How do you evaluate # e^( ( 7 pi)/6 i) - e^( ( 17 pi)/8 i)# using trigonometric functions?

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Answer 1 · 2017-07-22T15:02:51

The answer is #=-sqrt3/2-sqrt(2+sqrt2)/2-i(1/2+sqrt(2-sqrt2)/2)#

We need

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

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

We apply Euler's relation

#e^(itheta)=costheta+isintheta#

#e^(7/6pii)=cos(7/6pi)+isin(7/6pi)#

#e^(17/8pii)=cos(17/8pi)+isin(17/8pi)#

So,

#cos(7/6pi)=cos(pi+1/6pi)=-cos(1/6pi)=-sqrt3/2#

#sin(7/6pi)=sin(pi+1/6pi)=-sin(1/6pi)=-1/2#

#cos(17/8pi)=cos(2pi+1/8pi)=cos(1/8pi)=sqrt((1+cos(pi/4))/2)=sqrt((1+sqrt2/2)/2)=sqrt(2+sqrt2)/2#

#sin(17/8pi)=sin(2pi+1/8pi)=sin(1/8pi)=sqrt((1-cos(pi/4))/2)=sqrt((1-sqrt2/2)/2)=sqrt(2-sqrt2)/2#

Therefore,

#e^(7/6pii)-e^(17/8pii)=-sqrt3/2-1/2i-(sqrt(2+sqrt2)/2+sqrt(2-sqrt2)/2i)#

#=-sqrt3/2-sqrt(2+sqrt2)/2-i(1/2+sqrt(2-sqrt2)/2)#