How do you multiply #e^(( pi )/ 8 i) * e^( pi/2 i ) # in trigonometric form?

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Answer 1 · 2016-09-30T10:31:04

As #e^(itheta)=costheta+isintheta#

#e^(pi/8i)=cos(pi/8)+isin(pi/8)# and

#e^(pi/2i)=cos(pi/2)+isin(pi/2)#

and #e^(pi/8i)*e^(pi/2i)=(cos(pi/8)+isin(pi/8))(cos(pi/2)+isin(pi/2))#

= #cos(pi/8)cos(pi/2)+cos(pi/8)xxisin(pi/2)+isin(pi/8)xxcos(pi/2)+isin(pi/8)xxisin(pi/2)#

= #cos(pi/8)cos(pi/2)+icos(pi/8)sin(pi/2)+isin(pi/8)cos(pi/2)+i^2sin(pi/8)sin(pi/2)#

= #cos(pi/8)cos(pi/2)+icos(pi/8)sin(pi/2)+isin(pi/8)cos(pi/2)-sin(pi/8)sin(pi/2)#

= #[cos(pi/8)cos(pi/2)-sin(pi/8)sin(pi/2)]+i[cos(pi/2)sin(pi/8)+sin(pi/2)cos(pi/8)]#

= #cos(pi/8+pi/2)+isin(pi/8+pi/2)#

= #e^(pi/8+pi/2)#