# How do you evaluate  e^( ( 15 pi)/8 i) - e^( ( 11 pi)/6 i) using trigonometric functions?

Mar 28, 2016

${e}^{\frac{15 \pi}{8} i} - {e}^{\frac{11 \pi}{6} i} = - 1.7899 + 0.8827 i$

#### Explanation:

As ${e}^{i \theta} = \cos \theta + i \sin \theta$, we have

${e}^{\frac{15 \pi}{8} i} = \cos \left(\frac{15 \pi}{8}\right) + i \sin \left(\frac{15 \pi}{8}\right)$

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

= $- \cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right) = - 0.9239 + 0.3827 i$

${e}^{\frac{11 \pi}{6} i} = \cos \left(\frac{11 \pi}{6}\right) + i \sin \left(\frac{11 \pi}{6}\right)$

= $\cos \left(2 \pi - \frac{\pi}{6}\right) + i \sin \left(2 \pi - \frac{\pi}{6}\right)$

= $\cos \left(\frac{\pi}{6}\right) - i \sin \left(\frac{\pi}{6}\right)$

= $\frac{\sqrt{3}}{2} - i \cdot \frac{1}{2} = 0.8660 - 0.5 i$

Hence ${e}^{\frac{15 \pi}{8} i} - {e}^{\frac{11 \pi}{6} i} = \left(- 0.9239 + 0.3827 i\right) - \left(0.8660 - 0.5 i\right)$

= $\left(- 0.9239 - 0.8660\right) + i \left(0.3827 + 0.5\right)$

= $- 1.7899 + 0.8827 i$