How do you evaluate #e^( ( pi)/12 i) - e^( ( 9 pi)/8 i)# using trigonometric functions?

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Answer 1 · 2017-07-16T07:29:09

The answer is #=(sqrt2+sqrt6)/4+sqrt(2+sqrt2)/2+i((sqrt6-sqrt2)/4+sqrt(2-sqrt2)/2)#

We apply Euler's relation

#e^(itheta)=costheta+isintheta#

#e^(1/12pii)=cos(1/12pi)+isin(1/12pi)#

#e^(9/8pii)=cos(9/8pi)+isin(9/8pi)#

#cos(1/12pi)=cos(1/3pi-1/4pi)#

#=cos(1/3pi)cos(1/4pi)+sin(1/3pi)sin(1/4pi)#

#=1/2*sqrt2/2+sqrt3/2*sqrt2/2#

#=(sqrt2+sqrt6)/4#

#sin(1/12pi)=sin(1/3pi-1/4pi)#

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

#=sqrt3/2*sqrt2/2-sqrt2/2*1/2#

#=(sqrt6-sqrt2)/4#

#cos(9/8pi)=cos(pi+1/8pi)#

#=cos(pi)cos(1/8pi)-sin(pi)cos(1/8pi)#

#=-1*cos(1/8pi)-0*cos(1/8pi)#

#=-cos(1/8pi)#

#Cos(2x)=2cos^2x-1#

#cos^2x=(1+cos(2x))/2#

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

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

#cos(9/8pi)=-sqrt(2+sqrt2)/2#

#sin(9/8pi)=sin(pi+1/8pi)=sin(pi)cos(1/8pi)+sin(1/8)picos(pi)#

#=0-sin(1/8pi)#

#cos^2(1/8pi)+sin^2(1/8pi)=1#

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

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

#sin(9/8pi)=-sqrt(2-sqrt2)/2#

Therefore,

#e^(1/12pii)-e^(9/8pii)=cos(1/12pi)+isin(1/12pi)-cos(9/8pi)+isin(9/8pi)#

#=(sqrt2+sqrt6)/4+sqrt(2+sqrt2)/2+i((sqrt6-sqrt2)/4+sqrt(2-sqrt2)/2)#