# How do you evaluate e^( ( pi)/12 i) - e^( ( 3pi)/8 i) using trigonometric functions?

##### 1 Answer
Jul 20, 2017

See the explanation below

#### Explanation:

We apply Euler's relation

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

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

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

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

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

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

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

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

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

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

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

$\cos \left(\frac{3}{4} \pi\right) = 1 - 2 {\sin}^{2} \left(\frac{3}{8} \pi\right)$

$2 {\sin}^{2} \left(\frac{3}{8} \pi\right) = 1 - \cos \left(\frac{3}{4} \pi\right) = 1 + \frac{\sqrt{2}}{2}$

$\sin \left(\frac{3}{8} \pi\right) = \sqrt{\frac{1}{2} \left(1 + \frac{\sqrt{2}}{2}\right)} = \frac{1}{2} \sqrt{2 + \sqrt{2}}$

$\cos \left(\frac{3}{4} \pi\right) = 2 {\cos}^{2} \left(\frac{3}{8} \pi\right) - 1$

$2 {\cos}^{2} \left(\frac{3}{8} \pi\right) = 1 + \cos \left(\frac{3}{4} \pi\right) = 1 - \frac{\sqrt{2}}{2}$

$\cos \left(\frac{3}{8} \pi\right) = \frac{1}{2} \sqrt{2 - \sqrt{2}}$

Therefore,

${e}^{\frac{1}{12} \pi i} - {e}^{\frac{3}{8} \pi i} = \cos \left(\frac{1}{12} \pi\right) + i \sin \left(\frac{1}{12} \pi\right) - \cos \left(\frac{3}{8} \pi\right) - i \sin \left(\frac{3}{8} \pi\right)$

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