# How do you multiply e^(( 17 pi )/ 12 i) * e^( 3 pi/2 i )  in trigonometric form?

Mar 29, 2016

${e}^{i \frac{17}{12} \pi} . {e}^{i \frac{3}{2} \pi} = \cos \left(\frac{5}{12} \pi\right) + i \sin \left(\frac{5}{12} \pi\right)$

#### Explanation:

Recall the Trigonomic identities:

$\sin \left(x\right) = - \sin \left(- x\right)$
$2 \setminus \cos \setminus \theta \setminus \cos \setminus \varphi = \setminus \cos \left(\setminus \theta - \setminus \varphi\right) + \setminus \cos \left(\setminus \theta + \setminus \varphi\right)$
$2 \setminus \sin \setminus \theta \setminus \sin \setminus \varphi = \setminus \cos \left(\setminus \theta - \setminus \varphi\right) - \setminus \cos \left(\setminus \theta + \setminus \varphi\right)$
$2 \setminus \sin \setminus \theta \setminus \cos \setminus \varphi = \setminus \sin \left(\setminus \theta + \setminus \varphi\right) + \setminus \sin \left(\setminus \theta - \setminus \varphi\right)$

In Trigonomic form the following expression gives:

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

Using these in the given values we have:
${e}^{i \frac{17}{12} \pi} . {e}^{i \frac{3}{2} \pi} = \left(\cos \left(\frac{17}{12} \pi\right) + i \sin \left(\frac{17}{2} \pi\right)\right) \left(\cos \left(\frac{3}{2} \pi\right) + i \sin \left(\frac{3}{2} \pi\right)\right)$
$= \cos \left(\frac{17}{12} \pi\right) \cos \left(\frac{3}{2} \pi\right) + i \sin \left(\frac{17}{2} \pi\right) \cos \left(\frac{3}{2} \pi\right) + i \cos \left(\frac{17}{12} \pi\right) \sin \left(\frac{3}{2} \pi\right) - \sin \left(\frac{17}{2} \pi\right) \sin \left(\frac{3}{2} \pi\right)$

Using the identities this simplifies to:
$= \frac{1}{2} \left(\cos \left(\frac{53}{12} \pi\right) + \cos \left(\frac{17}{2} \pi\right) + i \left(\sin \left(\frac{53}{12} \pi\right) + \sin \left(- \frac{19}{12} \pi\right)\right) + i \left(\sin \left(\frac{53}{12} \pi\right) + \sin \left(\frac{19}{12} \pi\right)\right) - \cos \left(\frac{17}{2} \pi\right) + \cos \left(\frac{53}{12} \pi\right)\right)$
=cos(53/12pi)+isin(53/12pi))

Using the $2 \pi$ periodicity of the functions we have:
$= \cos \left(\frac{5}{12} \pi\right) + i \sin \left(\frac{5}{12} \pi\right)$