# How do you evaluate  e^( ( 11 pi)/6 i) - e^( ( 17 pi)/12 i) using trigonometric functions?

Oct 30, 2016

${e}^{11 \frac{\pi}{6} i} - {e}^{\frac{17 \pi}{12} i} = \frac{2 \sqrt{3} + \sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6} - 2}{4}$

#### Explanation:

Evaluating the expression is determined by using Euler's theorem

$\textcolor{b l u e}{{e}^{i x} = \cos x + i \sin x}$

e^(11pi/6i)=color(blue)(cos(11pi/6)+isin(11pi/6)
${e}^{11 \frac{\pi}{6} i} = \cos \left(\frac{12 \pi}{6} - \frac{\pi}{6}\right) + i \sin \left(\frac{12 \pi}{6} - \frac{\pi}{6}\right)$
${e}^{11 \frac{\pi}{6} i} = \cos \left(2 \pi - \frac{\pi}{6}\right) + i \sin \left(2 \pi - \frac{\pi}{6}\right)$
${e}^{11 \frac{\pi}{6} i} = \cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)$

Applying the trigonometric identities:

$\textcolor{g r e e n}{\cos \left(- \alpha\right) = \cos \alpha}$
$\textcolor{g r e e n}{\sin \left(- \alpha\right) = - \sin \alpha}$

Let us continue computing ${e}^{\frac{11 \pi}{6}}$

${e}^{11 \frac{\pi}{6} i} = \cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)$
${e}^{11 \frac{\pi}{6} i} = \textcolor{g r e e n}{\cos \left(\frac{\pi}{6}\right)} \textcolor{g r e e n}{- i \sin \left(\frac{\pi}{6}\right)}$
${e}^{11 \frac{\pi}{6} i} = \frac{\sqrt{3}}{2} - i \frac{1}{2}$

Let us compute ${e}^{\frac{17 \pi}{12} i}$

e^((17pi)/12i)=color(blue)(cos((17pi)/12)+isin((17pi)/12)
${e}^{\frac{17 \pi}{12} i} = \cos \left(\frac{12 \pi}{12} + \frac{5 \pi}{12}\right) + i \sin \left(\frac{12 \pi}{12} + \frac{5 \pi}{12}\right)$
${e}^{\frac{17 \pi}{12} i} = \cos \left(\pi + \frac{5 \pi}{12}\right) + i \sin \left(\pi + \frac{5 \pi}{12}\right)$

Applying the trigonometric identities

$\textcolor{b r o w n}{\cos \left(\pi + \alpha\right) = - \cos \alpha}$
$\textcolor{b r o w n}{\sin \left(\pi + \alpha\right) = - \sin \alpha}$

Let us continue computing ${e}^{\frac{17 \pi}{12} i}$
${e}^{\frac{17 \pi}{12} i} = \cos \left(\pi + \frac{5 \pi}{12}\right) + i \sin \left(\pi + \frac{5 \pi}{12}\right)$
e^((17pi)/12i)=color(brown)(-cos((5pi)/12)color(brown)-isin((5pi)/12)
${e}^{\frac{17 \pi}{12} i} = - \frac{\sqrt{6} - \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4}$
${e}^{\frac{17 \pi}{12} i} = \frac{- \sqrt{6} + \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4}$

hint:To evaluate $\cos \left(\frac{5 \pi}{12}\right) \mathmr{and} \sin \left(\frac{5 \pi}{12}\right)$
$\cos \left(\frac{5 \pi}{12}\right) = \cos \left(\frac{\pi}{6} + \frac{\pi}{4}\right)$
$\sin \left(\frac{5 \pi}{12}\right) = \sin \left(\frac{\pi}{6} + \frac{\pi}{4}\right)$

So,

color(blue)(e^(11pi/6i)-e^((17pi)/12i)

$= \frac{\sqrt{3}}{2} - i \frac{1}{2} - \left(\frac{- \sqrt{6} + \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4}\right)$

$= \frac{\sqrt{3}}{2} - i \frac{1}{2} - \left(\frac{- \sqrt{6} + \sqrt{2}}{4}\right) + i \frac{\sqrt{2} + \sqrt{6}}{4}$

$= \frac{\sqrt{3}}{2} - i \frac{1}{2} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4}$

$= \frac{2 \sqrt{3}}{4} - \frac{2 i}{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4}$

$= \frac{2 \sqrt{3}}{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4} - \frac{2 i}{4}$

$= \frac{2 \sqrt{3} + \sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6} - 2}{4}$