# How do you evaluate  e^( ( 13 pi)/8 i) - e^( ( 7 pi)/12 i) using trigonometric functions?

Mar 28, 2016

${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{7 \pi}{12} i} = \left(0.6415 + 0.042 i\right)$

#### Explanation:

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

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

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

Hence, ${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{7 \pi}{12} i}$

= $\left(\cos \left(\frac{13 \pi}{8}\right) + i \sin \left(\frac{13 \pi}{8}\right)\right) - \left(\cos \left(\frac{7 \pi}{12}\right) + i \sin \left(\frac{7 \pi}{12}\right)\right)$

As $\cos \left(13 \frac{\pi}{8}\right) = 0.3827$, $\sin \left(13 \frac{\pi}{8}\right) = - 0.9239$,

$\cos \left(7 \frac{\pi}{12}\right) = - 0.2588$ and $\sin \left(\frac{7 \pi}{12}\right) = - 0.9659$

${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{7 \pi}{12} i}$

= (0.3827+i(-0.9239))-((-0.2588)+i(-0.9659)

= $\left(0.6415 + 0.042 i\right)$