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

$- 2 i$
${e}^{i x} = \cos x + i \sin x$.
$\cos \left(3 \frac{\pi}{2}\right) = \cos \left(\frac{\pi}{2}\right) = 0$
$\sin \left(3 \frac{\pi}{2}\right) = \sin \left(\pi + \frac{\pi}{2}\right) = - \sin \left(\frac{\pi}{2}\right) = - 1$ and $\sin \left(\frac{\pi}{2}\right) = 1$
So the expression simplifies to $- i - i = - 2 i$.(