How do you evaluate e^( ( 23 pi)/12 i) - e^( ( 13 pi)/12 i) using trigonometric functions?

Apr 23, 2018

${e}^{i \left(\frac{23 \pi}{12}\right)} - {e}^{i \left(\frac{13 \setminus \pi}{12}\right)} = \frac{\sqrt{6} + \sqrt{2}}{2}$

Explanation:

${e}^{i \left(\frac{23 \pi}{12}\right)} - {e}^{i \left(\frac{13 \setminus \pi}{12}\right)}$

$= {e}^{- 2 \pi i} {e}^{i \left(\frac{23 \pi}{12}\right)} - {e}^{i \pi} {e}^{i \left(\frac{\setminus \pi}{12}\right)}$

$= {e}^{i \left(- \frac{\pi}{12}\right)} + {e}^{i \left(\frac{\setminus \pi}{12}\right)}$

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

$= 2 \setminus \cos \left(\frac{\pi}{12}\right)$

$\frac{\pi}{12}$ is ${15}^{\circ}$. We avoid the nested radical with the difference formula:

$\cos \left(\frac{\pi}{12}\right) = \cos \left({45}^{\circ} - {30}^{\circ}\right) = \cos {45}^{\circ} \cos {30}^{\circ} + \sin {45}^{\circ} \sin {30}^{\circ}$$= \setminus \frac{\sqrt{2}}{2} \left(\setminus \frac{\sqrt{3}}{2} + \frac{1}{2}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$

${e}^{i \left(\frac{23 \pi}{12}\right)} - {e}^{i \left(\frac{13 \setminus \pi}{12}\right)} = \frac{\sqrt{6} + \sqrt{2}}{2}$