# How do you evaluate  e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i) using trigonometric functions?

##### 1 Answer
Apr 9, 2016

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

#### Explanation:

${e}^{i x} = \cos x + i \sin x$.
$\cos \left(2 \pi - x\right) = \cos x , \sin \left(2 \pi - x\right) = - \sin x , \cos \left(\pi + x\right) = - \cos x \mathmr{and} \sin \left(\pi + x\right) = - \sin x$.

So, ${e}^{11 \frac{\pi}{6} i} = \cos \left(11 \frac{\pi}{6}\right) + i \sin \left(11 \frac{\pi}{6}\right) = \cos \left(2 \pi - \frac{\pi}{6}\right) + i \sin \left(2 \pi - \frac{\pi}{6}\right) = \cos \left(\frac{\pi}{6}\right) - i \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} - i \left(\frac{1}{2}\right)$.
${e}^{9 \frac{\pi}{8}} = \cos \left(9 \frac{\pi}{8}\right) + i \sin \left(9 \frac{\pi}{8}\right) = \cos \left(\pi + \frac{\pi}{8}\right) + i \sin \left(\pi + \frac{\pi}{8}\right) = - \cos \left(\frac{\pi}{8}\right) - i \sin \left(\frac{\pi}{8}\right)$.
$\cos A = \sqrt{\frac{1 + \cos 2 A}{2}} , \sin A = \sqrt{\frac{1 - \cos 2 A}{2}}$

So, $\cos \left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \cos \left(\frac{\pi}{4}\right)}{2}} = \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}}$.
$\sin \left(\frac{\pi}{8}\right) = \sqrt{\frac{1 - \cos \left(\frac{\pi}{4}\right)}{2}} = \sqrt{\frac{\sqrt{2} - 1}{2 \sqrt{2}}}$.

Npw, ${e}^{9 \frac{\pi}{8}} = \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}} - i \sqrt{\frac{\sqrt{2} - 1}{2 \sqrt{2}}}$.

So, the given expression =$\left(\frac{\sqrt{3}}{2} + \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}}\right) - i \left(\frac{1}{2} - \sqrt{\frac{\sqrt{2} - 1}{2 \sqrt{2}}}\right)$