# How do you evaluate e^( ( 23 pi)/12 i) - e^( ( 13 pi)/2 i) using trigonometric functions?

Mar 24, 2017

$\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} - i \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} - i$

#### Explanation:

Euler's Identity states that ${e}^{i x} = \cos x + i \sin x$, and from that ${e}^{\frac{23 \pi}{12} i} = \cos \left(\frac{23 \pi}{12}\right) + i \sin \left(\frac{23 \pi}{12}\right)$ and ${e}^{\frac{13 \pi}{2} i} = \cos \left(\frac{13 \pi}{2}\right) + i \sin \left(\frac{13 \pi}{2}\right)$.

Substituting these into your problem, we have $\cos \left(\frac{23 \pi}{12}\right) + i \sin \left(\frac{23 \pi}{12}\right) - \cos \left(\frac{13 \pi}{2}\right) - i \sin \left(\frac{13 \pi}{2}\right)$. If you can use a calculator then you can stop here to find the values of these functions.

If you need an exact answer there's a bit farther to go in order to put these functions in terms of special angles.

Since $\sin \left(\theta\right) = \sin \left(\theta + 2 n \pi\right)$ and $\cos \left(\theta\right) = \cos \left(\theta + 2 n \pi\right)$ where $n \in \mathbb{Z}$ we can say $n = \text{-} 6$ and change our expression to $\cos \left(\frac{23 \pi}{12}\right) + i \sin \left(\frac{23 \pi}{12}\right) - \cos \left(\frac{\pi}{2}\right) - i \sin \left(\frac{\pi}{2}\right)$

Since sin(2npi-theta)=sin("-"theta)="-"sin(theta) and $\cos \left(2 n \pi - \theta\right) = \cos \left(\text{-} \theta\right) = \cos \left(\theta\right)$ where $n \in \mathbb{Z}$ we can say $n = 1$ and change our expression to $\cos \left(\frac{\pi}{12}\right) - i \sin \left(\frac{\pi}{12}\right) - \cos \left(\frac{\pi}{2}\right) - i \sin \left(\frac{\pi}{2}\right)$.

Finally, since $\frac{\pi}{12} = \frac{\frac{\pi}{6}}{2}$, our expression is $\cos \left(\frac{\frac{\pi}{6}}{2}\right) - i \sin \left(\frac{\frac{\pi}{6}}{2}\right) - \cos \left(\frac{\pi}{2}\right) - i \sin \left(\frac{\pi}{2}\right)$ and we can use the half-angle formulas $\sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(\theta\right)}{2}}$ and $\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \left(\theta\right)}{2}}$ leaving us with $\pm \sqrt{\frac{1 + \cos \left(\frac{\pi}{6}\right)}{2}} - i \left(\pm \sqrt{\frac{1 - \cos \left(\frac{\pi}{6}\right)}{2}}\right) - \cos \left(\frac{\pi}{2}\right) - i \sin \left(\frac{\pi}{2}\right)$. Ugly, I know, but easy since $\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$, $\cos \left(\frac{\pi}{2}\right) = 0$, and $\sin \left(\frac{\pi}{2}\right) = 1$.

"+"sqrt((1+sqrt3/2)/2)-i("+"sqrt((1-sqrt3/2)/2))-0-i so the answer is $\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} - i \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} - i$