# How do you evaluate  e^( ( pi)/2 i) - e^( ( 23 pi)/3 i) using trigonometric functions?

Dec 17, 2016

With the help of Euler's formula we can take this right down to
0.5 + $i$ (1 + $\frac{\sqrt{3}}{2}$)

#### Explanation:

Start with Euler's formula: ${e}^{i x}$ = $\cos x + i S \in x$

cos $\frac{\pi}{2}$ + $i$ sin $\frac{\pi}{2}$ - cos $\frac{23 \pi}{3}$ - $i$ sin $\frac{23 \pi}{3}$

To simplify, note cos $\frac{\pi}{2}$ = 0, sin $\frac{\pi}{2}$ = 1

and since each complete revolution of a unit circle is 2$\pi$, which we will write as $\frac{6 \pi}{3}$, cos $\frac{23 \pi}{3}$ = cos $\frac{5 \pi}{3}$
and likewise for sin $\frac{23 \pi}{3}$ = sin $\frac{5 \pi}{3}$

So, we now have the original expression simplified down to

$i$ - cos $\frac{5 \pi}{3}$ - $i$ sin $\frac{5 \pi}{3}$

where cos $\frac{5 \pi}{3}$ = 0.5 and sin $\frac{5 \pi}{3}$ = - $\left(\frac{\sqrt{3}}{2}\right)$

So, finally, the result is

0.5 + $i$ (1 + $\frac{\sqrt{3}}{2}$)