How do you evaluate # e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12 i)# using trigonometric functions?

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Answer 1 · 2016-06-03T12:30:09

#=1/(2sqrt2)((sqrt3+1)-i(sqrt3-1))#

#e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12i)#

#=[cos((5pi)/4)+isin((5pi)/4)] -[cos((19pi)/12)+isin((19pi)/12)]#

#=[cos((5pi)/4)-cos((19pi)/12)]-i[sin((19pi)/12)-sin((5pi)/4)]#

#=[cos((15pi)/12)-cos((19pi)/12)]-i[sin((19pi)/12)-sin((15pi)/12)]#

#=[2sin((17pi)/12)sin((2pi)/12)]-i[2cos((17pi)/12)sin((2pi)/12)]#

#=[2sin((17pi)/12)xx1/2]-i[2cos((17pi)/12)xx1/2]#

#=sin((17pi)/12)-icos((17pi)/12)#

Now

#sin((17pi)/12)=sqrt(1/2(1-cos((17pi)/6)))=sqrt(1/2(1-cos(3pi-pi/6))#
#=sqrt(1/2(1+cos(pi/6)))=sqrt(1/2(1+cos(pi/6)))#

#=sqrt((2+sqrt3)/4)=sqrt((4+2sqrt3)/8)=(sqrt3+1)/(2sqrt2)#

and

#=sqrt(1/2(1-cos(pi/6)))=sqrt(1/2(1-cos(pi/6)))#

#=sqrt((2-sqrt3)/4)=sqrt((4-2sqrt3)/8)=(sqrt3-1)/(2sqrt2)#

Hence

#e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12i)#

#=sin((17pi)/12)-icos((17pi)/12)#

#=1/(2sqrt2)((sqrt3+1)-i(sqrt3-1))#