How do you evaluate #e^( ( 23 pi)/12 i) - e^( ( 7 pi)/12 i)# using trigonometric functions?

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Answer 1 · 2017-07-22T21:03:54

#e^((23pi)/(12)i)-e^((7pi)/(12)i)=sqrt(6)/2-isqrt(6)/2#

#e^(x*i)=cos(x)+isin(x)#

#e^((23pi)/(12)i)-e^((7pi)/(12)i)=(cos((23pi)/12)+isin((23pi)/12))-(cos((7pi)/12)+isin((7pi)/12))#

#cos((23pi)/12)=(sqrt(6)+sqrt(2))/4#
#sin((23pi)/12)=(-sqrt(6)+sqrt(2))/4#
#cos((7pi)/12)=(-sqrt(6)+sqrt(2))/4#
#sin((7pi)/12)=(sqrt(6)+sqrt(2))/4#

#e^((23pi)/(12)i)-e^((7pi)/(12)i)=((sqrt(6)+sqrt(2))/4+i(-sqrt(6)+sqrt(2))/4)-((-sqrt(6)+sqrt(2))/4+i(sqrt(6)+sqrt(2))/4)#

#(sqrt(6)+sqrt(2))/4-(-sqrt(6)+sqrt(2))/4=(sqrt(6)+sqrt(6)+sqrt(2)-sqrt(2))/4=(2sqrt(6))/4=sqrt(6)/2#

#(-sqrt(6)+sqrt(2))/4-(sqrt(6)+sqrt(2))/4=(-sqrt(6)-sqrt(6)+sqrt(2)-sqrt(2))/4=(-2sqrt(6))/4=-sqrt(6)/2#

#e^((23pi)/(12)i)-e^((7pi)/(12)i)=sqrt(6)/2-isqrt(6)/2#