How do you evaluate e^( ( 23 pi)/12 i) - e^( ( 17 pi)/12 i) using trigonometric functions?

Jan 20, 2017

Possibly it is $\left(\cos \alpha + \sin \alpha\right) + i \left(\cos \alpha - \sin \alpha\right)$ where $\alpha = \frac{\pi}{12}$

Explanation:

Noting that ${e}^{\frac{24 \pi i}{12}} = 1$ and ${e}^{\frac{18 \pi i}{12}} = - i$ we take out the common factor ${e}^{\frac{- \pi i}{12}}$:

${e}^{\frac{23 \pi}{12} i} - {e}^{\frac{27 \pi}{12} i}$
$= {e}^{\frac{24 - 1}{12} \pi i} - {e}^{\frac{18 - 1}{12} \pi i}$
$= {e}^{- \frac{\pi}{12} i} \left({e}^{\frac{24}{12} \pi i} - {e}^{\frac{18}{12} \pi i}\right)$
=e^(-pi/12 i)(1-i))
$= \left(\cos \left(\frac{\pi}{12}\right) + \sin \left(\frac{\pi}{12}\right)\right) + i \left(\cos \left(\frac{\pi}{12}\right) - \sin \left(\frac{\pi}{12}\right)\right)$
or something like that.