How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( 5 pi)/12 i)# using trigonometric functions?

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Answer 1 · 2017-08-22T17:22:12

The answer is #=1/4(3sqrt2-sqrt6)+i(1/4(3sqrt2+sqrt6))#

We apply Euler's formula

#e^(i x)=cosx +i sinx#

So,

#e^(7/4pi i)-e^(5/12pi i)=#

#cos(7/4pi)+isin(7/4pi)-cos(5/12pi)-isin(5/12pi)#

We calculate separately

#cos(7/4pi)=cos(pi+3/4pi)=cos(pi)cos(3/4pi)-sin(pi)sin(3/4pi)#

#=-1*-sqrt2/2-0=sqrt2/2#

#sin(7/4pi)=sin(pi+3/4pi)=sin(pi)cos(3/4pi)+sin(3/4pi)cos(pi)#

#=0+sqrt2/2*-1=-sqrt2/2#

#cos(5/12pi)=cos(3/12pi+2/12pi)=cos(1/4pi+1/6pi)=cos(1/4pi)cos(1/6pi)-sin(1/4pi)sin(1/6pi)#

#=sqrt2/2*sqrt3/2-sqrt2/2*1/2=1/4(sqrt6-sqrt2)#

#sin(5/12pi)=sin(3/12pi+2/12pi)=sin(1/4pi)cos(1/6pi)+sin(1/6pi)cos(1/4pi)#

#=sqrt2/2*sqrt3/2+sqrt2/2*1/2=1/4(sqrt6+sqrt2)#

#e^(7/4pi i)-e^(5/12pi i)=(sqrt2/2-1/4(sqrt6-sqrt2))-i(sqrt2/2+1/4(sqrt6+sqrt2))#

#=1/4(3sqrt2-sqrt6)+i(1/4(3sqrt2+sqrt6))#