# How do you evaluate e^( ( 23 pi)/12 i) - e^( ( 7 pi)/12 i) using trigonometric functions?

Jul 22, 2017

${e}^{\frac{23 \pi}{12} i} - {e}^{\frac{7 \pi}{12} i} = \frac{\sqrt{6}}{2} - i \frac{\sqrt{6}}{2}$

#### Explanation:

${e}^{x \cdot i} = \cos \left(x\right) + i \sin \left(x\right)$

${e}^{\frac{23 \pi}{12} i} - {e}^{\frac{7 \pi}{12} i} = \left(\cos \left(\frac{23 \pi}{12}\right) + i \sin \left(\frac{23 \pi}{12}\right)\right) - \left(\cos \left(\frac{7 \pi}{12}\right) + i \sin \left(\frac{7 \pi}{12}\right)\right)$

$\cos \left(\frac{23 \pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$
$\sin \left(\frac{23 \pi}{12}\right) = \frac{- \sqrt{6} + \sqrt{2}}{4}$
$\cos \left(\frac{7 \pi}{12}\right) = \frac{- \sqrt{6} + \sqrt{2}}{4}$
$\sin \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$

${e}^{\frac{23 \pi}{12} i} - {e}^{\frac{7 \pi}{12} i} = \left(\frac{\sqrt{6} + \sqrt{2}}{4} + i \frac{- \sqrt{6} + \sqrt{2}}{4}\right) - \left(\frac{- \sqrt{6} + \sqrt{2}}{4} + i \frac{\sqrt{6} + \sqrt{2}}{4}\right)$

$\frac{\sqrt{6} + \sqrt{2}}{4} - \frac{- \sqrt{6} + \sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{6} + \sqrt{2} - \sqrt{2}}{4} = \frac{2 \sqrt{6}}{4} = \frac{\sqrt{6}}{2}$

$\frac{- \sqrt{6} + \sqrt{2}}{4} - \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{- \sqrt{6} - \sqrt{6} + \sqrt{2} - \sqrt{2}}{4} = \frac{- 2 \sqrt{6}}{4} = - \frac{\sqrt{6}}{2}$

${e}^{\frac{23 \pi}{12} i} - {e}^{\frac{7 \pi}{12} i} = \frac{\sqrt{6}}{2} - i \frac{\sqrt{6}}{2}$