^{\frac{17}{12} \pi i} = - \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} - \left(\frac{\sqrt{2} - \sqrt{6}}{4} + i \frac{- \sqrt{6} - \sqrt{2}}{4}\right)$

$= \frac{\sqrt{6} - 3 \sqrt{2}}{4} - i \frac{\sqrt{6} - \sqrt{2}}{4}$