# How do you evaluate  e^( ( 13 pi)/8 i) - e^( (5 pi)/12 i) using trigonometric functions?

##### 1 Answer
Jan 25, 2018

${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{5 \pi}{12} i} \approx 0.12 - 1.89 i$

#### Explanation:

We can represent $a {e}^{i x}$ in trig form as $a {e}^{i x} = a \left(\cos x + i \sin x\right)$

Using this for ${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{5 \pi}{12} i}$ gives us:
$\left(\cos \left(\frac{13 \pi}{8}\right) + i \sin \left(\frac{13 \pi}{8}\right)\right) - \left(\cos \left(\frac{5 \pi}{12}\right) + i \sin \left(\frac{5 \pi}{12}\right)\right)$

=cos((13pi)/8)+isin((13pi)/8)-cos((5pi)/12)-isin((5pi)/12))

$= \cos \left(\frac{13 \pi}{8}\right) - \cos \left(\frac{5 \pi}{12}\right) - i \sin \left(\frac{5 \pi}{12}\right) + i \sin \left(\frac{13 \pi}{8}\right)$

$= \left(\cos \left(\frac{13 \pi}{8}\right) - \cos \left(\frac{5 \pi}{12}\right)\right) - i \left(\sin \left(\frac{5 \pi}{12}\right) - \sin \left(\frac{13 \pi}{8}\right)\right)$

$\approx 0.1238643873 - i \left(1.889805359\right)$

$= 0.1238643873 - 1.889805359 i$

$\approx 0.12 - 1.89 i$