# How do you evaluate  e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12 i) using trigonometric functions?

Jun 3, 2016

$= \frac{1}{2 \sqrt{2}} \left(\left(\sqrt{3} + 1\right) - i \left(\sqrt{3} - 1\right)\right)$

#### Explanation:

${e}^{\frac{5 \pi}{4} i} - {e}^{\frac{19 \pi}{12} i}$

$= \left[\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right] - \left[\cos \left(\frac{19 \pi}{12}\right) + i \sin \left(\frac{19 \pi}{12}\right)\right]$

$= \left[\cos \left(\frac{5 \pi}{4}\right) - \cos \left(\frac{19 \pi}{12}\right)\right] - i \left[\sin \left(\frac{19 \pi}{12}\right) - \sin \left(\frac{5 \pi}{4}\right)\right]$

$= \left[\cos \left(\frac{15 \pi}{12}\right) - \cos \left(\frac{19 \pi}{12}\right)\right] - i \left[\sin \left(\frac{19 \pi}{12}\right) - \sin \left(\frac{15 \pi}{12}\right)\right]$

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

$= \left[2 \sin \left(\frac{17 \pi}{12}\right) \times \frac{1}{2}\right] - i \left[2 \cos \left(\frac{17 \pi}{12}\right) \times \frac{1}{2}\right]$

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

Now

• sin((17pi)/12)=sqrt(1/2(1-cos((17pi)/6)))=sqrt(1/2(1-cos(3pi-pi/6))
$= \sqrt{\frac{1}{2} \left(1 + \cos \left(\frac{\pi}{6}\right)\right)} = \sqrt{\frac{1}{2} \left(1 + \cos \left(\frac{\pi}{6}\right)\right)}$

$= \sqrt{\frac{2 + \sqrt{3}}{4}} = \sqrt{\frac{4 + 2 \sqrt{3}}{8}} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

and

• cos((17pi)/12)=sqrt(1/2(1+cos((17pi)/6)))=sqrt(1/2(1+cos(3pi-pi/6))

$= \sqrt{\frac{1}{2} \left(1 - \cos \left(\frac{\pi}{6}\right)\right)} = \sqrt{\frac{1}{2} \left(1 - \cos \left(\frac{\pi}{6}\right)\right)}$

$= \sqrt{\frac{2 - \sqrt{3}}{4}} = \sqrt{\frac{4 - 2 \sqrt{3}}{8}} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

Hence

${e}^{\frac{5 \pi}{4} i} - {e}^{\frac{19 \pi}{12} i}$

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

$= \frac{1}{2 \sqrt{2}} \left(\left(\sqrt{3} + 1\right) - i \left(\sqrt{3} - 1\right)\right)$