# How do you evaluate  e^( ( 13 pi)/8 i) - e^( (17 pi)/12 i) using trigonometric functions?

Mar 26, 2016

${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{17 \pi}{12} i} = 0.0003 + 0.01142 i$#

#### Explanation:

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

${e}^{\frac{13 \pi}{8} i} = \cos \left(\frac{13 \pi}{8}\right) + i \sin \left(\frac{13 \pi}{8}\right)$

= $\cos \left(\frac{13 \pi}{8} - 2 \pi\right) + i \sin \left(\frac{13 \pi}{8} - 2 \pi\right)$

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

= $0.99979 - 0.02056 i$ (using scientific calculator)

Similarly ${e}^{\frac{17 \pi}{12} i} = \cos \left(\frac{17 \pi}{12}\right) + i \sin \left(\frac{17 \pi}{12}\right)$

= $\cos \left(\frac{17 \pi}{12} - 2 \pi\right) + i \sin \left(\frac{17 \pi}{12} - 2 \pi\right)$

= $\cos \left(\frac{- 7 \pi}{12}\right) + i \sin \left(\frac{- 7 \pi}{12}\right)$

= $0.99949 - 0.03198 i$ (using scientific calculator)

Hence ${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{17 \pi}{12} i}$

= $\left(0.99979 - 0.02056 i\right) - \left(0.99949 - 0.03198 i\right)$

= $\left(0.99979 - 0.99949\right) - \left(0.02056 - 0.03198\right) i$

= $0.0003 - \left(- 0.01142\right) i$

= $0.0003 + 0.01142 i$