# How do you evaluate  e^( (3 pi)/2 i) - e^( ( pi)/8 i) using trigonometric functions?

Dec 24, 2017

The answer is $= - \frac{1}{2} \sqrt{\sqrt{2} + 2} - \frac{i}{2} \sqrt{2 - \sqrt{2}} - i$

#### Explanation:

Apply Euler's Identity

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

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

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

Therefore,

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

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

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

$\cos \left(\frac{3}{2} \pi\right) = 0$

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

So,

${e}^{\frac{3}{2} \pi i} - {e}^{\frac{1}{8} \pi i} = 0 - i - \frac{1}{2} \sqrt{\sqrt{2} + 2} - i \frac{1}{2} \sqrt{2 - \sqrt{2}}$