How do you evaluate # e^( (3 pi)/2 i) - e^( ( pi)/8 i)# using trigonometric functions?

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Answer 1 · 2017-12-24T06:57:51

The answer is #=-1/2sqrt(sqrt2+2)-i/2sqrt(2-sqrt2)-i#

Apply Euler's Identity

#e^(itheta)=costheta+isintheta#

#cos2theta=2cos^2theta-1#, #=>#, #costheta=sqrt((cos2theta+1)/2)#

#cos2theta=1-2sin^2theta#,#=>#, #sintheta=sqrt((1-cos2theta)/2)#

Therefore,

#e^(3/2pii)-e^(1/8pii)=cos(3/2pi)+isin(3/2pi)-cos(1/8pi)-isin(1/8pi)#

#cos(1/8pi)=sqrt(((cos(pi/4))+1)/2)=1/2sqrt(sqrt2+2)#

#sin(1/8pi)=sqrt(((1-cos(pi/4)))/2)=1/2sqrt(2-sqrt2)#

#cos(3/2pi)=0#

#sin(3/2pi)=-1#

So,

#e^(3/2pii)-e^(1/8pii)=0-i-1/2sqrt(sqrt2+2)-i1/2sqrt(2-sqrt2)#