How do you evaluate # e^( ( pi)/4 i) - e^( ( 11 pi)/8 i)# using trigonometric functions?

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Answer 1 · 2018-08-13T13:34:13

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

Apply Euler's Identity

#e^(itheta)=costheta+isintheta#

#e^(ipi/4)=cos(pi/4)+isin(pi/4)#

#=sqrt2/2+isqrt2/2#

#e^(i11/8pi)=cos(11/8pi)+isin(11/8pi)#

#cos(2theta)=2cos^2theta-1#

#costheta=sqrt((1+cos2theta)/2)#

#cos(11/8pi)=sqrt((1+cos(11/4pi)/2)#

#=sqrt((1-sqrt2/2)/2)#

#=sqrt(2-sqrt2)/2#

#cos(2theta)=1-2sin^2theta#

#sintheta=sqrt((1-cos(2theta))/2)#

#sin(11/8pi)=sqrt((1-cos(11/4pi))/2)#

#=sqrt((1+sqrt2/2)/2)#

#=sqrt(2+sqrt2)/2#

Finally,

#e^(ipi/4)-e^(i11/8pi)#

#=sqrt2/2+isqrt2/2-sqrt(2-sqrt2)/2-isqrt(2+sqrt2)/2#

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