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

1 Answer
Mar 29, 2016

#e^(i17/12pi).e^(i3/2pi)=cos(5/12pi)+isin(5/12pi)#

Explanation:

Recall the Trigonomic identities:

#sin(x)=-sin(-x)#
#2\cos \theta \cos \varphi = \cos(\theta - \varphi) + \cos(\theta + \varphi)#
#2\sin \theta \sin \varphi = \cos(\theta - \varphi) - \cos(\theta + \varphi)#
#2\sin \theta \cos \varphi = \sin(\theta + \varphi) + \sin(\theta - \varphi)#

In Trigonomic form the following expression gives:

#e^(x+iy)=e^x(cosy+isinx)#
#=>e^(iy)=cosy+isiny#

Using these in the given values we have:
#e^(i17/12pi).e^(i3/2pi)=(cos(17/12pi)+isin(17/2pi))(cos(3/2pi)+isin(3/2pi))#
#=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:
#=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(53/12pi)+isin(53/12pi))#

Using the #2pi# periodicity of the functions we have:
#=cos(5/12pi)+isin(5/12pi)#