# How do you evaluate  e^( ( 7 pi)/4 i) - e^( ( 2 pi)/3 i) using trigonometric functions?

Aug 20, 2017

Use Euler's identity...

#### Explanation:

...which I will use, but not prove, here:

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

so, for the first term, we have $x = \frac{7 \pi}{4}$
so the first term can be re-written:

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

and the second can be rewritten:

$\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$

So, plugging it all back in:

$\left(\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)\right) - \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)$

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