How do you evaluate e^( ( 7 pi)/6 i) - e^( ( 19 pi)/8 i) using trigonometric functions?

1 Answer
Nov 4, 2016

2sin((29pi)/24)e^(((25pi)/24)i)

Explanation:

we know euler's theorem states that e^(itheta)=costheta+isintheta
so,e^(7pi/6i)-e^(19pi/8 i)
=cos(7pi/6)+isin(7pi/6)-(cos(19pi/8)+isin(19pi/8))
=cos(7pi/6)-cos(19pi/8)+i(sin(7pi/6)-sin(19pi/8))
now cosc-cosd=2sin((c+d)/2) sin((d-c)/2)
sinc-sind=2cos((c+d)/2) sin((c-d)/2)
so we get cos((7pi)/6)-cos((19pi)/8)
=2sin((85pi)/24)sin((29pi)/24)
again similarly we get i(sin((7pi)/6)-sin((19pi)/8))
=2icos((85pi)/24)sin((29pi)/24)
so,e^((7pi)/6)-e^((19pi)/8)=2sin((29pi)/24)(sin((85pi)/24)-icos((85pi)/24))
=2sin((29pi)/24)[sin(2pi+((37pi)/24))-icos(2pi+((37pi)/24))]
=2sin((29pi)/24)[sin((37pi)/24)-icos((37pi)/24)]
=2sin((29pi)/24)[cos((25pi)/24)+isin((25pi)/24)]
=2sin((29pi)/24)e^(((25pi)/24)i).