How do you multiply $e^{\frac{17 π}{12} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 17 pi )/ 12 i) * e^( 3 pi/2 i )$ in trigonometric form?

Question

How do you multiply $e^{\frac{17 π}{12} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 17 pi )/ 12 i) * e^( 3 pi/2 i )$ in trigonometric form?

1 Answer

Impact of this question

2058 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-29T14:05:28

$e^{i \frac{17}{12} π} . e^{i \frac{3}{2} π} = cos (\frac{5}{12} π) + i sin (\frac{5}{12} π)$ $e^(i17/12pi).e^(i3/2pi)=cos(5/12pi)+isin(5/12pi)$

Explanation:

Recall the Trigonomic identities:

$sin (x) = - sin (- x)$ $sin(x)=-sin(-x)$
$2 cos θ cos φ = cos (θ - φ) + cos (θ + φ)$ $2\cos \theta \cos \varphi = \cos(\theta - \varphi) + \cos(\theta + \varphi)$
$2 sin θ sin φ = cos (θ - φ) - cos (θ + φ)$ $2\sin \theta \sin \varphi = \cos(\theta - \varphi) - \cos(\theta + \varphi)$
$2 sin θ cos φ = sin (θ + φ) + sin (θ - φ)$ $2\sin \theta \cos \varphi = \sin(\theta + \varphi) + \sin(\theta - \varphi)$

In Trigonomic form the following expression gives:

$e^{x + i y} = e^{x} (cos y + i sin x)$ $e^(x+iy)=e^x(cosy+isinx)$
$\Rightarrow e^{i y} = cos y + i sin y$ $=>e^(iy)=cosy+isiny$

Using these in the given values we have:
$e^{i \frac{17}{12} π} . e^{i \frac{3}{2} π} = (cos (\frac{17}{12} π) + i sin (\frac{17}{2} π)) (cos (\frac{3}{2} π) + i sin (\frac{3}{2} π))$ $e^(i17/12pi).e^(i3/2pi)=(cos(17/12pi)+isin(17/2pi))(cos(3/2pi)+isin(3/2pi))$
$= cos (\frac{17}{12} π) cos (\frac{3}{2} π) + i sin (\frac{17}{2} π) cos (\frac{3}{2} π) + i cos (\frac{17}{12} π) sin (\frac{3}{2} π) - sin (\frac{17}{2} π) sin (\frac{3}{2} π)$ $=cos(17/12pi)cos(3/2pi)+isin(17/2pi)cos(3/2pi)+icos(17/12pi)sin(3/2pi)-sin(17/2pi)sin(3/2pi)$

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

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

Science

Math

Humanities

... and beyond

How do you multiply $e^{\frac{17 π}{12} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 17 pi )/ 12 i) * e^( 3 pi/2 i )$ in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you multiply e17π12i⋅e3π2ie^(( 17 pi )/ 12 i) * e^( 3 pi/2 i ) in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you multiply $e^{\frac{17 π}{12} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 17 pi )/ 12 i) * e^( 3 pi/2 i )$ in trigonometric form?