How do you evaluate # e^( ( 7 pi)/6 i) - e^( ( pi)/3 i)# using trigonometric functions?

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Answer 1 · 2017-12-18T11:13:39

#e^((7pi)/6i)-e^((pi)/3i)=i((1+sqrt(3))/2)-(1+sqrt(3))/2#

Using Euler's formula (#e^(x i)=cosx-isinx#) we get:
#e^((7pi)/6i)-e^((pi)/3i)=(cos((7pi)/6)-isin((7pi)/6))-(cos(pi/3)-isin(pi/3))#

#cos((7pi)/6)=-sqrt(3)/2#
#sin((7pi)/6)=-1/2#
#cos(pi/3)=1/2#
#sin(pi/3)=sqrt(3)/2#

#e^( ( 7 pi)/6 i) - e^( ( pi)/3 i)=(-sqrt(3)/2+i/2)-(1/2-sqrt(3)/2i)#

#=-sqrt(3)/2+i/2-1/2+sqrt(3)/2i#

#=-(1+sqrt(3))/2+i((1+sqrt(3))/2)#

#=i((1+sqrt(3))/2)-(1+sqrt(3))/2#