# How do you evaluate  e^( ( 19 pi)/12 i) - e^( ( pi)/4 i) using trigonometric functions?

May 19, 2016

${e}^{\frac{19 \pi}{12}} - {e}^{\frac{\pi}{4}} = - 0.4483 - i 1.673$

#### Explanation:

${e}^{\frac{19 \pi}{12} i} = \cos \left(\frac{19 \pi}{12}\right) + i \sin \left(\frac{19 \pi}{12}\right)$ and ${e}^{\frac{\pi}{4} i} = \cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)$

Hence, ${e}^{\frac{19 \pi}{12}} - {e}^{\frac{\pi}{4}} = \cos \left(\frac{19 \pi}{12}\right) + i \sin \left(\frac{19 \pi}{12}\right) - \cos \left(\frac{\pi}{4}\right) - i \sin \left(\frac{\pi}{4}\right)$

= $\cos \left(\frac{- 5 \pi}{12}\right) - \cos \left(\frac{\pi}{4}\right) + i \left(\sin \left(\frac{- 5 \pi}{12}\right) - \sin \left(\frac{\pi}{4}\right)\right)$

= $\cos \left(\frac{5 \pi}{12}\right) - \frac{1}{\sqrt{2}} + i \left(- \sin \left(\frac{5 \pi}{12}\right) - \frac{1}{\sqrt{2}}\right)$

= $0.2588 - 0.7071 + i \left(- 0.9659 - 0.7071\right)$

= $- 0.4483 - i 1.673$