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

1 Answer
1s2s2p
Dec 18, 2017

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

Explanation:

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#

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
1951 views around the world
You can reuse this answer
Creative Commons License