How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( 4 pi)/3 i)# using trigonometric functions?

1 Answer
IDKwhatName
Aug 6, 2017

#e^((7pi)/4i)-e^((4pi)/3i)=(1+sqrt(2))/2+i((-sqrt(2)+sqrt(3))/2)#

Explanation:

#e^(ix)=cosx+isinx#

Here, we have #e^((7pi)/4i)-e^((4pi)/3i)=cos((7pi)/4)+isin((7pi)/4)-(cos((4pi)/3)+isin((4pi)/3))=cos((7pi)/4)+isin((7pi)/4)-cos((4pi)/3)-isin((4pi)/3)#

#cos((7pi)/4)=sqrt(2)/2#
#sin((7pi)/4)=-sqrt(2)/2#
#cos((4pi)/3)=-1/2#
#sin((4pi)/3)=-sqrt(3)/2#

Therefore, #e^((7pi)/4i)-e^((4pi)/3i)=sqrt(2)/2-isqrt(2)/2+1/2+isqrt(3)/2=(1+sqrt(2))/2+i((-sqrt(2)+sqrt(3))/2)#

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