# How do you evaluate  e^( ( 7 pi)/4 i) - e^( ( 5 pi)/12 i) using trigonometric functions?

Aug 22, 2017

The answer is $= \frac{1}{4} \left(3 \sqrt{2} - \sqrt{6}\right) + i \left(\frac{1}{4} \left(3 \sqrt{2} + \sqrt{6}\right)\right)$

#### Explanation:

We apply Euler's formula

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

So,

${e}^{\frac{7}{4} \pi i} - {e}^{\frac{5}{12} \pi i} =$

$\cos \left(\frac{7}{4} \pi\right) + i \sin \left(\frac{7}{4} \pi\right) - \cos \left(\frac{5}{12} \pi\right) - i \sin \left(\frac{5}{12} \pi\right)$

We calculate separately

$\cos \left(\frac{7}{4} \pi\right) = \cos \left(\pi + \frac{3}{4} \pi\right) = \cos \left(\pi\right) \cos \left(\frac{3}{4} \pi\right) - \sin \left(\pi\right) \sin \left(\frac{3}{4} \pi\right)$

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

$\sin \left(\frac{7}{4} \pi\right) = \sin \left(\pi + \frac{3}{4} \pi\right) = \sin \left(\pi\right) \cos \left(\frac{3}{4} \pi\right) + \sin \left(\frac{3}{4} \pi\right) \cos \left(\pi\right)$

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

$\cos \left(\frac{5}{12} \pi\right) = \cos \left(\frac{3}{12} \pi + \frac{2}{12} \pi\right) = \cos \left(\frac{1}{4} \pi + \frac{1}{6} \pi\right) = \cos \left(\frac{1}{4} \pi\right) \cos \left(\frac{1}{6} \pi\right) - \sin \left(\frac{1}{4} \pi\right) \sin \left(\frac{1}{6} \pi\right)$

$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$

$\sin \left(\frac{5}{12} \pi\right) = \sin \left(\frac{3}{12} \pi + \frac{2}{12} \pi\right) = \sin \left(\frac{1}{4} \pi\right) \cos \left(\frac{1}{6} \pi\right) + \sin \left(\frac{1}{6} \pi\right) \cos \left(\frac{1}{4} \pi\right)$

$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right)$

${e}^{\frac{7}{4} \pi i} - {e}^{\frac{5}{12} \pi i} = \left(\frac{\sqrt{2}}{2} - \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)\right) - i \left(\frac{\sqrt{2}}{2} + \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right)\right)$

$= \frac{1}{4} \left(3 \sqrt{2} - \sqrt{6}\right) + i \left(\frac{1}{4} \left(3 \sqrt{2} + \sqrt{6}\right)\right)$