# How do you evaluate  e^( ( 3 pi)/8 i) - e^( ( 11 pi)/6 i) using trigonometric functions?

Feb 23, 2018

${e}^{\frac{3 \pi}{8}} - {e}^{\frac{11 \pi}{16}} = - 2 \sin \left(\frac{\pi}{32}\right) \sin \left(\frac{5 \pi}{32}\right) - 2 i \cos \left(\frac{\pi}{32}\right) \sin \left(\frac{5 \pi}{32}\right)$

#### Explanation:

${e}^{\frac{3 \pi}{8}} - {e}^{\frac{11 \pi}{16}}$
$= \cos \left(\frac{3 \pi}{8}\right) + i \sin \left(\frac{3 \pi}{8}\right) - \left(\cos \left(\frac{11 \pi}{16}\right) + i \sin \left(\frac{11 \pi}{16}\right)\right)$

$= \left(\cos \left(\frac{3 \pi}{8}\right) - \cos \left(\frac{11 \pi}{16}\right)\right) + i \left(\sin \left(\frac{3 \pi}{8}\right) - \sin \left(\frac{11 \pi}{16}\right)\right)$

$- 2 \sin \left(\frac{\frac{3 \pi}{8} + \frac{11 \pi}{16}}{2}\right) \sin \left(\frac{\frac{3 \pi}{8} - \frac{11 \pi}{16}}{2}\right) + i \left(2 \cos \left(\frac{\frac{3 \pi}{8} + \frac{11 \pi}{16}}{2}\right) \sin \left(\frac{\frac{3 \pi}{8} + \frac{11 \pi}{16}}{2}\right)\right)$

$\frac{\frac{3 \pi}{8} + \frac{11 \pi}{16}}{2} = \frac{\pi}{2} \left(\frac{3}{8} + \frac{11}{16}\right) = \frac{\pi}{2} \left(\frac{6 + 11}{16}\right) = 17 \frac{\pi}{32}$

$\frac{17 \pi}{32} = 2 \pi - \frac{\pi}{32}$

$\frac{\frac{3 \pi}{8} + \frac{11 \pi}{16}}{2} = 2 \pi - \frac{\pi}{32}$

$\frac{\frac{3 \pi}{8} - \frac{11 \pi}{16}}{2} = \frac{\pi}{2} \left(\frac{3}{8} - \frac{11}{16}\right) = \frac{\pi}{2} \left(\frac{6 - 11}{16}\right) = - \frac{5 \pi}{32}$

$\frac{- 5 \pi}{32} = 2 \pi - 5 \frac{\pi}{32}$

$\frac{\frac{3 \pi}{8} - \frac{11 \pi}{16}}{2} = 2 \pi - 5 \frac{\pi}{32}$

=-2sin(2pi-pi/32)sin(2pi-(5pi)/32)+i(2cos(2pi-pi/32)sin(2pi-(5pi)/32)

${e}^{\frac{3 \pi}{8}} - {e}^{\frac{11 \pi}{16}} = - 2 \sin \left(\frac{\pi}{32}\right) \sin \left(\frac{5 \pi}{32}\right) - 2 i \cos \left(\frac{\pi}{32}\right) \sin \left(\frac{5 \pi}{32}\right)$