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

Mar 22, 2016

Let us convert those complex numbers into their trigonometric forms.

#### Explanation:

Let us take the following equality:

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

Thus:

${e}^{\frac{13 \pi}{12} i} = \cos \left(\frac{13 \pi}{12}\right) + i \sin \left(\frac{13 \pi}{12}\right) =$
$= - \frac{1 + \sqrt{3}}{2 \sqrt{2}} - i \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \frac{- 1 - \sqrt{3} - i \sqrt{3} + i}{2 \sqrt{2}} =$
$= \frac{- \left(1 + \sqrt{3}\right) + i \left(1 - \sqrt{3}\right)}{2 \sqrt{2}}$

In the same way:

${e}^{\frac{13 \pi}{2} i} = \cos \left(\frac{13 \pi}{2}\right) + i \sin \left(\frac{13 \pi}{2}\right) =$
$= 0 + i \cdot 1 = i$

We sum it:

${e}^{\frac{13 \pi}{12} i} - {e}^{\frac{13 \pi}{2} i} = \frac{- \left(1 + \sqrt{3}\right) + i \left(1 - \sqrt{3}\right)}{2 \sqrt{2}} - i =$
$= \frac{- \left(1 + \sqrt{3}\right) + i \left(1 - \sqrt{3}\right) - i \cdot 2 \sqrt{2}}{2 \sqrt{2}} =$
$= \frac{- \left(1 + \sqrt{3}\right) + i \left(1 - \sqrt{3} - 2 \sqrt{2}\right)}{2 \sqrt{2}} \approx$
$\approx - 0.97 - 1.26 i$

Tip: in the same way that we can write exponentials as trigonometric functions, we can write trigonometric functions as exponentials.

$\cos \theta = \frac{{e}^{i \theta} + {e}^{- i \theta}}{2}$

$\sin \theta = \frac{{e}^{i \theta} - {e}^{- i \theta}}{2 i}$