# How do you evaluate  e^( ( pi)/2 i) - e^( ( 19 pi)/12 i) using trigonometric functions?

Apr 27, 2016

I get this.

Interestingly enough, you can use Euler's formula, which states:

$\setminus m a t h b f \left({e}^{i x} = i \sin x + \cos x\right)$

Thus, you can substitute $x = \frac{\pi}{2}$ or $\frac{19 \pi}{12}$ to give:

${e}^{i \cdot \pi \text{/"2) - e^(i*19pi"/} 12}$

$= \left[i \sin \left(\frac{\pi}{2}\right) + \cos \left(\frac{\pi}{2}\right)\right] - \left[i \sin \left(\frac{19 \pi}{12}\right) + \cos \left(\frac{19 \pi}{12}\right)\right]$

We know that $\frac{19 \pi}{12} = {\left[\frac{19 \cdot 180}{12}\right]}^{\circ} = {285}^{\circ}$, so...

$= \textcolor{g r e e n}{\left[i \sin \left({90}^{\circ}\right) + \cos \left({90}^{\circ}\right)\right] - \left[i \sin \left({285}^{\circ}\right) + \cos \left({285}^{\circ}\right)\right]}$

Okay, so what is $\sin$ or $\cos \left({285}^{\circ}\right)$? We could use some trig relationships to get:

$\sin \left({285}^{\circ}\right) = \sin \left(- {75}^{\circ}\right) = - \sin \left({75}^{\circ}\right) = - \sin \left({30}^{\circ} + {45}^{\circ}\right)$

Using the addition identity $\sin \left(u + v\right) = \sin u \cos v + \cos u \sin v$,

$\implies - \left[\sin {30}^{\circ} \cos {45}^{\circ} + \cos {30}^{\circ} \sin {45}^{\circ}\right]$

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

$= \frac{- \sqrt{2} - \sqrt{6}}{4} \approx - 0.9659$

Then, consider $\cos \left({285}^{\circ}\right)$ to be:

$\cos \left({285}^{\circ}\right) = \cos \left(- {285}^{\circ}\right) = \cos \left({75}^{\circ}\right) = \cos \left({30}^{\circ} + {45}^{\circ}\right)$

Using the addition identity $\cos \left(u + v\right) = \cos u \cos v - \sin u \sin v$,

$\implies \cos {30}^{\circ} \cos {45}^{\circ} - \sin {30}^{\circ} \sin {45}^{\circ}$

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

$= \frac{\sqrt{6} - \sqrt{2}}{4} \approx 0.2588$

So overall, we have:

$\textcolor{b l u e}{{e}^{i \cdot \pi \text{/"2) - e^(i*19pi"/} 12}}$

$= \left[i \cancel{\sin \left({90}^{\circ}\right)} + {\cancel{\cos \left({90}^{\circ}\right)}}^{0}\right] - \left[i \sin \left({285}^{\circ}\right) + \cos \left({285}^{\circ}\right)\right]$

$= i - i \left[\frac{- \sqrt{2} - \sqrt{6}}{4}\right] - \left[\frac{\sqrt{6} - \sqrt{2}}{4}\right]$

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

$= \textcolor{b l u e}{- \frac{\sqrt{6} - \sqrt{2}}{4} + \left(1 + \frac{\sqrt{2} + \sqrt{6}}{4}\right) i}$

$\approx \textcolor{b l u e}{- 0.2588 + 1.9659 i}$