How do you evaluate #e^( ( pi)/12 i) - e^( ( 3pi)/8 i)# using trigonometric functions?

Question

1383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-20T06:41:18

See the explanation below

We apply Euler's relation

#e^(itheta)=costheta+isintheta#

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

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

#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(3/4pi)=1-2sin^2(3/8pi)#

#2sin^2(3/8pi)=1-cos(3/4pi)=1+sqrt2/2#

#sin(3/8pi)=sqrt(1/2(1+sqrt2/2))=1/2sqrt(2+sqrt2)#

#cos(3/4pi)=2cos^2(3/8pi)-1#

#2cos^2(3/8pi)=1+cos(3/4pi)=1-sqrt2/2#

#cos(3/8pi)=1/2sqrt(2-sqrt2)#

Therefore,

#e^(1/12pii)-e^(3/8pii)=cos(1/12pi)+isin(1/12pi)-cos(3/8pi)-isin(3/8pi)#

#=((sqrt2+sqrt6)/4)-1/2sqrt(2-sqrt2)+i(((sqrt6-sqrt2)/4)-1/2sqrt(2+sqrt2))#